Answer:
Magnitude of the electrostatic force on the +32 µC charge, [tex]F_{net} = 12 N[/tex]
Explanation:
Let q₁ = +32 µC, x₁ = 0
q₂ = +20 µC, x₂ = 40 cm = 0.4 m
q₃ = -60 µC, x₃ = 60 cm = 0.6 m
Let magnitude of the electrostatic force on the +32 µC charge due to the + 20 µC charge = F₁ (i.e force on q₁ due to q₂)
[tex]F_{2} = \frac{kq_{1}q_{2} }{x_{2}^2 }[/tex]
[tex]F_{2} = \frac{9 * 10^{9} * 32 * 10^{-6} * 20 * 10^{-6} }{0.4^2 }\\F_{2} = 36 N[/tex]
Let magnitude of the electrostatic force on the +32 µC charge due to the -60 µC charge = F₂ (i.e force on q₁ due to q₃)
[tex]F_{3} = \frac{kq_{1}q_{3} }{x_{3}^2 }[/tex]
[tex]F_{3} = -\frac{9 * 10^{9} * 32 * 10^{-6} * 60 * 10^{-6} }{0.6^2 }\\F_{3} =-48 N[/tex]
The electrostatic force on the 32 µC charge, [tex]F_{net} = |F_{2} + F_{3}|[/tex]
[tex]F_{net} =| 36 + (-48)| \\F_{net} =|- 12 N| \\ F_{net} = 12 N[/tex]
How is the particle displacement related to the direction of wave movement in a longitude wave?
Answer:
The displacement of particles is perpendicular to the direction of wave motion.
Which is the correct representation of the right-hand rule for a current flowing to the right?
Answer:
The third image
Explanation:
The one with the thumb pointing to the right
Answer:
3, correct on Edge 2020
does work done by the electric force depends on the path taken.
Answer:
yes but not sure
Explanation:
i don't know so much but I just guessed I'm not sure
A long solid conducting cylinder with radius a = 12 cm carries current I1 = 5 A going into the page. This current is distributed uniformly over the cross section of the cylinder. A cylindrical shell with radius b = 21 cm is concentric with the solid cylinder and carries a current I2 = 3 A coming out of the page. 1)Calculate the y component of the magnetic field By at point P, which lies on the x axis a distance r = 41 cm from the center of the cylinders.
Answer:
Explanation:
We shall use Ampere's circuital law to find magnetic field at required point.
The point is outside the circumference of two given wires so whole current will be accounted for .
Ampere's circuital law
B = ∫ Bdl = μ₀ I
line integral will be over circular path of radius r = 41 cm .
Total current I = 5A -3A = 2A .
∫ Bdl = μ₀ I
2π r B = μ₀ I
2π x .41 B = 4π x 10⁻⁷ x 2
B = 2 x 10⁻⁷ x 2 / .41
= 9.75 x 10⁻⁷ T . It will be along - ve Y - direction.
Forests have been completely wiped out in some parts the world. In areas that are economically depressed, people use the cut wood to fuel fires for cooking and for making shelters or homes. How would deforestation most likely impact Earth's biogeochemical cycles?
Group of answer choices
Less carbon dioxide will be present in the atmosphere because the people living in the area still breath in oxygen.
An increase in carbon dioxide will occur in the atmosphere because there are less trees to use the gas in the process of photosynthesis.
Carbon dioxide levels will not change, since plants do not have a role in the carbon cycle.
More water will be released into the atmosphere because transpiration and condensation will produce more precipitation.
Answer: An increase in carbon dioxide will occur in the atmosphere because there are less trees to use the gas in the process of photosynthesis.
Explanation: Photosynthesis can be simply defined as a process where green plants uses minerals from the soil, water and carbon dioxide to produce their own food (energy-rich organic compound/starch) while oxygen is released into the environment.
On the other hand, Human beings and other animals inhales the oxygen released into the environment by green plants and breath out carbon dioxide.
Deforestation is the permanent removal trees either for commercial or household use or in some cases, to make room or space for other activities or development.
Now when these trees (green plants) are permanently cut down without replacement, humans and other animals continuously exhale carbon dioxide and there will be little or no green plants (trees) available to convert this carbon dioxide into oxygen.
Hence, there will be an increased amount of carbon dioxide in the atmosphere as their are little or no tress to use the gas (carbon dioxide) in the process of photosynthesis.
A Texas cockroach of mass 0.157 kg runs counterclockwise around the rim of a lazy Susan (a circular disk mounted on a vertical axle) that has a radius 14.9 cm, rotational inertia 5.92 x 10-3 kg·m2, and frictionless bearings. The cockroach's speed (relative to the ground) is 2.92 m/s, and the lazy Susan turns clockwise with angular velocity ω0 = 3.89 rad/s. The cockroach finds a bread crumb on the rim and, of course, stops. (a) What is the angular speed of the lazy Susan after the cockroach stops? (b) Is mechanical energy conserved as it stops?
Answer:
-7.23 rad/s
Explanation:
Given that
Mass of the cockroach, m = 0.157 kg
Radius of the disk, r = 14.9 cm = 0.149 m
Rotational Inertia, I = 5.92*10^-3 kgm²
Speed of the cockroach, v = 2.92 m/s
Angular velocity of the rim, w = 3.89 rad/s
The initial angular momentum of rim is
Iw = 5.92*10^-3 * 3.89
Iw = 2.3*10^-2 kgm²/s
The initial angular momentum of cockroach about the axle of the disk is
L = -mvr
L = -0.157 * 2.92 * 0.149
L = -0.068 kgm²/s
This means that we can get the initial angular momentum of the system by summing both together
2.3*10^-2 + -0.068
L' = -0.045 kgm²/s
After the cockroach stops, the total inertia of the spinning disk is
I(f) = I + mr²
I(f) = 5.92*10^-3 + 0.157 * 0.149²
I(f) = 5.92*10^-3 + 3.49*10^-3
I(f) = 9.41*10^-3 kgm²
Final angular momentum of the disk is
L'' = I(f).w(f)
L''= 9.41*10^-3w(f)
Using the conservation of total angular momentum, we have
-0.068 = 9.41*10^-3w(f) + 0
w(f) = -0.068 / 9.41*10^-3
w(f) = -7.23 rad/s
Therefore, the speed of the lazy Susan after the cockroach stops is -7.23 and is directed in the opposite direction of the initial lazy Susan angular speed
b)
The mechanical energy of the cockroach is not converted as it stops
A ball with a mass of 4 kg is initially traveling at 2 m/s and has a 5 N force applied for 3 s. What is the initial momentum of the ball?
Answer:
The initial momentum of the ball is 8 kg-m/s.
Explanation:
Given that,
Mass of the ball is 4 kg
Initial speed of the ball is 2 m/s
Force applied to the ball is 5 N for 3 seconds
It is required to find the initial momentum of the ball. Initial momentum means that the product of mass and initial velocity of the ball. It is given as :
[tex]p_i=mu\\\\p_i=4\ kg\times 2\ m/s\\\\p_i=8\ kg-m/s[/tex]
So, the initial momentum of the ball is 8 kg-m/s.
How the musculoskeletal and nervous system develop as a human grows
Answer:
Explanation:
A fan pushes hot air out of a vent and into a room. The hot air displaces cold air in the room, causing the cold air to move closer to the floor.
The hot air displacing the cold air is an example of transfer by
A. A PH202 student lives next to a construction site and sees a crane with a wrecking ball demolish the building next door. The wrecking ball swings along the wall between her house and the neighbor’s house. In an effort to determine the length of the cable on the wrecking ball the student builds a pendulum using a random rock and a string. Her pendulum turns out to be 0.500m long. While she plays with her pendulum she realizes that the wrecking ball swings back and forth in the same amount of time that it takes the rock to complete 5 full oscillations. What is the length of the cable on the wrecking ball?
Answer:
The length of cable is 12.5 m
Explanation:
Since, the wrecking ball completes 1 oscillation, in the same time, as it takes for the rock to complete 5 oscillations.
Therefore,
Time Period of Wrecking Ball = 5 (Time Period of Rock)
Since,
Time Period of Pendulum = 2π√(L/g)
Therefore,
2π√(L₁/g) = 5[2π√(L₂/g)]
√L₁ = 5√L₂
Squaring on both sides:
L₁ = 25 L₂
where,
L₁ = Length of Cable = ?
L₂ = Length of string = 0.5 m
Therefore,
L₁ = 25 (0.5 m)
L₁ = 12.5 m
A solid cylinder of mass m and radius R rolls down a ramp, starting from rest at a height h above a nearby horizontal surface. The coefficients of kinetic and static friction and are non-zero, and sufficiently large that the cylinder rolls down the ramp without slipping. Assume that the coefficient of rolling friction is zero. As the cylinder leaves the ramp, it continues along a horizontal surface (with the same frictional coefficients as the ramp).
Required:
What is the speed V of the cylinder after it has traveled a distance D along the horizontal surface?
Answer:
the volocity is 50
Explanation:
A steam engine takes in superheated steam at 270 °C and discharges condensed steam from its cylinder at 50 °C. The engine has an efficiency of 30%, and taken in 50 kJ from the hot steam per cycle. If a Carnot engine takes in the same amount of heat per cycle and operates at these temperatures, the work it can turn into is most likely to be:a) 15 kJ. b) 20 kJ. c) 10 kJ. d) 50 kJ.
Answer:
b) 20 kJ
Explanation:
Efficiency of carnot engine = (T₁ - T₂ ) / T₁ Where T₁ is temperature of hot source and T₂ is temperature of sink .
T₁ = 270 + 273 = 543K
T₂ = 50 + 273 = 323 K
Putting the given values of temperatures
efficiency = (543 - 323) / 543
= .405
heat input = 50 KJ
efficiency = output work / input heat energy
.405 = output work / 50
output work = 20.25 KJ.
= 20 KJ .
Based on the simple blackbody radiation model described in class, answer the following question. The planets Mars and Venus have albedo values of 0.15 and 0.75, and observed surface temperatures of approximately 220 K and 700 K, respectively. The average distance of Mars from the sun is 2.28 x 108 km, and the average distance of Venus is 1.08 x 108 km. Given that the radius of the sun is 7 x 108 m, and the energy flux at the surface of the sun is 6.28 x 107 W/m2 , what is the extent of the greenhouse effect for Mars
Answer:
The extent of greenhouse effect on mars is [tex]G_m = 87 K[/tex]
Explanation:
From the question we are told that
The albedo value of Mars is [tex]A_1 = 0.15[/tex]
The albedo value of Mars is [tex]A_2 = 0.15[/tex]
The surface temperature of Mars is [tex]T_1 = 220 K[/tex]
The surface temperature of Venus is [tex]T_2 = 700 K[/tex]
The distance of Mars from the sun is [tex]d_m = 2.28*10^8 \ km = 2.28*10^8* 1000 = 2.28*10^{11} \ m[/tex]
The distance of Venus from sun is [tex]d_v = 1.08 *10^{8} \ km = 1.08 *10^{8} * 1000 = 1.08 *10^{11} \ m[/tex]
The radius of the sun is [tex]R = 7*10^{8} \ m[/tex]
The energy flux is [tex]E = 6.28 * 10^{7} W/m^2[/tex]
The solar constant for Mars is mathematically represented as
[tex]T = [\frac{E R^2 (1- A_1)}{\sigma d_m} ][/tex]
Where [tex]\sigma[/tex] is the Stefan's constant with a value [tex]\sigma = 5.6*10^{-8} \ Wm^{-2} K^{-4}[/tex]
So substituting values
[tex]T = \frac{6.28 *10^{7} * (7*10^8)^2 * (1-0.15)}{(5.67 *10^{-8}) * (2.28 *10^{11})^2)}[/tex]
[tex]T = 307K[/tex]
So the greenhouse effect on Mars is
[tex]G_m = T - T_1[/tex]
[tex]G_m = 307 - 220[/tex]
[tex]G_m = 87 K[/tex]
A particle is projected at an angle 60 degrees to the horizontal with a speed of 20m/s. (i) calculate total time of flight of the particle. (i) speed of the particle at its maximum height
Answer:
Time of flight=3.5 seconds
Speed at maximum height is 0
Explanation:
Φ=60°
initial velocity=u=20m/s
Acceleration due to gravity=g=9.8 m/s^2
Total time of flight=T
Final speed=v
question 1:
T=(2 x u x sinΦ)/g
T=(2 x 20 x sin60)/9.8
T=(2 x 20 x 0.8660)/9.8
T=34.64/9.8
T=3.5 seconds
Question 2
Speed at maximum height is 0
What is the gravitational force related to the distance between two objects?
Answer:
Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces. So as two objects are separated from each other, the force of gravitational attraction between them also decreases.
answer: It is inversely proportional to the square of the distance.
Explanation:
for the team
The shaft of a motor has an angular displacement θ that is a function of time given by the equation: θ(t) = 4.40 t 3 rad/s3 + 1.40 t2 rad/s2 . At time t = 0.00 s the wheel is at rest and is oriented at θ = 0.00 rad. a) Derive the equation that specifies the angular velocity of the shaft as a function of time. b) Derive the equation that specifies the angular acceleration as a function of time.
Answer:
a) [tex]\omega = 13.2t^2\frac{rad}{s^3}+2.80t\frac{rad}{s^2}[/tex]
b) [tex]\alpha=26.4t\frac{rad}{s^3}+2.80\frac{rad}{s^2}[/tex]
Explanation:
You have that the angular displacement is given by:
[tex]\theta=4.40t^3\frac{rad}{s^3}+1.40t^2\frac{rad}{s^2}[/tex]
a) the angular velocity is given by the derivative in time, of the angular displacement, that is:
[tex]\omega=\frac{d\theta}{dt}=\frac{d}{dt}[4.40 t^3 rad/s^3 + 1.40 t^2 rad/s^2]\\\\\omega=\frac{d\theta}{dt}=13.2t^2\frac{rad}{s^3}+2.80t\frac{rad}{s^2}[/tex]
b) the angular acceleration is the derivative, in time, of the angular velocity:
[tex]\alpha=\frac{d\omega}{dt}=\frac{d}{dt}[13.2t^2\frac{rad}{s^3}+2.80t\frac{rad}{s^2}]\\\\\alpha=26.4t\frac{rad}{s^3}+2.80\frac{rad}{s^2}[/tex]
Einstein developed much of his understanding of relativity through the use of gedanken, or thought, experiments. In a gedanken experiment, Einstein would imagine an experiment that could not be performed because of technological limitations, and so he would perform the experiment in his head. By analyzing the results of these experiments, he was led to a deeper understanding of his theory. In each the following gedanken experiments, Albert is in the exact center of a glass-sided freight car speeding to the right at a very high speed vvv relative to you. Albert has a flashlight in each hand and directs them at the front and rear ends of the freight car. Albert switches the flashlights on at the same time.
In Albert's frame of reference, which beam of light travels at a greater speed, the one directed toward the front or the one toward the rear of the train, or do they travel at the same speed? Which beam travels faster in your frame of reference? Enter the answers for Albert's frame of reference and your frame of reference separated by a comma using the terms front, rear, and same. For example, if in Albert's frame of reference the beam of light directed toward the front of the train travels at a greater speed and in your frame of reference the two beams travel at the same speed, then enter front,same.
Answer:
For eintein's frame of reference, both beam travel at the same speed.
For my own frame of reference, both beams travel at the same speed.
Explanation:
According to special relativity, the speed of light is the same in all direction on all reference frame. If not for this law we will assume the from beam will have a relative speed that will be the speed of light plus the speed of the fright car. This is not so and it violates the speed limit of light which according to the first law is the highest speed possible and nothing can go beyond that.
8. At temperature 15°C, aluminum rivets have a diameter of 0.501 cm, and holes drilled in a titanium sheet have a diameter of 0.500 cm. If both the aluminum rivets and the titanium sheet are cooled together, at what temperature will the rivets just fit into the appropriate holes in the titanium sheet? Use 25x10-6 (°C)-1 for the coefficient of linear expansion for aluminum, and 8.5x10-6 (°C)-1 for titanium
Answer:
The temperature is [tex]T = -106 ^oC[/tex]
Explanation:
From the question we are told that
The temperature is [tex]T_1 = T_t= T_a=15^oC[/tex]
The diameter is [tex]d_1 = 0.5001 cm[/tex]
The diameter of the hole [tex]d_2 = 0.500 \ cm[/tex]
The coefficient of linear expansion for aluminum is [tex]\alpha _1 = 25 *10^{-6} \ ^oC^{-1}[/tex]
The coefficient of linear expansion for titanium is [tex]\alpha _2 = 8.5 *10^{-6} \ ^o C^{-1}[/tex]
According to the law of linear expansion
[tex]d = d_o (1 + \alpha \Delta T )[/tex]
Where [tex]d_o[/tex] represents the original diameter
So for aluminum
[tex]d_a = d_1 (1 + \alpha_1 (T- T_a) )[/tex]
Where [tex]d_a[/tex] is the new diameter of aluminum
[tex]T_a[/tex] is the new temperature of the aluminum
So for titanium
[tex]d_t = d_2 (1 + \alpha_1 (T- T_t) )[/tex]
Where [tex]d_t[/tex] is the new diameter of titanium
[tex]T_t[/tex] is the new temperature of the aluminum
So for the aluminum rivets to fit into the holes
[tex]d_a = d_t[/tex]
=> [tex]d_1 (1 + \alpha_1 (T- T_a) ) = d_2 (1 + \alpha_2 (T- T_t) )[/tex]
Making T the subject of the formula
[tex]T = \frac{(d_1 - d_2 ) + (d_2 *\alpha_2 T_t) - d_1 \alpha_1 * T_a }{d_2 \alpha_2 - d_1 \alpha_1 }[/tex]
Substituting values
[tex]T = \frac{(0.501 - 0.500 ) + (0.500 *(8.5*10^{-6}) * 15) - 0.500* (25*10^{-6}) * 15 }{0.500 * (8.5 *10^{-6}) - 0.501 * (25 *10^{-6}) }[/tex]
[tex]T = -106 ^oC[/tex]
What do you think will be different about cars in the future? Describe a change that is already being developed or that you think should be invented.
Answer:
Flying cars.
Explanation:
A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes and the car comes to a rest uniformly in a distance of 160 m. What are the magnitude and direction of the net force applied to the car to bring it to rest?
Answer:
Force applied to stop the car = 1,250 N
Explanation:
Given:
Mass of car (M) = 1,000 kg
Initial velocity (U) = 20 m/s
Final velocity (V) = 0 m/s
Distance (S) = 160 m
Find:
Force applied to stop the car.
Computation:
[tex]v^2 = u^2 + 2as\\\\0^2=20^2+2(a)(160)\\\\0=400+320(a)\\\\Acceleration = a = -1.25m/s^2\\\\Force = ma \\\\Force= 1,000(1.25)\\\\Force = 1,250 N[/tex]
Force applied to stop the car = 1,250 N
A bike travels 15.0 km in 45.0 min. Its average speed in km/h is .
The average speed of a bus traveling a distance of 15.0 km in 45.0 min is 20 km/hour.
What is speed?The speed of an object, also known as v in kinematics, is the size of the change in that object's position over time or the size of the change in that object's position per unit of time, making it a scalar quantity.
The distance travelled by an object in a time interval is divided by the length of the interval to determine its average speed.
Distance travelled by the bike = 15.0 km
Time taken by the object = 45.0 minute = (45.0 ÷ 60) hour = 0.75 hour
Hence, the average speed of the object = distance travelled / time taken
= 15.0 km/0.75 hour
= 20 km/hour,
Therefore, the average speed of a bus traveling a distance of15.0 km in 45.0 min is 20 km/hour.
Learn more about speed here:
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A piston with stops containing water goes through an expansion process through the addition of heat. State 1 the pressure is 200 kPa and the volume is 2 m3. After half of the heat has been delivered the piston hits the stops corresponding to a volume of 5 m3. After all the heat has been delivered, state 2, the pressure is 1000 kPa with the piston resting on the stops. What is the work?
Answer:
The work will be "600 kJ/kg".
Explanation:
(1-a) ⇒ Constant Pressure
(a-2) ⇒ Constant Volume
The given values are:
In state 1,
Pressure, P₁ = 200 kPa
Volume, V₁ = 2m³
In state 2,
Pressure, P₂ = 1000 kPa
Volume, V₁ = 5m³
Now,
In process (1-a), work will be:
⇒ W₁₋ₐ = P₁(Vₐ - V₁)
On putting the values, we get
⇒ W₁₋ₐ = 200(5-2)
⇒ = 200(3)
⇒ = 600 kJ/kg
In process (a-2), work will be:
⇒ Wₐ₋₂ = 0
∴ (The change in the volume will be zero.)
So,
⇒ Total work = (W₁₋ₐ) + (Wₐ₋₂)
⇒ = 600 + 0
⇒ = 600 kJ/kg
Modern wind turbines generate electricity from wind power. The large, massive blades have a large moment of inertia and carry a great amount of angular momentum when rotating. A wind turbine has a total of 3 blades. Each blade has a mass of m = 5500 kg distributed uniformly along its length and extends a distance r = 46 m from the center of rotation. The turbine rotates with a frequency of f = 11 rpm.
Required:
a. Calculate the total moment of inertia of the wind turbine about its axis, in units of kilogram meters squared.
b. Calculate the angular momentum of the wind turbine, in units of kilogram meters squared per second.
Answer:
Explanation:
moment of inertia of each blade which is similar to rod rotating about its one end
= 1/3 ml²
moment of inertia of 3 blades = ml²
= 5500 x 46²
I = 11638 x 10³ kg m²
angular velocity = 2πn where n is rotation per second
n = 11 / 60
angular velocity = 2π x 11/60
= 1.1513 rad /s
angular momentum
= moment of inertia x angular velocity
= 11638 x 10³ x 1.1513
= 13399 x 10³ kg m² per second.
A student performs an experiment that involves the motion of a pendulum. The student attaches one end of a string to an object of mass M and secures the other end of the string so that the object is at rest as it hangs from the string. When the student raises the object to a height above its lowest point and releases it from rest, the object undergoes simple harmonic motion. As the student collects data about the time it takes for the pendulum to undergo one oscillation, the student observes that the time for one swing significantly changes after each oscillation. The student wants to conduct the experiment a second time. Which two of the following procedures should the student consider when conducting the second experiment?
a) Make sure that the length of the string is not too long.
b) Make sure that the mass of the pendulum is not too large.
c) Make sure that the difference in height between the pendulum's release position and rest position is not too large.
d) Make sure that the experiment is conducted in an environment that has minimal wind resistance.
Answer:
the answers the correct one is cη
Explanation:
In this simple pendulum experiment the student observes a significant change in time between each period. This occurs since an approximation used is that the sine of the angle is small, so
sin θ = θ
with this approach the equation will be surveyed
d² θ / dt² = - g / L sin θ
It is reduced to
d² θ / dt² = - g / L θ
in which the time for each oscillation is constant, for this approximation the angle must be less than 10º so that the difference between the sine and the angles is less than 1%
The angle is related to the height of the pendulum
sin θ = h / L
h = L sin θ.
Therefore the student must be careful that the height is small.
When reviewing the answers the correct one is cη
Considering the approximation of simple harmonic motion, the correct option is:
(c) Make sure that the difference in height between the pendulum's release position and rest position is not too large.
Simple Harmonic MotionAccording to Newton's second law in case of rotational motion, we have;
[tex]\tau = I \alpha[/tex]
Applying this, in the case of a simple pendulum, we get;
[tex]-mg\,sin\,\theta =mL^2 \,\frac{d^2 \theta}{dt^2}[/tex]
On, rearranging the above equation, we get;
[tex]mL^2 \,\frac{d^2 \theta}{dt^2} + mg\,sin\,\theta=0\\\\\implies \frac{d^2 \theta}{dt^2} +\frac{g}{L} sin \,\theta=0[/tex]
Now, if angular displacement is very small, i.e.; the bob of the pendulum is only raised slightly.
Then, [tex]sin\, \theta \approx \theta[/tex]
[tex]\implies \frac{d^2 \theta}{dt^2} +\frac{g}{L} \,\theta=0[/tex]
This is now in the form of the equation of a simple harmonic motion.
[tex]\frac{d^2 \theta}{dt^2} +\omega^2 \,\theta=0[/tex]
Comparing both these equations, we can say that;
[tex]\omega = \sqrt{\frac{g}{L}}[/tex]
[tex]T=2\pi\sqrt{\frac{L}{g}}[/tex]
This relation for the time period can only be obtained if the angular displacement is very less.
So, the correct option is;
Option (c): Make sure that the difference in height between the pendulum's release position and rest position is not too large.
Learn more about simple harmonic motion here:
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5.00 kg of liquid water is heated to 100.0 °C in a closed system. At this temperature, the density of liquid water is 958 kg/m3 . The pressure is maintained at atmospheric pressure of 1.01 x 105 Pa. A moveable piston of negligible weight rests on the surface of the water. The water is then converted to steam by adding an additional amount of heat to the system. When all of the water is converted, the final volume of the steam is 8.50 m3 . The latent heat of vaporization of water is 2.26 x 106 J/kg. Calculate how much work is done and the change in the internal energy during this isothermal process.
Answer:
1.04 x 107 J.
Explanation:
We can use the following method to do the calculation
Total energy given to water to convert intosteam
dQ = m* l
dQ = 5.00* 2.26 * 106
= 1.13* 107 J
Work done at constantpressure dW = P* dV
Initialvolume V1 = 5.00kg / 958
= 5.22* 10-3 m3
Finalvolume = 8.50 m3
=> dW = 1.01* 105 * ( 8.50 - 5.22 * 10-3)
= 8.58* 105 J
First law of thermodynamicsis dQ = ΔU + dW
Change in internalenergy ΔU = 1.13* 107 - 8.58 *105
= 1.04 x 107 J as our answer
A bicycle coasting downhill reaches its maximum speed at the bottom of the
hill.
This speed would be even greater if some of the bike's
energy had
not been transformed into
energy
A) kinetic; heat
OB) heat; potential
C) kinetic; potential
OD) potential; kinetic
OB
mmnjnjlkdhfutydjfyiudtkcgvyftdcgvjyiluftgyiuyu ( had to do that cuz it wouldn't let through)
Electric fields are MOST associated with ________.
Help ill give you brainliest !!!
Answer:
1. B
2. A
3. C
4. B
5. A
6. Muscular strength is different than muscular endurance because of the fact that muscular strength is the amount of force that can be exerted in one instance. Muscular endurance is how long that you can exert that force without being completely exhausted.
7. Some benefits to strength training is the increase in muscular endurance. There is also the benefit of better muscular strength.
Explanation:
By which process does the heat from the Sun reach the Earth? (AKS 4b DOK 1) *
10) Two students want to use a 12-meter long rope to create standing waves. They first measure the speed at which a single wave pulse moves from one end of the rope to another and find that it is 36 m/s. What frequency must they vibrate the rope at to create the second harmonic
Answer:
To create a second harmonic the rope must vibrate at the frequency of 3 Hz
Explanation:
First we find the fundamental frequency of the rope. The fundamental frequency is the frequency of the rope when it vibrates in only 1 loop. Therefore,
f₁ = v/2L
where,
v = speed of wave = 36 m/s
L = Length of rope = 12 m
f₁ = fundamental frequency
Therefore,
f₁ = (36 m/s)/2(12 m)
f₁ = 1.5 Hz
Now the frequency of nth harmonic is given in general, as:
fn = nf₁
where,
fn = frequency of nth harmonic
n = No. of Harmonic = 2
f₁ = fundamental frequency = 1.5 Hz
Therefore,
f₂ = (2)(1.5 Hz)
f₂ = 3 Hz
the speed of sound is 343m/s. dezeirey is positioned 5m behind her. how many seconds will it take for the echo from the wall to reach her
Answer:
t = 0.029 s
Explanation:
We have,
Speed of sound is 343 m/s.
Dezeirey is positioned 5 m behind her.
It is required to find the time taken for the echo from the wall to reach her. The total distance covered by the echo when it reaches her is 2d or 10 m.
Time taken,
[tex]t=\dfrac{d}{v}\\\\t=\dfrac{10\ m}{343\ m/s}\\\\t=0.029\ s[/tex]
So, it will take 0.029 seconds for the echo from the wall to reach her.
