The average distance between Earth and the Sun is 1.5 x 1011m.
(a) Calculate the average speed of Earth in its orbit (assumed to be circular) in meters per second. m/s
(b) What is this speed in miles per hour? mph

Answers

Answer 1

The average speed of the earth in its orbit is 2.98 x 104 m/s or 6.67 x 104 mph.

The average distance between the earth and the sun is 1.5 x 1011m.

This can be done using the formula for the speed of an object in circular motion:Speed = distance/time

For the earth's orbit around the sun, we know that the distance covered is the circumference of the circle with radius equal to the average distance between the earth and the sun.

Circumference = 2πr, where r is the radius of the circle.

So the distance covered by the earth in one orbit is:Distance covered = 2πrwhere r = 1.5 x 1011mTherefore, distance covered = 2π(1.5 x 1011)m = 9.42 x 1011m

We also know that the time taken for one complete orbit is one year or 365 days, or 3.154 x 107 seconds.

Therefore:Time taken for one orbit = 3.154 x 107 seconds

Now we can use the formula for speed to find the average speed of the earth in its orbit:

Speed = distance/timeSpeed = (9.42 x 1011m)/(3.154 x 107s)Speed = 2.98 x 104m/s

Therefore, the average speed of the earth in its orbit is 2.98 x 104m/s.

Convert m/s to miles/hour

We can convert m/s to miles/hour by using the conversion factor: 1 mile = 1609.34m and 1 hour = 3600s

Therefore, 1 mile/hour = 1609.34/3600 m/s = 0.44704 m/s

So to convert the speed of the earth from m/s to miles/hour, we need to divide by 0.44704:

Speed in miles/hour = (2.98 x 104 m/s)/0.44704Speed in miles/hour = 6.67 x 104 mph

Therefore, the average speed of the earth in its orbit is 2.98 x 104 m/s or 6.67 x 104 mph.

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Related Questions

A charge -5.5 nC is placed at (-3.1.-3) m and another charge 9.3 nC is placed at (-2,3,-2) m. What is the electric field at (1,0,0)m?

Answers

The electric field at (1,0,0) m due to the given charges is -1.2 x 10^5 N/C, directed towards the left.

Let's first calculate the electric field at point P due to the first charge:q1 = -5.5 nC, r1 = (-3.1, -3, 0) m and r = (1, 0, 0) m

The distance between charge 1 and point P is:r = √((x2 - x1)² + (y2 - y1)² + (z2 - z1)²)r = √((1 - (-3.1))² + (0 - (-3))² + (0 - 0)²)r = √(4.1² + 3² + 0²)r = 5.068 m

Therefore, the electric field at point P due to charge 1 is:

E1 = kq1 / r1²E1 = (9 x 10^9 Nm²/C²) x (-5.5 x 10^-9 C) / (5.068 m)²E1 = -4.3 x 10^5 N/C (towards left, as the charge is negative)

Now, let's calculate the electric field at point P due to the second charge:

q2 = 9.3 nC, r2 = (-2, 3, -2) m and r = (1, 0, 0) m

The distance between charge 2 and point P is:

r = √((x2 - x1)² + (y2 - y1)² + (z2 - z1)²)

r = √((1 - (-2))² + (0 - 3)² + (0 - (-2))²)

r = √(3² + 3² + 2²)r = √22 m

Therefore, the electric field at point P due to charge 2 is:

E2 = kq2 / r2²

E2 = (9 x 10^9 Nm²/C²) x (9.3 x 10^-9 C) / (√22 m)²

E2 = 3.1 x 10^5 N/C (towards right, as the charge is positive)

Now, the total electric field at point P due to both charges is:

E = E1 + E2

E = -4.3 x 10^5 N/C + 3.1 x 10^5 N/C

E = -1.2 x 10^5 N/C

Therefore, the electric field at (1,0,0) m due to the given charges is -1.2 x 10^5 N/C, directed towards the left.

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The electric field at point P (1, 0, 0)m is (-2.42 × 10⁶) î + 6.91 × 10⁶ ĵ N/C.

The given charges are -5.5 nC and 9.3 nC. The position vectors of these charges are (-3.1, -3, 0)m and (-2, 3, -2)m. We need to find the electric field at (1, 0, 0)m.

Let's consider charge q1 (-5.5 nC) and charge q2 (9.3 nC) respectively with position vectors r1 and r2. Electric field due to q1 at point P (1,0,0)m is given by:r1 = (-3.1, -3, 0)mq1 = -5.5 nC

Position vector r from q1 to P = rP - r1 = (1, 0, 0)m - (-3.1, -3, 0)m = (4.1, 3, 0)m

Using the formula of electric field, the electric field due to q1 at point P will be given by:

E1 = kq1 / r²

where k is the Coulomb constantk = 9 × 10⁹ N m² C⁻²

Electric field due to q1 at point P isE1 = 9 × 10⁹ × (-5.5) / (4.1² + 3²) = -2.42 × 10⁶ N/C

Now, let's consider charge q2. The position vector of q2 is given by:r2 = (-2, 3, -2)mq2 = 9.3 nC

Position vector r from q2 to P = rP - r2 = (1, 0, 0)m - (-2, 3, -2)m = (3, -3, 2)m

Electric field due to q2 at point P will be given by:

E2 = kq2 / r²

Electric field due to q2 at point P is

E2 = 9 × 10⁹ × 9.3 / (3² + (-3)² + 2²) = 6.91 × 10⁶ N/C

Now, we can get the total electric field due to the given charges by adding the electric fields due to q1 and q2 vectorially.

The vector addition of electric fields E1 and E2 is given by the formula:

E = E1 + E2

Let's consider charge q1 (-5.5 nC) and charge q2 (9.3 nC) respectively with position vectors r1 and r2. Electric field due to q1 at point P (1,0,0)m is given by:r1 = (-3.1, -3, 0)mq1 = -5.5 nC

Position vector r from q1 to P = rP - r1 = (1, 0, 0)m - (-3.1, -3, 0)m = (4.1, 3, 0)m

Using the formula of electric field, the electric field due to q1 at point P will be given by:E1 = kq1 / r²

where k is the Coulomb constant

k = 9 × 10⁹ N m² C⁻²

The magnitude of the electric field due to q1 at point P is given by|E1| = 9 × 10⁹ × |q1| / r²= 9 × 10⁹ × 5.5 / (4.1² + 3²) N/C= 2.42 × 10⁶ N/C

The direction of the electric field due to q1 at point P is towards the charge q1.

Now, let's consider charge q2. The position vector of q2 is given by:r2 = (-2, 3, -2)mq2 = 9.3 nC

Position vector r from q2 to P = rP - r2 = (1, 0, 0)m - (-2, 3, -2)m = (3, -3, 2)m

The magnitude of the electric field due to q2 at point P will be given by:

E2 = kq2 / r²= 9 × 10⁹ × 9.3 / (3² + (-3)² + 2²) N/C= 6.91 × 10⁶ N/C

The direction of the electric field due to q2 at point P is away from the charge q2.

Now, we can get the total electric field due to the given charges by adding the electric fields due to q1 and q2 vectorially. The vector addition of electric fields E1 and E2 is given by the formula:E = E1 + E2E = (-2.42 × 10⁶) î + 6.91 × 10⁶ ĵ N/C

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what is the wavelength, in nm , of a photon with energy 0.30 ev ?

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The wavelength of  0.3 eV of photon is 4136 nm.

Thus, There is a wavelength and a frequency for every photon. The distance between two electric field peaks with the same vector is known as the wavelength. The number of wavelengths a photon travels through each second is what is known as its frequency.

A photon cannot truly have a colour, unlike an EM wave. Instead, a photon will match a specific colour of light. A single photon cannot have colour since it cannot be recognized by the human eye, which is how colour is defined.  

0.3 ev= 0.3 x 1.602 x 10⁻¹⁹ J

λ = 4136 x 10⁻⁹ m

λ = 4136 nm → infrared.

Thus, The wavelength of  0.3 eV of photon is 4136 nm.

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E11: Please show complete solution and explanation. Thank
you!
11. Discuss the physical interpretation of any one Maxwell relation.

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One of the Maxwell's relations that has a significant physical interpretation is the relation between the partial derivatives of entropy with respect to volume and temperature in a thermodynamic system. This relation is given by:

([tex]∂S/∂V)_T = (∂P/∂T)_V[/tex]

Here, (∂S/∂V)_T represents the partial derivative of entropy with respect to volume at constant temperature, and (∂P/∂T)_V represents the partial derivative of pressure with respect to temperature at constant volume.

The physical interpretation of this relation is that it relates the response of a system's entropy to changes in volume and temperature, while keeping one of these variables constant.

It shows that an increase in temperature at constant volume leads to an increase in entropy per unit volume. Conversely, an increase in volume at constant temperature results in an increase in entropy per unit temperature.

This Maxwell relation helps to establish a connection between the thermodynamic properties of a system and provides insights into the behavior of entropy in response to changes in temperature and volume.

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A particale's velocity function is given by V=3t³+5t²-6 with X in meter/second and t in second Find the velocity at t=2s
A particale's velocity function is given by V=3t³+5t²-6 with X in meter/se

Answers

The velocity of the particle at t=2s is 38 m/s.

The velocity function of the particle is given by V = 3t³ + 5t² - 6, where V represents the velocity in meters per second (m/s), and t represents time in seconds (s). This equation is a polynomial function that describes how the velocity of the particle changes over time.

The velocity function of the particle is V = 3t³ + 5t² - 6, we need to find the velocity at t=2s.

Substituting t=2 into the velocity function, we have:

V = 3(2)³ + 5(2)² - 6

V = 3(8) + 5(4) - 6

V = 24 + 20 - 6

V = 38 m/s

It's important to note that the velocity of the particle can be positive or negative depending on the direction of motion. In this case, since we are given the velocity function without any information about the initial conditions or the direction, we can interpret the velocity as a magnitude. Thus, at t=2s, the particle has a velocity of 38 m/s, regardless of its direction of motion.

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The displacement of a wave traveling in the negative y-direction
is D(y,t)=(9.0cm)sin(45y+70t+π)D(y,t)=(9.0cm)sin⁡(45y+70t+π), where
y is in m and t is in s.
What is the frequency of this wave?
Wh

Answers

The displacement of a wave traveling in the negative y-direction depends on the amplitude and frequency of the wave.

The displacement of a wave traveling in the negative y-direction is a combination of factors. The first factor is the amplitude, which is the maximum distance that a particle moves from its rest position as a wave passes through it. The second factor is the frequency, which is the number of waves that pass a fixed point in a given amount of time. The displacement of a wave is given by the formula y = A sin(kx - ωt + ϕ), where A is the amplitude, k is the wave number, x is the position, ω is the angular frequency, t is the time, and ϕ is the phase constant. This formula shows that the displacement depends on the amplitude and frequency of the wave.

These variables have the same fundamental meaning for waves. In any case, it is useful to word the definitions in a more unambiguous manner that applies straightforwardly to waves: Amplitude is the distance between the wave's maximum displacement and its resting position. Frequency is the number of waves that pass by a particular point every second.

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information to answer the next two questions: A Nerf ball is launched horizontally from a rooftop and lands on the ground, 3.50 m from the base of the building, in a time of 2.20 s. Question 32 (1 point) The horizontal speed of the ball is 21.6 m/s 1.59 m/s 07.70 m/s 00.0629 m/s Projectile Motion Characteristics Component of Motien 11. Vertical 1 2. Affected by gravity Exhibits form motion 3. Exhibits form accelerated motion 4. Component of initial velocity is v, sind Component of initial velocity is v, cus 5. Question 29 (1 point) ✓ Saved The characteristics that apply to the horizontal component of projectile motion are 3 and 5 1,3 and 4 O2 and 5 1,2 and 4 The correct values for I, II, III, and IV, respectively are Components of Vectors x componet Ad 1 II IV. 20 m, 0 m, 26 m, and 15 m -20 m, 0 m, 26 m, and -15 m 20 m, 0 m, -26 m, and 15 m 0 m, -20 m, 26 m, and 15 m O. Question 23 (1 point) ✓ Saved The magnitude of the resultant displacement is 7.1 m 1.3 x 10³ m 36 m 22 m

Answers

32. The horizontal speed of the ball is 7.70 m/s.

29. The characteristics that apply to the horizontal component of projectile motion are 1, 3, and 4.

23. The magnitude of the resultant displacement is 7.1 m.

32. To find the horizontal speed of the ball, we use the formula: horizontal speed = horizontal distance ÷ time. In this case, the horizontal distance is given as 3.50 m and the time is given as 2.20 s. Plugging in the values, we get: horizontal speed = 3.50 m ÷ 2.20 s = 1.59 m/s.

29. The characteristics of projectile motion are as follows:

1. Vertical motion: A projectile experiences vertical motion due to the influence of gravity.

3. Exhibits uniform motion: The horizontal component of projectile motion is uniform since there is no acceleration in the horizontal direction.

4. Exhibits accelerated motion: The vertical component of projectile motion is accelerated due to the force of gravity.

5. Component of initial velocity is v, sinθ: The vertical component of the initial velocity is v multiplied by the sine of the launch angle θ.

23. The resultant displacement of the ball refers to the straight-line distance from the initial point to the final point. To calculate the magnitude of the resultant displacement, we use the Pythagorean theorem. Since the horizontal and vertical components of displacement are given as 3.50 m and 2.20 m respectively, the magnitude of the resultant displacement is: √((3.50 m)² + (2.20 m)²) = 4.18 m.

Therefore,

32. The horizontal speed of the ball is 7.70 m/s.

29. The characteristics that apply to the horizontal component of projectile motion are 1, 3, and 4.

23. The magnitude of the resultant displacement is 7.1 m.

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Write short notes on
Forced circulation evaporation
Agitated thin film evaporation

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Agitated thin film evaporation is a process used to separate components from liquid mixtures. It is particularly useful for heat-sensitive materials that need to be processed at low temperatures.

The process involves heating the liquid mixture in a vessel while simultaneously exposing it to a vacuum. The heat and vacuum cause the mixture to evaporate, and the resulting vapors are condensed back into a liquid, which can be collected separately. The process is typically carried out in a thin film evaporator, which consists of a heated cylindrical vessel with a rotating blade that agitates the mixture as it evaporates. This helps to increase the rate of evaporation and improve the quality of the separated components.

When a liquid becomes a gas, this is known as evaporation. When puddles of rain "disappear" on a hot day or when wet clothes dry in the sun, it is easy to imagine. In these models, the fluid water isn't really disappearing — it is dissipating into a gas, called water fume. Global evaporation takes place.

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suppose the previous forecast was 30 units, actual demand was 50 units, and ∝ = 0.15; compute the new forecast using exponential smoothing.

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By using the formula of exponential smoothing, we can get the new forecast. Hence, the new forecast using exponential smoothing is 33 units.

Given:

Previous forecast = 30 units

Actual demand = 50 unitsα = 0.15Formula used:

New forecast = α(actual demand) + (1 - α)(previous forecast)

New forecast = 0.15(50) + (1 - 0.15)(30)New forecast = 7.5 + 25.5

New forecast = 33 units

Therefore, the new forecast using exponential smoothing is 33 units.

In exponential smoothing, the new forecast is computed by using the actual demand and previous forecast. In this question, the previous forecast was 30 units and actual demand was 50 units, with α = 0.15. By using the formula of exponential smoothing, we can get the new forecast. Hence, the new forecast using exponential smoothing is 33 units.

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a lens has a refractive power of -1.50. what is its focal length?

Answers

It has been determined that the focal length of the lens is -0.6667 m.

Given: The refractive power of a lens is -1.50We are supposed to find the focal length of the given lens

Solution:The formula to find the focal length of a lens is given by:1/f = (n-1) (1/R1 - 1/R2)

Given: Refractive power (P) = -1.50

As we know that, P = 1/f (Where f is the focal length)

Hence, -1.50 = 1/fOr, f = -1/1.5= -0.6667 m

Therefore, the focal length of the given lens is -0.6667 m.

From the above calculations, it has been determined that the focal length of the lens is -0.6667 m.

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take the radius of the earth to be 6,378 km. (a) what is the angular speed (in rad/s) of a point on earth's surface at latitude 65° n?

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The angular speed of a point on Earth's surface at latitude 65° N is approximately 7.292 × 10^(-5) rad/s.

To calculate the angular speed, we need to consider the rotational motion of the Earth. The angular speed (ω) is defined as the change in angular displacement per unit of time. At any latitude on Earth's surface, the angular speed can be calculated using the formula ω = v / r, where v is the linear velocity and r is the radius of the Earth.

The linear velocity can be found using the formula v = R * cos(latitude), where R is the rotational speed of the Earth and latitude is the given latitude. The rotational speed of the Earth is approximately 2π radians per 24 hours. By substituting the given values into the formulas, we can calculate the angular speed.

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the magnitude of the electric field at a point p for a certain electromagnetic wave is 570 n/c. what is the magnitude of the magnetic field for that wave at p? group of answer choices

Answers

The magnitude of the magnetic field for that wave at point p is 1.9 × 10-6 T.

The given question is related to the electromagnetic wave. The magnitude of the electric field at a point p for a certain electromagnetic wave is 570 N/C.

We need to determine the magnitude of the magnetic field for that wave at p.

So, we know that an electromagnetic wave consists of an electric field and a magnetic field perpendicular to each other.

We can use the formula to find the relation between the electric and magnetic fields for an electromagnetic wave.c = E/B Where,c is the speed of light (3 x 108 m/s)E is the electric field intensityB is the magnetic field intensity

Using the above equation, we can find the magnetic field for that wave at point P.

Magnitude of the electric field, E = 570 N/CMagnitude of the speed of light, c = 3 x 108 m/s

Putting values in the above formula;570 = B x 3 x 108B = 570/3 x 108B = 1.9 × 10-6 T

Therefore, the magnitude of the magnetic field for that wave at point p is 1.9 × 10-6 T.

Thus, the magnitude of the magnetic field for that wave at point p is 1.9 × 10-6 T.

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how fast are the ions moving when they emerge from the velocity selector?

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The ions are moving at a constant velocity when they emerge from the velocity selector.

When ions emerge from the velocity selector, they are moving at a constant velocity. The velocity selector is a device used to filter and control the speed of charged particles, such as ions, in scientific experiments. It consists of crossed electric and magnetic fields that exert forces on the ions, allowing only those with a specific velocity to pass through unaffected. As a result, the ions that emerge from the velocity selector have their velocities adjusted to match the desired value. This constant velocity allows for accurate measurements and control of the ions' movement in further experiments or applications.

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why did the masses of the objects have to be very small to be able to get the objects very close to each other?

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The masses of the objects have to be very small to be able to get the objects very close to each other because of the gravitational force.

Gravitational force is the force of attraction between any two objects with mass. It is an attractive force that acts between all objects with mass. The strength of the gravitational force depends on the masses of the objects involved and the distance between them. When the objects are close to each other, the gravitational force between them becomes stronger. If the masses of the objects are very large, the gravitational force between them becomes very strong. This means that it is very difficult to get the objects very close to each other because of the strong force of gravity. However, if the masses of the objects are very small, the gravitational force between them becomes very weak. This means that it is much easier to get the objects very close to each other because there is less gravitational force pushing them apart.

Gravitational force is one of the fundamental forces in nature. It is an attractive force that acts between any two objects with mass. The strength of the gravitational force depends on the masses of the objects involved and the distance between them. When the objects are close to each other, the gravitational force between them becomes stronger. If the masses of the objects are very large, the gravitational force between them becomes very strong. This means that it is very difficult to get the objects very close to each other because of the strong force of gravity. However, if the masses of the objects are very small, the gravitational force between them becomes very weak. This means that it is much easier to get the objects very close to each other because there is less gravitational force pushing them apart. In general, the strength of the gravitational force between two objects is given by the formula F = Gm1m2/r^2, where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them. As you can see from this formula, the strength of the gravitational force decreases as the distance between the objects increases.

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how far is the motorcycle from the car when it reaches this speed?

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The motorcycle is approximately 17.97 meters away from the car when it reaches the same speed as the car.

To find the distance between the car and the motorcycle when the motorcycle reaches the same speed as the car, we can use the equations of motion. Let's assume the initial position of both the car and the motorcycle is 0.For the car:
Initial velocity, u1 = 83 km/h
Final velocity, v1 = 83 km/h
Acceleration, a1 = 0 (since the car is traveling at a steady speed)
Time, t1 = ?
For the motorcycle:
Initial velocity, u2 = 0 (since it starts from rest)
Final velocity, v2 = 83 km/h
Acceleration, a2 = 7.4 m/s^2
Time, t2 = ?
Using the equation v = u + at, we can find the time it takes for the motorcycle to reach the same speed as the car:v2 = u2 + a2t2
83 km/h = 0 + (7.4 m/s^2) * t2
Converting the velocities to meters per second:
83 km/h = (83 * 1000 m) / (3600 s) = 23.06 m/s23.06 m/s = 7.4 m/s^2 * t2
t2 = 23.06 m/s / 7.4 m/s^2
t2 ≈ 3.12 seconds
Now, we can find the distance traveled by the motorcycle using the equation:
s2 = u2t2 + (1/2) * a2 * t2^2
s2 = 0 + (1/2) * (7.4 m/s^2) * (3.12 s)^2s2 ≈ 17.97 meters
Therefore, the motorcycle is approximately 17.97 meters away from the car when it reaches the same speed as the car.

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Following is the complete answer: A car is traveling at a steady 83 km/h in a 50 km/h zone. A police motorcycle takes off at the instant the car passes it, accelerating at a steady 7.4m/s2 . How far is the motorcycle from the car when it reaches this speed?

Find the rest energy, in terajoules, of a 18.5 g piece of chocolate. 1 TJ is equal to 1012 J. rest energy: TJ

Answers

The rest energy of an 18.5 g piece of chocolate is 1.6601 x 10⁻³ TJ. Answer: 1.6601 x 10⁻³ TJ.

The rest energy, in terajoules, of an 18.5 g piece of chocolate can be found using the equation: E=mc², where E is energy, m is mass, and c is the speed of light squared. Given that 1 TJ is equal to 10¹² J, we can convert the final answer to terajoules (TJ).Here's how to solve the problem:

Convert the mass of chocolate to kilograms. There are 1000 grams in a kilogram, so 18.5 g = 0.0185 kg.

Plug the mass into the equation E=mc²: E = (0.0185 kg) x (299792458 m/s)².

Simplify and solve: E = (0.0185 kg) x (8.98755178736818 x 10¹⁶ m²/s²).

E = 1.6601 x 10¹⁵ J.4.

Convert to terajoules: 1 TJ = 10¹² J, so 1.6601 x 10¹⁵ J = 1.6601 x 10⁻³ TJ.

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a metal sphere has a net negative charge of 1.1 × 10-6 coulomb. approximately how many more elec- trons than protons are on the sphere? 1. 1.8 × 1012 2. 5.7 × 1012 3. 6.9 × 1012 4. 9.9 × 1012

Answers

The correct option is 3. 6.9 × 10¹². More electrons than protons are present on the metal sphere.

An electron carries a negative charge of 1.6 × 10⁻¹⁹ C.A proton carries a positive charge of 1.6 × 10⁻¹⁹ C.The total charge on the sphere is -1.1 × 10⁻⁶ C.So, the total number of electrons present on the sphere will be more than the total number of protons present on it.

To calculate the number of excess electrons, divide the total charge on the sphere by the charge on each electron.n= Total charge on the sphere / Charge carried by one electron n = 1.1 × 10⁻⁶ C / 1.6 × 10⁻¹⁹ C = 6.875 × 10¹²6.875 × 10¹² electrons more than the number of protons present on the sphere. 6.9 × 10¹² electrons are more than protons present on the sphere. Therefore, the correct option is 3. 6.9 × 10¹².

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how to calculate the distance between a sensor and an electric harge

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In order to calculate the distance between a sensor and an electric charge, you need to know the electric field strength produced by the charge and the sensitivity of the sensor to that field strength. The calculation involves using Coulomb's Law to find the electric field strength and then using the inverse square law to determine the distance.

Coulomb's Law states that the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The formula for Coulomb's Law is:F = k * (q1 * q2) / d^2where F is the force between the charges, k is Coulomb's constant (9 x 10^9 N m^2/C^2), q1 and q2 are the charges, and d is the distance between the charges.The electric field strength produced by the charge is given by:E = F / q2where E is the electric field strength and q2 is the test charge (the charge on the sensor).To calculate the distance between the sensor and the charge, you can use the inverse square law, which states that the intensity of a field (in this case, the electric field) is inversely proportional to the square of the distance from the source. The formula for the inverse square law is:I = I0 * (d0 / d)^2where I is the intensity of the field at distance d, I0 is the intensity of the field at distance d0, and d0 is a reference distance (usually chosen to be 1 meter). Rearranging this equation, we get:d = sqrt(I0 / I) * d0So to calculate the distance between the sensor and the charge, you need to first find the electric field strength at the sensor and the electric field strength at a reference distance (e.g. 1 meter). Then you can use the inverse square law to calculate the distance between the sensor and the charge.

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what is the pressure on the sample if f = 340 n is applied to the lever? express your answer to two significant figures and include the appropriate units.

Answers

The amount of pressure exerted on the sample due to the applied force is 4.25 x 10⁷ Nm.

The force applied physically to an object per unit area is referred to as pressure. Per unit area, the force is delivered perpendicularly to the surfaces of the objects.

The diameter of the large cylinder, d₁ = 10 cm = 0.1 m

The diameter of the small cylinder, d₂ = 2 cm = 0.02 m

The area of the given sample, A = 4 cm² = 4 x 10⁻⁴m²

So, the force acting on the small cylinder is given by,

(F x 2L) - (F₂ x L) = 0

2FL - F₂L = 0

So,

F₂L = 2FL

Therefore, F₂ = 2 x F

F₂ = 2 x 340 N

F₂ = 680 N

In order to calculate the force acting on the large cylinder,

We know that, P₁ = P₂

So, we can write that,

F₁/A₁ = F₂/A₂

F₁/d₁² = F₂/d₂²

Therefore,

F₁ = F₂d₁²/d₂²

F₁ = 680 x (0.1/0.02)²

F₁ = 680 x 100/4

F₁ = 17000 N

Therefore, the pressure exerted on the sample is,

P = F₁/A

P = 17000/(4 x 10⁻⁴)

P = 4.25 x 10⁷ Nm

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A 60 kg astronaut in a full space suit (mass of 130 kg) presses down on a panel on the outside of her spacecraft with a force of 10 N for 1 second. The spaceship has a radius of 3 m and mass of 91000 kg. Unfortunately, the astronaut forgot to tie herself to the spacecraft. (a) What velocity does the push result in for the astronaut, who is initially at rest? Be sure to state any assumptions you might make in your calculation.(b) Is the astronaut going to remain gravitationally bound to the spaceship or does the astronaut escape from the ship? Explain with a calculation.(c) The quick-thinking astronaut has a toolbelt with total mass of 5 kg and decides on a plan to throw the toolbelt so that she can stop herself floating away. In what direction should the astronaut throw the belt to most easily stop moving and with what speed must the astronaut throw it to reduce her speed to 0? Be sure to explain why the method you used is valid.(d) If the drifting astronaut has nothing to throw, she could catch something thrown to her by another astronaut on the spacecraft and then she could throw that same object.Explain whether the drifting astronaut can stop if she throws the object at the same throwing speed as the other astronaut.

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a. Push does not result in any initial velocity for the astronaut .b. The astronaut will not remain gravitationally bound to the spaceship. c. To stop herself from floating away, the astronaut can use the principle of conservation of momentum again.  

(a) To determine the velocity acquired by the astronaut, we can use the principle of conservation of momentum. Since no external forces are acting on the system (astronaut + spacecraft), the total momentum before and after the push must be equal.

Let's assume the positive direction is defined as the direction in which the astronaut pushes the panel. The initial momentum of the system is zero since both the astronaut and the spacecraft are at rest.

Initial momentum = Final momentum

0 = (mass of astronaut) * (initial velocity of astronaut) + (mass of spacecraft) * (initial velocity of spacecraft)

Since the astronaut is initially at rest, the equation becomes:

0 = (mass of astronaut) * 0 + (mass of spacecraft) * (initial velocity of spacecraft)

Solving for the initial velocity of the spacecraft:

(initial velocity of spacecraft) = -[(mass of astronaut) / (mass of spacecraft)] * 0

However, the mass of the astronaut is given as 60 kg and the mass of the space suit is given as 130 kg. We need to use the total mass of the astronaut in this case, which is 60 kg + 130 kg = 190 kg.

(initial velocity of spacecraft) = -[(190 kg) / (91000 kg)] * 0

The negative sign indicates that the spacecraft moves in the opposite direction of the push.

Therefore, the push does not result in any initial velocity for the astronaut.

(b) The astronaut will not remain gravitationally bound to the spaceship. In this scenario, the only force acting on the astronaut is the gravitational force between the astronaut and the spacecraft. The force of gravity is given by Newton's law of universal gravitation:

F_ gravity = (G * m1 * m2) / r^2

Where:

F_ gravity is the force of gravity

G is the gravitational constant

m1 is the mass of the astronaut

m2 is the mass of the spacecraft

r is the distance between the astronaut and the spacecraft (the radius of the spaceship in this case)

Using the given values:

F_ gravity = (6.67430 x 10^-11 N m^2/kg^2) * (60 kg) * (91000 kg) / (3 m)^2

Calculating the force of gravity, we find that it is approximately 3.022 N.

The force applied by the astronaut (10 N) is greater than the force of gravity (3.022 N), indicating that the astronaut will escape from the ship. The astronaut's push is strong enough to overcome the gravitational attraction.

(c) To stop herself from floating away, the astronaut can use the principle of conservation of momentum again. By throwing the toolbelt, the astronaut imparts a backward momentum to it, causing herself to move forward with an equal but opposite momentum, ultimately reducing her speed to zero.

Let's assume the positive direction is defined as the direction opposite to the astronaut's initial motion.

The momentum before throwing the toolbelt is zero since the astronaut is initially drifting with a certain velocity.

Initial momentum = Final momentum

0 = (mass of astronaut) * (initial velocity of astronaut) + (mass of toolbelt) * (initial velocity of toolbelt)

Since we want the astronaut to reduce her speed to zero, the equation becomes:

0 = (mass of astronaut) * (initial velocity of astronaut) + (mass of toolbelt) * (initial velocity of toolbelt)

The direction of the initial velocity of the toolbelt should be opposite to the astronaut's initial motion, while its magnitude should be such that the astronaut's total momentum becomes zero.

Therefore, to stop moving, the astronaut should throw the toolbelt in the direction opposite to her initial motion with a velocity equal to her own initial.

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the ball in the figure rotates counterclockwise in a circle of radius 3.39 m with a constant angular speed of 8.00 rad/s. at t = 0, its shadow has an x coordinate of 2.00 m and is moving to the right.

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To determine the position of the shadow at a specific time, we can use the concept of angular velocity and the relationship between angular displacement and linear displacement.

Given:

Radius of the circle (r) = 3.39 m

Angular speed (ω) = 8.00 rad/s

Initial x-coordinate of the shadow (x) = 2.00 m The ball rotates counterclockwise, which means the shadow moves to the right initially. We can use the equation: x = r * cos(θ) At t = 0, the angular displacement (θ) is 0, and the x-coordinate of the shadow is 2.00 m. We can solve for θ using the inverse cosine function:

θ = cos^(-1)(x/r)

θ = cos^(-1)(2.00 m / 3.39 m)

Calculating the value of θ: θ ≈ 55.40 degrees. Since the ball rotates counterclockwise at a constant angular speed, we can determine the angular displacement at any given time using the equation: θ = ω * tmNow, let's find the angular displacement at t = 0. We substitute the values:θ = 8.00 rad/s * 0 s θ = 0 rad. Therefore, the shadow is initially at an angular displacement of 55.40 degrees, and the angular displacement remains 0 at t = 0.

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According to solubility rules, which compound should dissolve in water? Select one: ОКРО, 0 MgCO3 O Caso O AgBI

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MgCO₃ is the only compound that should dissolve in water according to the given solubility rules. Solubility rules predict the solubility of various ionic compounds based on their cation and anion constituents.

These rules are helpful for predicting what substances will dissolve in water and which will not, among other things. According to solubility rules, MgCO₃ should dissolve in water. MgCO₃ is a salt that contains Mg²⁺ cation and CO₃²⁻ anion. When MgCO₃ is added to water, the Mg²⁺ and CO₃²⁻ ions separate, or dissociate, from one another and are surrounded by water molecules.

This separation process, referred to as hydration, occurs because water molecules are polar, meaning they have a partially positive and partially negative charge. When an ionic compound is added to water, the water molecules surround the positively and negatively charged ions and dissolve the salt into the water.

The other compounds, K₃PO₄, CaSO₄, and AgBr are not very soluble in water according to solubility rules. Hence, MgCO₃ is the only compound that should dissolve in water according to the given solubility rules.

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Which kind of force and motion causes a pencil that is dropped to fall to the floor?

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The force of gravity causes a pencil that is dropped to fall to the floor. The time it takes for an object to fall from a certain height depends on its initial velocity and the acceleration due to gravity.

When an object falls, it is because gravity is acting on it. The force of gravity is the force of attraction between any two objects with mass. Gravity causes the objects to be pulled toward each other. The strength of gravity depends on the mass of the objects and the distance between them.The motion of a falling object is called free fall. Free fall occurs when an object falls under the influence of gravity alone, with no other forces acting on it. The acceleration of an object in free fall is constant, and is equal to the acceleration due to gravity, which is approximately 9.8 meters per second squared (m/s²) near the surface of the Earth.

When an object is dropped, it begins to fall because of the force of gravity. Gravity is a force that exists between any two objects that have mass. The force of gravity depends on the mass of the objects and the distance between them. The force of gravity acts on the object from the moment it is dropped until it hits the floor.The motion of an object that is falling under the influence of gravity alone is called free fall. In free fall, the object is accelerating because of gravity. The acceleration of an object in free fall is constant, and is equal to the acceleration due to gravity, which is approximately 9.8 meters per second squared (m/s²) near the surface of the Earth.When an object is in free fall, the only force acting on it is gravity. This means that there is no air resistance or other force to slow it down. As a result, the object falls faster and faster until it hits the ground.

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In this classic example of momentum conservation we’ll see why a rifle recoils when it is fired. A marksman holds a 3.00 kg rifle loosely, so that we can ignore any horizontal external forces acting on the rifle–bullet system. He fires a bullet of mass 5.00 g horizontally with a speed vbullet=300m/s . What is the recoil speed vrifle of the rifle? What are the final kinetic energies of the bullet and the rifle?

Question:

The same rifle fires a bullet with mass 7.7 g at the same speed as before. For the same idealized model, find the ratio of the final kinetic energies of the bullet and rifle.

Answers

The ratio of final kinetic energies of the bullet to the rifle is: Kf/Kr = 346.5 J/0.375 J = 924.

The momentum of the rifle before firing the bullet is zero. The bullet is fired horizontally with a speed of 300 m/s. The direction of recoil of the rifle will be opposite to that of the bullet. Let the recoil velocity of the rifle be vr. Then according to the law of conservation of momentum, the momentum of the rifle-bullet system after firing is zero. We can express this mathematically as:0 = -5 x 10^-3 kg x 300 m/s + (3 + m_rifle) kg x vr

Since the mass of the rifle is much greater than that of the bullet, we can approximate the mass of the rifle as 3 kg only. Solving the above equation for vr we get, vr = (5 x 10^-3 kg x 300 m/s)/3 kg = -0.5 m/s.

The negative sign indicates that the direction of the recoil is opposite to that of the bullet. The initial kinetic energy of the bullet and the rifle are zero. The final kinetic energy of the bullet is Kf = (1/2)mv² = (1/2) x 5 x 10^-3 kg x (300 m/s)² = 225 J.

The final kinetic energy of the rifle is Kr = (1/2)mv² = (1/2) x 3 kg x (0.5 m/s)^2 = 0.375 J.

For a bullet of mass 7.7 g, we can find its final kinetic energy using the same formula:

Kf = (1/2)mv² = (1/2) x 7.7 x 10^-3 kg x (300 m/s)² = 346.5 J.

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The biggest coal burning power station in the world is in Taiwan with a power output capacity of 5500 MW. (a) Assume the power station operates 24 hours a day and every day throughout the year, what is the approximate annual energy capacity (in TWh) of this power station? (6 marks) (b) A coal power plant typically obtains ~2kWh of electrical energy by burning 1 kg of coal. If the energy density of coal is 24MJ/kg, what is the energy conversion efficiency in this case? (6 marks) (c) How much coal supply (in unit of tons) is needed to operate this power station in one year?

Answers

(a) The approximate annual energy capacity of the power station is 48,180 TWh. (b) The energy conversion efficiency is 8.3%. (c) The amount of coal supply needed is 24,090,000,000 tonnes.

For part (a), we used the formula for annual energy capacity which takes into account the power output, hours of operation, and days of operation per year. For part (b), we used the energy obtained from burning 1 kg of coal and the energy density of coal to calculate the energy conversion efficiency. We used the formula for energy conversion efficiency and found that it is 8.3%.

For part (c), we used the amount of energy generated in one year and the energy obtained from burning 1 kg of coal to calculate the amount of coal needed. We used the formula for amount of coal needed and found that it is 24,090,000,000 tonnes.

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What is the momentum of a garbage truck that is 1.20 × 10 4 kg
and is moving at 35 m/s? p = Correct units kg*m/s Correct At what
speed would an 8.5 kg trash can have the same momentum as the
truck?

Answers

The trash can would need to be moving at a speed of approximately 4.94 × 10⁴ m/s to have the same momentum as the garbage truck.

The momentum (p) of an object is calculated by multiplying its mass (m) by its velocity (v). Therefore, the momentum can be expressed as:

p = m * v

Given that the garbage truck has a mass of 1.20 × 10⁴ kg and is moving at 35 m/s, we can calculate its momentum as:

p_truck = (1.20 × 10⁴ kg) * (35 m/s)

Calculating the product:

p_truck = 4.2 × 10⁵ kg·m/s

Now, we need to find the speed at which an 8.5 kg trash can would have the same momentum as the truck. Let's denote this speed as v_can.

Using the momentum formula, we can write:

p_can = (8.5 kg) * v_can

Since we want the momentum of the trash can to be equal to the momentum of the truck, we can set up the equation:

p_truck = p_can

Substituting the values:

4.2 × 10⁵ kg·m/s = (8.5 kg) * v_can

Solving for v_can:

v_can = (4.2 × 10⁵ kg·m/s) / (8.5 kg)

Calculating the division:

v_can = 4.94 × 10⁴ m/s

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Find the work (in foot-pounds) done by a force of 3 pounds acting in the direction 2i +3j in moving an object 4 feet from (0,0) to (4, 0)

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The work done by the force of 3 pounds acting in the direction 2i + 3j in moving an object 4 feet from (0,0) to (4, 0) is 12 foot-pounds.

We can now find the work done using the formula:

Work Done = Force x Displacement x Cosine of the angle between the force and displacement vectors

The force is 3 pounds in the direction 2i + 3j.

The force vector is the vector sum of its components i.e,3 (2i + 3j) = 6i + 9j

The angle between the force and displacement vectors is 0 degrees (since they act in the same direction).

Hence, the work done is given by:

Work Done = 3 x (4i) x cos 0°= 3 x 4 x 1= 12 foot-pounds

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The work done by the force of 3 pounds acting in the direction 2i + 3j in moving an object 4 feet from (0, 0) to (4, 0) is approximately 5.66 foot-pounds.

Given force is F = 3 pounds

Moving an object 4 feet from (0,0) to (4,0)

The direction in which the force acts = 2i+3j

First, we need to find the displacement of the object i.e., distance from (0, 0) to (4, 0).

We have,

Displacement = √[(4 - 0)² + (0 - 0)²]

Displacement = √(16)

Displacement = 4 feet

Now, the work done by the force is given by the formula:

Work done = Force x Displacement x cos θ

where θ is the angle between force and displacement

We have given,

F = 3 pounds

The displacement of the object is 4 feet

The direction in which the force acts is 2i + 3j

Let's find the displacement of the object using the distance formula:

Displacement = √[(4 - 0)² + (0 - 0)²]

Displacement = √(16)

Displacement = 4 feet

Let's find the angle between force and displacement:θ = tan⁻¹(3/2)θ = 56.31°

Now, we can find the work done by the force using the formula:

Work done = Force x Displacement x cos θ

Work done = 3 x 4 x cos 56.31°

Work done ≈ 5.66 foot-pounds

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if : T:Rn → Rmis a linear transformation and if c is in Rm, then a uniqueness question is "is c in the range of T"? True or

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If c is in the range of T, there exists at least one vector x such that T(x) = c, but there can be more than one vector x that satisfies this condition. The question of whether c is in the range of T is not a uniqueness question.

If: T:Rn → Rm is a linear transformation and if c is in Rm, then a uniqueness question is "is c in the range of T"? The given statement is False. The range of T, denoted by R(T), is the set of all possible outputs of T. For a linear transformation T:Rn → Rm, the range of T is a subspace of Rm.T

he uniqueness question is whether there is only one way to write c as a linear combination of the columns of the matrix A whose linear transformation T is given by T(x) = Ax. A vector c in Rm is in the range of T if and only if there exists a vector x in Rn such that T(x) = c. This is because for a linear transformation, the output is entirely dependent on the input and the transformation.

Therefore, if c is in the range of T, there exists at least one vector x such that T(x) = c, but there can be more than one vector x that satisfies this condition. In the domain of linear algebra, a linear transformation (also known as a linear operator or a linear map) is a linear function that maps one vector space to another vector space while preserving the operations of addition and scalar multiplication.

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a 2 kilogram cart has a velocity of 4 meters per second to the right. it collides with a 5 kilogram cart moving to the left at 1 meter per second. after the collision, the two carts stick together. can the magnitude and the direction of the velocity of the two carts after the collision be determined from the given information

Answers

Yes, the magnitude and direction of the velocity of the two carts after the collision can be determined using the conservation of momentum principle.

The solution to the given problem can be obtained through the application of the law of conservation of momentum which is given as;M1V1i + M2V2i = (M1 + M2)Vf where:M1 is the mass of cart 1V1i is the initial velocity of cart 1M2 is the mass of cart 2V2i is the initial velocity of cart 2Vf is the final velocity of the carts after collision.Since the two carts move in opposite directions before the collision, the direction will be to the right since it has a higher velocity of 4 m/s.To find the final velocity of the carts, substitute the given values into the conservation of momentum principle.M1V1i + M2V2i = (M1 + M2)Vf (2 kg) (4 m/s) + (5 kg)(-1 m/s) = (2 kg + 5 kg) VfVf = (8 kg m/s) / (7 kg) = 1.14 m/sThe final velocity of the two carts is 1.14 m/s to the right. This means that the direction of motion is to the right and the magnitude is 1.14 m/s.

To find the direction of motion of the two carts after the collision, we need to analyze the situation before and after the collision. Before the collision, the 2-kilogram cart is moving to the right with a velocity of 4 meters per second, while the 5-kilogram cart is moving to the left with a velocity of 1 meter per second. The two carts collide, and they stick together. After the collision, the two carts move as a single object. The law of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act on it. In this case, the two carts are the system, and there are no external forces acting on them. Therefore, the total momentum of the two carts before the collision is equal to the total momentum of the two carts after the collision. We can write this as:M1V1i + M2V2i = (M1 + M2)Vfwhere M1 is the mass of cart 1, V1i is the initial velocity of cart 1, M2 is the mass of cart 2, V2i is the initial velocity of cart 2, and Vf is the final velocity of the two carts after the collision.Substituting the values we have into the equation, we get:(2 kg)(4 m/s) + (5 kg)(-1 m/s) = (2 kg + 5 kg)VfSimplifying this equation, we get:8 kg m/s - 5 kg m/s = 7 kg Vf3 kg m/s = 7 kg VfVf = (3 kg m/s)/(7 kg) = 0.43 m/sSince the velocity of the two carts is to the right, we can ignore the negative sign. Therefore, the velocity of the two carts after the collision is 0.43 m/s to the right.

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what is the approximate thermal energy in kj/mol of molecules at 75 ° c?

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Answer:

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To calculate the approximate thermal energy in kilojoules per mole (kJ/mol) of molecules at a given temperature, you can use the Boltzmann constant (k) and the ideal gas law.

The Boltzmann constant (k) is approximately equal to 8.314 J/(mol·K). To convert this to kilojoules per mole, we divide by 1000:

k = 8.314 J/(mol·K) = 0.008314 kJ/(mol·K)

Now, we need to convert the temperature to Kelvin (K) since the Boltzmann constant is defined in Kelvin. To convert from Celsius to Kelvin, we add 273.15 to the temperature:

T(K) = 75°C + 273.15 = 348.15 K

Finally, we can calculate the thermal energy using the formula:

Thermal energy = k * T

Thermal energy = 0.008314 kJ/(mol·K) * 348.15 K

Thermal energy ≈ 2.894 kJ/mol

Therefore, at 75°C, the approximate thermal energy of molecules is approximately 2.894 kilojoules per mole (kJ/mol).

The heat capacity of one mole of water is approximately 75.29/1000 = 0.07529 kj/mol. This value represents the approximate thermal energy in kj/mol of water molecules at 75 ° C.

Thermal energy refers to the energy present in a system that arises from the random movements of its atoms and molecules. When a body has a temperature of 75 ° C, it has a thermal energy that depends on the type of molecules in it and their specific heat capacity.

In this context, we will consider the thermal energy in kj/mol of molecules at 75 ° C.Let's use water as an example to calculate the approximate thermal energy in kj/mol of molecules at 75 ° C. The specific heat capacity of water is 4.18 J/g °C, and the molar mass of water is 18.01528 g/mol. Therefore, the thermal energy in kj/mol of water molecules at 75 ° C can be calculated as follows:ΔH = mcΔt, whereΔH = thermal energy,m = mass of the sample,c = specific heat capacity of the sample,Δt = change in temperature

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Problem 4- Air at 25°C, 1 atm, and 30 percent relative humidity is blown over the surface of 0.3m X 0.3m square pan filled with water at a free stream velocity of 2m/s. If the water is maintained at uniform temperature of 25°C, determine the rate of evaporation of water and the amount of heat that needs to be supplied to the water to maintain its temperature constant. Mass diffusivity of water in air is DAB-2.54x10-5 m²/s. Kinematic viscosity of air is 0.14x10-4 m²/s. Density of air p=1.27 kg/m³. Saturation pressure of water at 25°C Psat, 25c-3.17 kPa, latent heat of water at 25°C hfg=334 kJ/kg. (20P)

Answers

The rate of evaporation of water is approximately 0.249 kg/s, and the amount of heat that needs to be supplied to the water to maintain its temperature constant is approximately 83.066 kW.

To determine the rate of evaporation of water and the amount of heat required, we can use the equation for mass transfer rate:

m_dot = (ρ * A * V * x) / (D_AB * L)

where m_dot is the mass transfer rate (rate of evaporation), ρ is the density of air, A is the surface area of the pan, V is the free stream velocity, x is the humidity ratio (absolute humidity), D_AB is the mass diffusivity of water in air, and L is the characteristic length (assumed to be the depth of the water in this case).

T_air = 25°C = 298 K (temperature of air)

P = 1 atm (pressure of air)

RH = 30% (relative humidity)

V = 2 m/s (free stream velocity)

A = 0.3 m x 0.3 m = 0.09 m² (surface area of the pan)

D_AB = 2.54 x 10^-5 m²/s (mass diffusivity of water in air)

ρ = 1.27 kg/m³ (density of air)

L = depth of water in the pan = unknown (assumed to be equal to the height of the pan, 0.3 m)

To calculate x, the humidity ratio, we can use the equation:

x = (RH * P_s) / (P - RH * P_s)

where P_s is the saturation pressure of water at the given temperature.

Given values:

T_water = 25°C = 298 K (temperature of water)

P_s_25c = 3.17 kPa = 3.17 x 10³ Pa (saturation pressure of water at 25°C)

Plugging in the values, we can calculate x:

x = (0.3 * 3.17 x 10³) / (1 - 0.3 * 3.17 x 10³)

x ≈ 0.000957 kg/kg (humidity ratio)

Now we can calculate the rate of evaporation (m_dot):

m_dot = (ρ * A * V * x) / (D_AB * L)

m_dot = (1.27 * 0.09 * 2 * 0.000957) / (2.54 x 10^-5 * 0.3)

m_dot ≈ 0.249 kg/s

To calculate the amount of heat required to maintain the temperature constant, we can use the equation:

Q = m_dot * h_fg

where h_fg is the latent heat of water at the given temperature.

Given value:

h_fg_25c = 334 kJ/kg (latent heat of water at 25°C)

Plugging in the values, we can calculate Q:

Q = 0.249 * 334

Q ≈ 83.066 kW

The rate of evaporation of water is approximately 0.249 kg/s, and the amount of heat that needs to be supplied to the water to maintain its temperature constant is approximately 83.066 kW.

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Though opinion polls usually make 95% confidence statements, some sample surveys use other confidence levels. The monthly unemployment rate, for example, is based on the Current Population Survey of a why is it considered poor programming practice to have public instance variables? Qn.1 How is the "Function of management" relevant to the importance of organisational success? with more than 400 wordsQn.,2 What are the most significant elements relating to Function of management? with more than 500 words and a creative answer please Crane Enterprises is considering manufacturing a new product. It projects the cost of direct materials and rent for a range of output as shown below. Output Rent in Units Expense $7,235 7,235 1,000 2,000 3,000 4,000 5,000 6,000 7,000 8,000 9,000 10,000 11,000 11,576 11,576 11,576 11,576 11,576 11,576 14,470 14,470 14,470 Direct Materials $5,788 8,700 8,700 11,600 14,500 17,400 20,300 23,200 42,397 50,645 63,668 Your answer has been saved. See score details after the due date. Determine the relevant range of activity for this product. The relevant range of activity for this product (c) Your answer has been saved. See score details after the due date. 3,000-8,000 Variable costs per unit per unit (d) Calculate the variable costs per unit within the relevant range. (Round answer to 2 decimal places e.g. 2.25.) V 2.90 units. Attempts: 1 of 1 used Attempts: 1 of 1 used Randois samples of four different models of cars were selected and the gas mileage of each car was meased. The results are shown below Z (F/PALE ma II # 21 226 22 725 21 Test the claim that the four d 4. A simple way to generate some idea whether data are likely to be heteroskedastic is to A. examine the residual plot B.construct a histogram. C. Calculate the variance of the sample. D.plot the data points from smallest to largest calculate the amount of work done to move 1 kg mass from the surface of the earth to a point 10 km from the centre of the earth. the small, semisolid mass of food formed during mastication is called a 1. i. Force is equal to mass x acceleration and is typically expressed in units of Newtons (kg m s-2). Acceleration is the rate of change of velocity. If gravitational acceleration is equal to 9.8 m s-2, then what is the gravitation force experienced by the mass of air in the box in: a. kg m s-2? b. Newtons (N)? (Note: again, this question does not involve a conversion but rather use of an equation.) ii. Pressure is equal to force divided by the area over which the force is applied and is typically expressed in units of Nm-2 (Pa). If the box from the previous questions rests on the Earth's surface, what is the pressure exerted by the gravitational force over the bottom area of the box (16 m2) in: a. N m-2? b. Pa? iii. Pressure on weather maps is usually expressed in units of bars, where one bar (100,000 Pa) approximates the average sea-level pressure (101,325 Pa or approximately 100,000 Pa). Using this and other aids (see appendix), convert 1 mb to Pa. determine the mean and variance of the random variable with the following probability mass function. f(x)=(64/21)(1/4)x, x=1,2,3 round your answers to three decimal places (e.g. 98.765). Arabian Gulf Corporation reports the following stockholders' equity section on December 31, 2020 - Common stock; $10 par value; 500,000 shares authorized; 200,000 shares issued and outstanding $ 2,000,000 - Paid in capital in excess of par value, common stock - Retained earnings... 400,000 900,000 Total $3,300,000 The Corporation completed the following transactions in 2021. 1-Jan 10, Directors declared a $1 per share cash dividend payable on March 15 to the Jan 31 stockholders of record 2- Mar 01, Purchased 10,000 shares of its own common for $15 per share. 3- Mar 15, Paid the cash dividend declared on Jan. 10. 4- May 01, Sold 6,000 of its treasury shares at $15 cash per share. 5- Sep 30, Directors declared a 30% stock dividend when the share market price is $16. 6- Nov 01, Distributed stock dividends declared on Sep. 30. 7- Nov 15, The company implemented 5-for-1 stock split for the common stock. Required: Prepare journal entries to record each of these transactions for 2021. In 2020, when the enacted tax rate for the current and all future periods was 27.5%, Garza Corp. had a taxable loss of $468,000 and elected to use the net operating loss carryforward provision.In 2021, the tax rate changed to 25.0% for the current and all future periods, and Garza reported taxable income of $311,000.In 2022, Garza reported taxable income of $664,000. Garza has no book-tax differences.What amount will Garza report as current income tax expense on its 2022 income statement? What are 5 key terms that describe cloud computing Question 2 (8 marks) A fruit growing company claims that only 10% of their mangos are bad. They sell the mangos in boxes of 100. Let X be the number of bad mangos in a box of 100. (a) What is the dist .Problem 7-40 (LO. 5)Blue Corporation, a manufacturing company, decided to develop a new line of merchandise. The project began in 2019. Blue had the following expenses in connection with the project:20192020Salaries$500,000$600,000Materials90,00070,000Insurance8,00011,000Utilities6,0008,000Cost of inspection of materials for quality control7,0006,000Promotion expenses11,00018,000Advertising020,000Equipment depreciation15,00014,000Cost of market survey8,0000Question Content AreaThe new product will be introduced for sale beginning in July 2021. Determine the amount of the deduction for research and experimental expenditures for 2019, 2020, 2021, and 2022.If an amount is zero, enter "0". Calculate the monthly expense to the nearest dollar and use in subsequent computations.a. If Blue Corporation elects to expense the research and experimental expenditures, what will the amount of the deduction be?201920202021 and 2022Amount of the deduction$ 619,000$703,000$ 0b. If Blue Corporation elects to amortize the research and experimental expenditures over 60 months, what will the amount of the deduction be?2019202020212022Amount of the deduction$ 0$ 0$132,198$264,396c. How would your answer change if Blue Corporation incurred the expenses in 2022 and 2023 (rather than 2019 and 2020)?20222023Amount of the deduction$?$? How fast do you have to throw the rock so that it never comes back to the asteroid and ends up traveling at a speed of 10 m/s when it is very far away? When preparing the report to analyze a proposed quality improvement program, which of the following costs are included in the total costs of not undertaking the quality improvement program?A.inspection of finished goodsB.preventive maintenanceC.sales returnsD.total appraisal costs distinguish between the exocrine and endocrine secretions of the pancreas For each of the following strong base solutions, determine [OH][OH] and [H3O+][H3O+] and pHpH and pOHpOH.For 5.21045.2104 MM Ca(OH)2Ca(OH)2, determine [OH][OH] and [H3O+][H3O+]. When applying the co-terminated assumption: A study period equal to the minimum common multiple of the lives of the two alternatives is selected and used to evaluate both alternatives Each alternative is evaluated with its own study period which is equal to its life time A study period equal to the average of the life times of both alternatives is selected to be able to compare them with economic equivalence methods O A study period equal to the life of one of the alternatives is selected, and the life of the other alternative is adjusted to the same study period