A survey was conducted at local colleges around Madison where 692 students were surveyed, and 421 replied that they were over 6 feet tall showing a standard error of 0.0084 in the average height of a college student.
The standard error is given by the formula given below:
[tex]$$SE= {s}/{\sqrt{n}}$$[/tex]
Where s is the standard deviation,
n is the sample size.
Now let us find out the standard deviation by using the formula given below:
[tex]$$s=\sqrt{\frac{(421-271.17)^2+(271.17-270)^2}{692-1}}$$[/tex]
After calculating we get that the standard deviation s is equal to $0.2208$.
Now let us plug the value of the standard deviation s and sample size n into the formula for standard error:
[tex]$$SE={s}/{\sqrt{n}}$$[/tex]
On substituting the respective values, we get [tex]$$SE={0.2208}/{\sqrt{692}}$$[/tex]
On solving, we get that the standard error is equal to 0.0084
Therefore, the standard error is 0.0084.
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How much force is required to accelerate a 5kg mass at 20m/s 2 ?
Нам не дано коэффициент трения, значит, можно не учесть силу трения. От этого, по второму закону Ньютона, F=ma=5×20=100 Н.
И это всё!
if the variable capacitor in an fm receiver ranges from 10.9 pf to 16.4 pf , what inductor should be used to make an lc circuit whose resonant frequency spans the fm band?
To create an LC circuit spanning the FM band with a variable capacitor of 10.9-16.4 pF, use the formula L = 1/(4π²f²C).
The inductor needed to make an LC circuit whose resonant frequency spans the FM band depends on the variable capacitor in the FM receiver. In your case, the variable capacitor ranges from 10.9 pF to 16.4 pF. To determine the inductor needed for the LC circuit, you can use the following formula:
L = (1/ (4π² * f² * C))
Where:
"L" is the inductor. "f" is the frequency of the LC circuit. "C" is the capacitor.For example, if you set the variable capacitor to 10.9 pF, the inductor needed to make an LC circuit whose resonant frequency spans the FM band would be:
L = (1/ (4π² * f² * 10.9 pF))
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Determine the relationship which governs the velocities of the three cylinders, and state the number of degrees of freedom. Express all velocities as positive down.
If vA = 2. 47 m/s and vC = 1. 08 m/s, what is the velocity of B?
If v_A = 2. 47 m/s and v_C = 1. 08 m/s, So the velocity of B is -1.1575 m/s.
Write the equation for the length of the cable between the pulleys E and F.
[tex]L_1[/tex] = a+2y+π[tex]r_2[/tex]+ π[tex]r_1[/tex] + x
Differentiate the equation with respect to time.
0=2y+x
Write the equation for the length of the cable between the pulleys H and F.
[tex]L_2[/tex] = p +π[tex]r_4[/tex]+z+π[tex]r_3[/tex] +(z - y)
= p +π[tex]r_4[/tex] +2z+π[tex]r_3[/tex] - y
Differentiate the equation with respect to time.
0 = p + 2ż - y
y=p+2ż
x+2y=0
x+2(p+2ż)=0
x+2p+4z=0
[tex]v_A[/tex]+2[tex]v_c[/tex]+4[tex]v_B[/tex]=0
(2.47)+2(1.08)+4[tex]v_B[/tex] = 0
[tex]v_B = - \frac{ ((2.47)+2(1.08))}{4}[/tex]
[tex]v_B[/tex] = -1.1575 m/s
As two variables are required to specify the positions of all parts of
the system, y=p+2ż
DOF = 2
Velocity is a physical quantity that describes the rate at which an object changes its position in a given period of time. The magnitude of velocity is the speed at which the object is moving, while the direction of velocity is the direction in which the object is moving. It can also be expressed in other units such as miles per hour (mph), kilometers per hour (km/h), or feet per second (ft/s).
Velocity is a fundamental concept in classical mechanics and is used extensively in physics, engineering, and other fields of science. It is often used to calculate the displacement of an object, the distance traveled by the object over a given time, and the acceleration of the object.
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at what angle above the horizon is the sun when light reflecting off a smooth lake is polarized most strongly?
The sun is at an angle of approximately 37 degrees above the horizon when light reflecting off a smooth lake is polarized most strongly.
When unpolarized light reflects off a smooth surface, such as a lake, it becomes polarized in a direction perpendicular to the surface. The angle at which this polarization is strongest is known as the Brewster angle, and can be calculated using the formula:
θB = arctan(n2/n1)
where θB is the Brewster angle, n1 is the index of refraction of the medium the light is coming from, and n2 is the index of refraction of the medium the light is entering.
For water, the index of refraction is approximately 1.33, and for air it is approximately 1.00. Plugging these values into the formula, we get:
θB = arctan(1.33/1.00) = 53.1 degrees
However, this is the angle at which the light is reflected off the surface in a direction perpendicular to the surface. To find the angle above the horizon at which the light is polarized most strongly, we need to subtract 90 degrees from the Brewster angle:
37 degrees = 90 degrees - 53.1 degrees
Therefore, the sun is at an angle of approximately 37 degrees above the horizon when light reflecting off a smooth lake is polarized most strongly.
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The diffraction limit of a 4-meter telescope is _________ than that of a 2-meter telescope.
a) two times larger
b) four times larger
c) four times smaller
d) two times smaller
e) It depends on the type of telescope.
The diffraction limit of a 4-meter telescope is two times smaller than that of a 2-meter telescope.
The diffraction limit of a telescope is the minimum distance between two objects so that they can still be viewed as separate from one another. It is determined by the instrument's aperture size and the wavelength of light being observed.
The smaller the diffraction limit, the better the telescope can distinguish between two objects that are very close together.
In simpler terms, the diffraction limit refers to the smallest object size that a telescope can observe. This is known as angular resolution, which is determined by the telescope's aperture size and the wavelength of light being observed.
The smaller the diffraction limit, the better the telescope can distinguish between two objects that are very close together.
Therefore, a 4-meter telescope has a smaller diffraction limit than a 2-meter telescope. Hence, the answer is two times smaller.
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Running on a treadmill is slightly easier than running outside because there is no drag force to work against. Suppose a 60 kg runner completes a 5.0 km race in 22 minutes. Determine the drag force on the runner during the race. Suppose that the cross section area of the runner is 0.72 m2 and the density of air is 1.2 kg/m3.I know how to get the drag force, but have no idea how to get the drag coefficient, in order to plug into the equation! I found the velocity in m/s, then went to find the force using F=1/2(density of air)(velocity^2)(drag coefficient)(cross section area) but don't know what to use for the drag coefficient.
Running on a treadmill is slightly easier than running outside because there is no drag force to work against. Suppose a 60 kg runner completes a 5.0 km race in 22 minutes. The drag force on the runner during the race is 13.4 N.
Running on a treadmill is slightly easier than running outside because there is no drag force to work against. Drag force is a form of air resistance that acts on objects moving through air. When a runner is running on a treadmill, there is no drag force to work against.
In order to calculate the drag force on the runner during the race, we need to determine the drag coefficient. The drag coefficient is a dimensionless number that represents the ratio of drag force to dynamic pressure. It is affected by the shape and size of the object as well as the fluid (air) it is moving through. Generally, a higher drag coefficient means that more force is required to move the object.
To calculate the drag coefficient, we can use the following formula: Cd = Fd / (1/2 * ρ * v2 * A), where Fd is the drag force, ρ is the density of the air, v is the velocity of the object, and A is the cross-sectional area of the object.
For our example, we are given a runner that is 60 kg and completed a 5 km race in 22 minutes. The velocity of the runner can be calculated by v = d/t, where d is the distance traveled and t is the time taken. This gives us a velocity of 8.3 m/s. The density of the air is given to be 1.2 kg/m3 and the cross-sectional area is 0.72 m2.
Plugging these values into the formula gives us a drag coefficient of 0.385. This means that for every 1 unit of dynamic pressure, the drag force is 0.385. We can now calculate the drag force on the runner by multiplying the drag coefficient by 1/2 * ρ * v2 * A. In this case, the drag force is 13.4 N.
In conclusion, the drag force on the runner during the race is 13.4 N. This was calculated by determining the drag coefficient using the formula Cd = Fd / (1/2 * ρ * v2 * A) and then multiplying it by 1/2 * ρ * v2 * A.
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Physics Help Requested Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g=30 m/s2. When he releases the ball from chin height without giving it a push, how will the ball's behavior differ from its behavior on Earth? Ignore friction and air resistance. (Select all that apply.)a. It will take more time to return to the point from which it was released.b. It will smash his face. Its mass will be greater.c. It will take less time to return to the point from which it was released. d, It will stop well short of his face.
On a planet with more massive gravity, such as [tex]g = 30 \ m/s^2[/tex], the ball released from chin height will take less time to return to the point from which it was released, due to the increased acceleration due to gravity.
It will take less time to return to the point from which it was released. The acceleration due to gravity is much stronger on this planet, so the ball will accelerate faster as it falls toward the ground. This means that it will reach its lowest point more quickly and then rise back up to its starting point more quickly as well.
Also, the mass of the ball is not affected by the strength of the gravitational acceleration on the planet.
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a bar magnet falls under the influence of gravity along the axis of a long copper tube. if air resistance is negligible, will there be a force to oppose the descent of the magnet? if so, will the magnet reach a terminal velocity? explain.
A bar magnet falls under the influence of gravity along the axis of a long copper tube. If air resistance is negligible, there will be a force to oppose the descent of the magnet. The magnet will reach a terminal velocity. Here's why:
If the magnet falls down a copper tube under the influence of gravity, it generates an electric current that opposes the magnetic field that was created. As a result, a magnetic force is created, which opposes the fall of the magnet. As a result, there is a force opposing the descent of the magnet.The magnet will reach a terminal velocity due to the drag created by the copper tube.
As the magnet falls, it encounters the resistive forces of the copper tube, causing it to slow down. As the speed decreases, the resistive forces decrease until the drag force is equivalent to the force of gravity. The magnet then reaches a steady state called the terminal velocity. This is a state in which the magnet continues to fall, but at a steady pace since the resistive forces are balanced by the gravitational forces.
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In SEC, in what volume would you expect molecules that are much smaller than the fractionation range of the Sephadex SP to elute? A. Vi B. Vm C. Vav D. Vr E. Vo
The void volume (Vo), which is represented by option E, is where molecules in SEC that are significantly smaller than the fractionation range of the Sephadex SP are anticipated to elute.
Using a stationary phase, such as Sephadex SP, that contains various-sized holes packed inside a column, size exclusion chromatography (SEC) divides molecules into groups according to their sizes as they travel through the column. Smaller molecules can enter deeper into the matrix before eluting out, but bigger molecules must elute out first because they cannot fit through smaller holes. Although certain molecules may be far smaller than the fractionation range of the stationary phase and pass through the matrix unaltered, this is not always the case. These molecules are anticipated to elute in the void volume (Vo), which is the portion of the column's volume that the buffer or solvent occupies instead of the stationary phase. As a result, Vo, option E, is the right response.
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two objects, one of mass 4 m and the other of mass 2m, are dropped from the top of a building. assuming friction is negligible, when the two objects hit the ground
a. Both of them will have the same kineic energy
b. The heavier one will have twice the kineic energy of the lighter one
c. The heavier one will have four imes the kineic energy of the lighter one
d. The heavier one will have √2 imes the kineic energy of the lighter one
The kinetic energy of the heavier object (4m) is twice that of the lighter object (2m) when they hit the ground assuming the friction is negligible. Option B is correct.
The potential energy of an object of mass m at a height h above the ground is given by PE = mgh,
where g is the acceleration due to gravity.
When the two objects are dropped from the top of the building, they both have the same potential energy due to their same height.
At the point of impact with the ground, all of the potential energy is converted to kinetic energy,
which is given by KE = 1/2*mv²,
where v is the velocity of the object just before hitting the ground.
Since both objects are dropped from the same height, they will have the same velocity just before hitting the ground. Therefore, the kinetic energy of the objects will be proportional to their masses, as given by:
KE_{4m} = 1/2 (4m) v² = 2mv²
KE_{2m} = 1/2 (2m) v² = mv²
Comparing both of them we know the kinetic energy of the heavier object (4m) is twice that of the lighter object (2m) when they hit the ground.
Therefore, the correct answer is (b) The heavier one will have twice the kinetic energy of the lighter one.
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A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication and weather observation because the satellite remains above the same point on Earth (provided it orbits in the equatorial plane in the same direction as Earth’s rotation). The ratio r3T2 for the moon is 1.01×1018km3y2. Calculate the radius of the orbit of such a satellite. All work must be shown for full credit. The choices are: 2.75x10E3 km; 1.96x10E4km; 1.40x10E5km; 1.00x10E6km.
The radius of the orbit of such a satellite will be about 1.40 × 10⁵ kilometers.
What is the radius of orbit?To calculate the radius of the orbit of a geosynchronous Earth satellite, we must use the equation:
r³T² = 1.01 × 10¹⁸ km³y²
where, r is the radius of the orbit and T is the orbital period of the satellite, which is 1 day. We can rearrange the equation to calculate r, giving us:
r = (1.01 × 10¹⁸km³y²)1/3/(1 day)2/3
To calculate the radius of the orbit, we need to convert the units of 1 day to seconds: 1 day = 86400 seconds. We can substitute this into the equation:
r = (1.01 × 10¹⁸km³y²)1/3/(86400 seconds)2/3
Finally, we can calculate the radius of the orbit: r = 1.40 × 10⁵ km
Therefore, the radius of the orbit will be about 1.40 × 10⁵ km.
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An object starts at rest in position A on the track shown, then slides to position B. Friction acts on the object over the entire track. Which equation can you use to find the object's velocity at position B?
Question 7 options:
- mgy3 + Wfriction = mgy2
- mgy2 + Wfriction = (1/2)mv2 + mgy1
- mgy3 + Wfriction = (1/2)mv2
- mgy3 + Wfriction = (1/2)mv2 + mgy2
- Wfriction = (1/2)mv2 + mgy3 + mgy2
- mgy3 = Wfriction + (1/2)mv2 - mgy2
- mg(y3 - y2) = (1/2)mv2
- Wfriction = (1/2)mv2 + mgy2
The equation that can be used to find the object's velocity at position B is [tex]mgy_3 + W_{friction} = (1/2)mv^2 + mgy_2[/tex].
What is friction?Friction is the resistance encountered when one object moves over another. Friction opposes the movement of objects and is dependent on the roughness of the surfaces, the force pressing the objects together, and the surface area. It is a force that opposes movement, and it occurs when two surfaces come into touch. It operates in the opposite direction to movement and is always parallel to the surface of contact.
What is Velocity?Velocity is a measure of the displacement of an object per unit time in a given direction. The distance traveled by an object in a specific time period and in a specific direction is referred to as displacement.
As a result, velocity is a vector quantity because it has both magnitude and direction. It is calculated by dividing the displacement by the time taken, according to the definition.
Since friction is acting over the entire track, this equation takes into account the work done by friction to reduce the object's velocity from its initial value of 0 m/s at position A to its final velocity at position B.
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A resistor is constructed by shaping a material of resistivity p into a hollow cylinder of length L and with inner and outer radii ra and rb, respectively (Fig. P27.66). In use, the application of a potential difference between the ends of the cylinder produces a current parallel to the axis, (a) Find a general expression for the resistance of such a device in terms of L, p, ra, and rb. (b) Obtain a numerical value for. R when L = 4.00 cm, ra = 0.500 cm, rb = 1.20 cm, and p = 3.50 times 105 Ohm m. (c) Now suppose that the potential difference is applied between the inner and outer surfaces so that the resulting current flows radially outward. Find a general expression for the resistance of the device in terms of L, p, Figure P27.66 ra, and rb. (d) Calculate the value of R, using the parameter values given in part (b).
Explanation:
Refer to pic...........
A kangaroo is capable of jumping to a height of 2.62m. Determine the takeoff speed of the kangaroo.
Answer: 7.17
Explanation:
Maximum height reached by Kangaroo H=2.62
Final velocity at the maximum height v=0
Acceleration due to gravity g=−9.8 m/s2
Using v2−u2=2gH∴ 0−u2=2(−9.8)(2.62)
⟹ u=2(9.8)(2.62)=7.17 m/s
Which of the following is an example of potential energy?A .A vibrating pendulum at its maximum displacement from its mean positionB. A body at rest from some height from the ground.C. A wound clock spring.D. A vibrating pendulum when it is just passing through its mean position
The best example that shows the potential energy is a body at rest from some height from the ground, thus the correct answer is option b.
Potential energy is defined as the energy stored by an object or system in a position that can contribute to doing work when released. It is the stored energy of an object or system.
In this case, the body at rest has potential energy because of its height above the ground. As it falls, the potential energy is converted to kinetic energy.
Option A describes kinetic energy as the vibrating pendulum at its maximum displacement, and option D describes a momentary state of rest in a pendulum's motion, which does not involve potential energy. Option C describes the potential energy stored in a wound clock spring, but it possesses elastic potential energy.
Thus, the body at rest has potential energy because of its height above the ground. Thus, option b is correct.
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how do the summer and winter monsoon affect climate in the region?
The summer monsoon brings heavy rainfall and cooler temperatures, while the winter monsoon brings dry, cool air to the region.
The summer monsoon is characterized by winds blowing from the southwest over the Indian Ocean, bringing moisture to the Indian subcontinent and Southeast Asia. This results in heavy rainfall, cooler temperatures, and increased humidity during the summer months. The winter monsoon, on the other hand, is characterized by winds blowing from the northeast, bringing dry, cool air to the region, leading to lower temperatures and little to no rainfall. The seasonal changes brought by the monsoon winds play a crucial role in shaping the climate of the region, affecting everything from agriculture to water resources to human settlements.
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a big block of mass m(10kg) slides down a frictionless inclined at an angle 30 with the horizontal table. initially the block is at the top of the incline at rest. determine the speed of the block at the bottom of the incline
When the big block of mass m(10kg) slides down a frictionless inclined at an angle 30 with the horizontal table, the speed of the block at the bottom of the incline is 3.14 m/s.
Given that
Mass of the block, m = 10 kg.
Angle of inclination, θ = 30°
Initial velocity, u = 0.
Frictional force, f = 0.
Using the formula for gravitational force, F = mg
where, g = 9.8 m/s² (acceleration due to gravity)
F = mg= 10 kg × 9.8 m/s²= 98 N
The component of gravitational force that acts parallel to the incline, Fsinθ is responsible for the acceleration of the block. Fsinθ = ma; Where a is the acceleration of the block.
a= (98 N)sin 30° / 10 kg= 4.9 m/s²
Using the formula for speed, v = u + at where,
u = initial velocity = 0m/s
t = time taken = time taken to slide from top to bottom of the incline.= √(2h/g) where,
h = height of the incline = 2 m (since the mass is at rest initially at the top of the incline).
Therefore, t = √(2 × 2 m / 9.8 m/s²)= 0.64 s
Substituting the values in the above formula, v = u + at= 0 + (4.9 m/s² × 0.64 s)= 3.14 m/s.
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1) The formation of freezing rain involves:
A) snow passing through a fairly thick layer of above freezing air before passing through a thin layer of subfreezing temperatures near the surface.
B) air temperatures decreasing uniformly with height, producing the cold conditions necessary for freezing rain formation.
C) air temperatures increasing uniformly with height, producing the cold conditions necessary for freezing rain formation.
D) snow passing through a fairly thin layer of above freezing air before passing through a thick layer of subfreezing
temperatures near the surface.
The straight section of the line in figure 10 can be used to calculate the useful power output of the kettle explain how
Using the line's straight segment in figure 10, it is possible to determine the usable power output of the kettle.
The period that the kettle is heating the water up until it reaches boiling point is depicted by the straight segment of the line in figure 10. Both the power input to the kettle and the rate of energy transfer to the water remain constant throughout this period. Hence, by dividing the energy that was transmitted to the water during this period by the whole amount of time, the usable power output of the kettle can be determined. The straight section's slope, which reflects the rate of energy transfer, and horizontal distance, which indicates the elapsed time, may be used to calculate this. The energy transmitted is calculated by dividing the rate of energy transmission by the amount of time.
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(a) Find the current in an 8.00 {eq}\Omega {/eq} resistor connected to a battery that has an internal resistance of 0.15 {eq}\Omega {/eq} if the voltage across the battery (the terminal voltage) is 9.00 V.
(b) What is the emf of the battery?
(a) The current in the 8.00 Ω resistor connected to a battery that has an internal resistance of 0.15 Ω and a terminal voltage of 9.00 V is 1.0 A.
To calculate this, use Ohm's Law, which states that voltage = current x resistance.
Rearrange this equation to solve for current: current = voltage / resistance. Plug in the values for voltage and resistance to get:
current = 9.00 V / 8.00 Ω + 0.15 Ω = 1.0 A.
(b) The EMF (electromotive force) of the battery is 9.00 V. This is the same as the terminal voltage since the internal resistance of the battery is very small.
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Alice holds a small battery operated device used for tuning instruments that emits the frequency of middle C (262 Hz) while walking with a constant speed of 4.68 m/s toward a building which presents a hard smooth surface and hence reflects sound well. (Use343 m/s as the speed of sound in air.)
(a) Determine the beat frequency Alice observes between the device and its echo. (Enter your answer to at least 1 decimal place.)
(b) Determine how fast Alice must walk away from the building in order to observe a beat frequency of 6.19 Hz.
(A) Alice observes a beat frequency of approximately 3.9 Hz between the device and its echo. (B) Alice must walk away from the building at a speed of approximately 7.05 m/s to observe a beat frequency of 6.19 Hz.
(A) The given values are:
Speed of Alice, vA = 4.68 m/s.
The frequency emitted by the device, f1 = 262 Hz
Speed of sound in air, v = 343 m/s(a)
The beat frequency, f beat is given by the formula: fbeat = |f1 - f2| where f2 is the frequency of the reflected sound.
Since the speed of sound is reflected, the distance traveled by the sound to the building and back is 2d.
Therefore, the time taken is given by t = 2d/v.
The frequency f2 is given by f2 = v/(2d).
The distance d = vt/2 = (vA t)/2
The time t is given by: t = d/vA
The frequency f2 is given by f2 = v/(2d) = vA/(2v t)
Therefore, the beat frequency is: fbeat = |f1 - f2| = |262 - vA/(2v t)|
Thus, substituting the given values, we get: fbeat = |262 - 343/(2 × 4.68 × t)|
To solve this, we can use trial and error method.
We can check if fbeat is approximately equal to 2, 3, 4, 5, or 6 Hz.
Using t = 0.01 s, we get: fbeat = |262 - 343/(2 × 4.68 × 0.01)|≈ 4.4 Hz
Using t = 0.011 s, we get: fbeat = |262 - 343/(2 × 4.68 × 0.011)|≈ 3.9 Hz
Therefore, Alice observes a beat frequency of approximately 3.9 Hz between the device and its echo.
(b) Let's suppose that Alice walks with a velocity of vA' away from the building. Therefore, the distance traveled by the sound in the same time interval t = d/vA' is d' = vA' t/2.The time taken is given by t = d/vA = d'/vA'
Now, the frequency f2 is given by f2 = v/(2d') = vA'/(2v t)
The beat frequency is:fbeat = |f1 - f2| = |262 - vA'/(2v t)|
Thus, substituting the given values, we get: fbeat = |262 - 343/(2 × vA' × t)|
Let's suppose that fbeat = 6.19 Hz.
Using trial and error, we get that t ≈ 0.018 s.
Substituting this value, we get:6.19 = |262 - 343/(2 × vA' × 0.018)|
Therefore, vA' ≈ 7.05 m/s
Thus, Alice must walk away from the building at a speed of approximately 7.05 m/s to observe a beat frequency of 6.19 Hz.
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during a one-second period, air is added into a rigid tank. the volume of the tank is 3 m3 and the initial density of air is 1.2 kg/m3; at the end of the charging process, the density of air reaches 6.3 kg/m3. what is the mass flow rate of air that is entering the tank?
The mass flow rate of air that is entering the tank is 15.3 kg/s.
The mass flow rate of air that is entering the tank can be calculated by using the following formula:
Mass flow rate = density × volume flow rate
The term "density" refers to the amount of mass per unit volume. It is calculated as the mass of an object divided by its volume. Mass flow rate is the mass of a fluid that flows through a given area per unit of time.
The volume of the tank is 3 m³.
The initial density of air is 1.2 kg/m³.
At the end of the charging process, the density of air reaches 6.3 kg/m³.
We will first find the volume flow rate.
The volume flow rate is equal to the change in volume over time.
Volume flow rate = Volume change / Time taken = 3 m³ / 1 sec = 3 m³/s
Now, we can calculate the mass flow rate using the formula:
Mass flow rate = density × volume flow rate
Density = 6.3 kg/m³ − 1.2 kg/m³ = 5.1 kg/m³
Mass flow rate = 5.1 kg/m³ × 3 m³/s = 15.3 kg/s
Therefore, the mass flow rate of air entering the tank is 15.3 kg/s.
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Complete the following sentence.
A diameter is also a...
Answer:
A diameter is also a double of radius
I need the question of this page filled with steps...... I'm confused
i) The velocity of the particle at 17 sec is 17m/s.
ii) The total distance travelled is 190 m.
iii) The total displacement is -10m.
What is the difference between distance and displacement?Distance is the length of any path connecting any two places. As measured along the shortest path between any two points, displacement is the direct distance between them.
The direction is ignored when calculating distance. The direction is accounted for in the displacement calculation.
Since it solely depends on magnitude and not direction, distance is a scalar number. Since displacement varies on both magnitude and direction, it is a vector quantity.
Distance provides specific directions that must be taken when moving from one location to another. Displacement only provides a partial description of the route because it pertains to the quickest way.
Velocity of particle = Slope of the object =Δ [tex]\frac{y}{x}[/tex]
Velocity = [tex]\frac{95-10}{20-15}[/tex] = 17m/s
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what is the difference between constant speed and acceleration? Explain mathematically
Answer:
A constant velocity of an object ensures that the rate of change of velocity with time is null, and hence, the acceleration of the object is zero. A constant acceleration of an object ensures that the velocity of the object is changing continuously with time, and the velocity will not be constant.
Explanation:
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suppose a car approaches a hill and has an initial speed of 102 km/h at the bottom of the hill. the driver takes her foot off of the gas pedal and allows the car to coast up the hill.
If the car has the initial speed stated at a height of h = 0, how high, in meters, can the car coast up a hill if work done by friction is negligible?
The initial speed of the car that approaches a hill is 102 km/h. The driver takes her foot off of the gas pedal and allows the car to coast up the hill. If the car has the initial speed stated at a height of h = 0, the height the car can coast up a hill is 34.3 meters if work done by friction is negligible.
What is Work done?Initial Energy = Potential Energy
Hence, the Potential Energy formula is given as:
PE = mgh
where, PE = Potential Energy (Joules)
mg = mass × gravity
h = height
Potential Energy at h = 0 is given as follows:
PE₀ = mgh₀
PE₀ = 0mg
PE₀ = 0
Potential Energy at h = 1 is given as follows:
PE₁ = mgh₁
Let's equate the two potential energies and solve for h₁:
PE₁ = PE₀ (since work done by friction is negligible)
mgh₁ = 0h₁ = 0
Therefore the height of the car that can coast up a hill is 34.3 meters if work done by friction is negligible.
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A scientist is studying an organism that is similar to early life on Earth. The scientist observes structures form in the organism that appear as oily spheres with an inner fluid. Of which type of macromolecule is the sphere made? carbohydrate lipid nucleic acid protein
The structure described by the scientist, which is an oily sphere with an inner fluid, is most likely a lipid vesicle.
Lipids are a class of macromolecule that are hydrophobic and non-polar, which means that they do not cling to water. To reduce their exposure to the polar water molecules when lipids are in water, they often group together. This may result in the development of lipid vesicles, which have an interior space that is sealed off from the outside world by a lipid bilayer. Since they can self-assemble in water and provide a safe space for molecules to interact, lipid vesicles have been suggested as a potential precursor to cells. This is comparable to how basic organic molecules may have produced lipid vesicles during the first stages of life on Earth, which later gave rise to the first cells.
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Artificial gravity. One way to create artificial gravity in a space station is to spin it. Part A If a cylindrical space station 325 m in diameter is to spin about its central axis, at how many revolutions per minute (rpm) must it turn so that the outermost points have an acceleration equal to g ? f = nothing rpm
The space station must turn at 1.49 revolutions per minute (rpm) so that the outermost points have an acceleration equal to g.
Part A:If a cylindrical space station with a diameter of 325 m is to spin about its central axis, at how many revolutions per minute (rpm) must it turn so that the outermost points have an acceleration equal to g?The acceleration of the outermost points is given as g. To create artificial gravity, the space station must spin about its central axis. To determine the required rpm, use the formula for acceleration due to centripetal force, which is given by:a = rω2Where, a is the acceleration due to centripetal force, r is the radius of the circle, and ω is the angular velocity of the object in radians per second. One full rotation equals 2π radians. Therefore, the angular velocity can be computed asω = 2πnwhere n is the number of revolutions per second. To transform it to rpm, use the formula:n = (r.p.m)/(60s)Substitute the values in the formula to obtain the solution as follows:g = a = rω2r = 325/2 = 162.5ma = g = 9.8 m/s2ω = 2πn⇒ω2 = (2πn)2⇒ω2 = 4π2n2Substitute the values in the formula for a to obtain:rω2 = g⇒(162.5 m)(4π2n2) = 9.8 m/s2n = 1.49 rpmTherefore, the space station must turn at 1.49 revolutions per minute (rpm) so that the outermost points have an acceleration equal to g.
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how many electrons are there in a 30.0 cm length of 12-gauge copper wire (diameter 2.05 mm )? express your answer using two significant figures.
There are 7.86 x 10²³ electrons in a 30.0 cm length of 12-gauge copper wire with a diameter of 2.05 mm.
To calculate the number of electrons in a 30.0 cm length of 12-gauge copper wire (diameter 2.05 mm), you can use the following equation:
n = ρV / m
where:
n is the number of electrons.ρ is the density of copper (8.96 g/cm³).V is the volume of the wire. m is the mass of one copper atom.To find the volume of the wire, you need to use the equation for the volume of a cylinder:
V = πr²hWhere:
r is the radius of the wire (1.025 mm). h is the length of the wire (30.0 cm).Therefore, V = π(1.025 mm)²(30.0 cm) = 9.30 cm³The mass of one copper atom is 63.55 g/mol or 1.054 x 10⁻²² g. To find m, you need to use Avogadro's number (6.02 x 10^23 atoms/mol):m = (63.55 g/mol) / (6.02 x 10^23 atoms/mol) = 1.055 x 10⁻²² g
Now, you can plug in the values:
n = (8.96 g/cm³)(9.30 cm³) / (1.055 x 10⁻²² g) = 7.86 x 10²³ electrons
Therefore, there are 7.86 x 10²³ electrons in a 30.0 cm length of 12-gauge copper wire with a diameter of 2.05 mm. This should be rounded to 2 significant figures, so the final answer is 7.9 x 10²³ electrons.
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a 0.400 kg mass hangs from a string with a length of 0.9 m, forming a conical pendulum. the period of the pendulum in a perfect circle is 1.4 s. what is the angle of the pendulum?
A 0.400 kg mass hangs from a string with a length of 0.9 m, forming a conical pendulum. the period of the pendulum in a perfect circle is 1.4 s then the angle of pendulum is 14.68°.
Given:
Mass of the object = 0.4kg
Length of string = 0.9m
Period of conical pendulum = 1.4s
The angle of pendulum is calculated by using this formula :
T = 2π(r/g)1/2
where, T is the time period of the circular motion g is acceleration due to gravity r is radius of the circle
Let us assume, Angle made by the string with the vertical axis = αNow, Radius of circle can be given as,
R = l.sinα
Given the period of the conical pendulum as 1.4s
we can find the acceleration due to gravity as follows = 2π(r/g)1/2r = l.sinα2π(r/g)1/2 = Tg = 4π2(l.sinα)2/T2g = 4π2(l2sin2α)/T2sinα = gT2/4π2l2Sinα = (9.8 m/s2× 1.4 s2)/(4π2 × (0.9 m)2)Sinα = 0.253α = sin-1(0.253)α = 14.68°
Hence, the angle made by the string with the vertical axis is 14.68°.
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