1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
RUDIKE [14]
3 years ago
12

A child is swinging back and forth with a constant period and amplitude. Somewhere in front of the child, a stationary horn is e

mitting a constant tone of frequency fs Five points are labeled in the drawing to indicate positions along the arc as the child swings. At which position(s) will the child hear the lowest frequency for the sound from the whistle?
Physics
1 answer:
Amanda [17]3 years ago
3 0

Answer:

Explanation:

  We shall apply concept of Doppler's effect of apparent frequency to this problem . Here observer is moving sometimes towards and sometimes away from the source . When observer moves towards the source , apparent frequency is more than real frequency and when the observer moves away from the source , apparent frequency is less than real frequency . The apparent frequency depends upon velocity of observer . The formula for apparent frequency when observer is going away is as follows .

f = f₀ ( V - v₀ ) / V , f is apparent , f₀ is real frequency , V is velocity of sound and v is velocity of observer .

f will be lowest when v₀ is highest .

velocity of observer is highest when he is at the equilibrium position or at middle point .

So apparent frequency is lowest when observer is at the middle point and going away from the source  while swinging to and from before the source of sound .

You might be interested in
For each of the following pairs of gas properties, describe the relationship between the properties, describe a simple system th
asambeis [7]

Answer:

For  each pair of properties of a gas, the relationships are (see the explanation for the description of the systems):

  1. (a) Volume and pressure: The relationship between them is inversely proportional.
  2. (b) Pressure and temperature: They have a directly proportional relationship.
  3. (c) Volume and temperature: They relationship is directly proportional.
  4. (d) Number of gas particles and pressure: The relationship is directly proportional between them.

Explanation:

1. Volume and pressure (temperature and amount of particles constant):

They have an inversely proportional relationship, because <em>if volume is reduced, the pressure increases, or if the volume increases, the pressure decreases</em>.

A simple system could be one similar to the one used by Boyle to test this relationship:

  • Seal the short extreme of a translucent J tube. It could be glass or plastic.
  • Put some water on it. As much as needed to have both sides of the tube filled.
  • Using a syringe, and a flexible small tube,inject a determined volume of air in the bottom in a way that the bubble is trapped in the seal side of the J tube.
  • Then if more water is added to the tube, it will increase the pressure (from the pressure definition is possible to in the trapped air, and is possible to measure the compression of the air bubble. The same is possible if using the syringe, and the flexible tube, some water is removed, and the increasing of volume could be observed.

2. Pressure and temperature (volume and amount of particles of the gas remains constant)

They have a directly proportional relationship, because <em>if temperature is reduced, the pressure decreases, or if the temperature increases, the pressure would increase, also</em>.

A simple system to show this is two cans of soda.

  • The can is rigid, so the volume is always constant, and the amount of gas inside the soda is the same.
  • Put one can under the sun, and the other in the cooler.
  • After a while, take it out the can in the cooler, and open both cans.
  • The one that was under the sun will "explode", in other words, it will liberate a lot of foam of gas and soda, meaning that the pressure inside the can was high.
  • The one that was in te cooler, won't liberate any foam, meaning that the pressure was low.

3. Volume and temperature (pressure and amount of particles of the gas remains constant)

They have a directly proportional relationship, because <em>if temperature is reduced, the pressure decreases, or if the temperature increases, the pressure will increase, also</em>.

A simple system to show this is a party balloon.

  • Fill the party balloon with some air, not enough to be close to explode, but enough to have it of a medium size. Tie the filling hole of the balloon.
  • The air inside the balloon would be at the same pressure than the atmosphere around it, so always will be at this pressure, and the close hole ensure that it has always the same amount of air inside.
  • Now is possible to use some heat source, for example as a hair dryer to increase the temperature of the balloon and its contents. The size of the balloon will increase. Then using water is possible to cool it down and watch how its size decreases.

4. Number of gas particles and pressure (volume and temperature of gas remains constant)

They have a directly proportional relationship, because <em>if the amount of gas particles is reduced, the pressure decreases, or if quantity of gas particles increases, the pressure will increase, also</em>.

A simple system to show this would be a bicycle tire:

  • The tire is rigid, so its volume is essentially constant, and the temperature would remains the same if not moving or driving it.
  • Using a tire gauge, it is possible to know the manometric pressure inside the tire, that is the difference between the actual pressure inside the tire and the atmospheric pressure.
  • Then each time that using an air pump some air is injected in the tire, it si possible to check the pressure inside it using the gauge, and observe how is increasing.
  • Also, is possible to open the valve, to allow some air to escape, then use the gauge to observe how the pressure decreases.

7 0
4 years ago
What causes seasons on Earth?
Bezzdna [24]

Explanation:

A. answer is the correct answer

6 0
3 years ago
Read 2 more answers
A football quarterback runs 15.0 m straight down the playing field in 3.00 s. He is then hit and pushed 3.00 m straight backward
algol [13]

Answer:

a) v_{1}=14.29m/s\\v_{2}=9.25m/s\\v_{3}=6.36m/s

b) v=+9.97m/s

Explanation:

From the exercise we know that

x_{1} =15m, t_{1}=3s

x_{2} =-3m, t_{1}=1.74s

x_{3} =29m, t_{3}=5.20s

From dynamics we know that the formula for average velocity is:

v=\frac{x_{2}-x_{1}  }{t_{2}-x_{1}  }

a) For the three intervals:

v_{1}=\frac{x_{2}-x_{1}  }{t_{2}-t_{1}  }=\frac{(-3-15)m}{(1.74-3)s}=14.29m/s

v_{2}=\frac{x_{3}-x_{2}  }{t_{3}-t_{2}  }=\frac{(29-(-3))m}{(5.20-1.74)s}=9.25m/s

v_{3}=\frac{x_{3}-x_{1}  }{t_{3}-t_{1}  }=\frac{(29-15)m}{(5.20-3)s}=6.36m/s

b) The average velocity for the entire motion can be calculate by the following formula:

v=\frac{v_{1}+v_{2}+v_{3}   }{n} =\frac{(14.29+9.25+6.36)m/s}{3}=+9.97m/s

8 0
3 years ago
Why don't we see other stars during day?
Monica [59]

Answer:

Stars aren't visible during the sunlit hours of daytime because the light-scattering properties of our atmosphere spread sunlight across the sky. Seeing the dim light of a distant star in the blanket of photons from our Sun becomes as difficult as spotting a single snowflake in a blizzard.

Explanation:

5 0
3 years ago
Read 2 more answers
5) Consider pushing a 50.0 kg box through a 5.00 m displacement on both a flat surface and up a
Svetach [21]

a) The work done by the gravitational force on the flat surface is zero

b) The work done by the gravitational force on the ramp is -634 J

c) The work done by the applied force on the flat surface is 500 J

d) The work done by the applied force on up along the ramp is 500 J

Explanation:

a)

The work done by a force is given by the equation

W=Fdcos \theta

where

F is the magnitude of the force

d is the dispalcement of the object

\theta is the angle between the direction of the force and of the displacement

In this problem, we want to calculate the work done by the gravitational force as the box is pushed across the flat ground.

We immediately notice that the gravitational force acts downward, while the displacement is horizontal: therefore, the angle between force and displacement is 90^{\circ}; this means that cos 90^{\circ}=0, and therefore, the work done is zero:

W=0

b)

In this case, the box is pushed along the ramp. We have:

F=mg=(50.0)(9.8)=490 N is the magnitude of the force of gravity, where

m = 50.0 kg is the mass of the box

g=9.8 m/s^2 is the acceleration of gravity

d = 5.00 m is the displacement of the box along the ramp

The ramp is inclined to the horizontal by 15.0^{\circ}, therefore the angle between the force of gravity and the displacement of the box (moving up along the ramp) is:

\theta=90^{\circ}+15^{\circ}=105^{\circ}

Therefore, the work done by gravity in this case is:

W=(490)(5.00)(cos 105^{\circ})=-634 J

c)

In this case, we want to calculate the work done by the force you apply as the box is pushed across the flat ground.

Here we have:

F = 100.0 N (force applied)

d = 5.00 m (displacement of the box)

\theta=0^{\circ} (the force is applied parallel to the flat surface, therefore force and displacement have same direction)

Therefore, the work done by the force you apply on the flat ground is:

W=(100.0)(5.00)(cos 0^{\circ})=500 J

d)

In this last case, we want to calculate the work done by the force you apply as the box is pushed up along the ramp.

This time we have:

F = 100.0 N (force applied is the same)

d = 5.00 m (displacement of the box is also the same)

\theta=0^{\circ} (the force is applied parallel to the ramp, therefore force and displacement have again same direction)

Therefore, the work done by the force you apply while pushing the box along the ramp is:

W=(100.0)(5.00)(cos 0^{\circ})=500 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

8 0
4 years ago
Other questions:
  • What does energy give an object the ability to do
    6·1 answer
  • Technician A says that the device that controls shift points on an automatic transmission is the valve body. Technician B says t
    10·1 answer
  • I have a big test coming up and these two questions i don't understand: Choose the best answer: A device that transforms mechani
    7·1 answer
  • If a pressure gauge measure
    13·1 answer
  • PLZ HELP, GIVING BRAINLIEST!!
    6·1 answer
  • Turning up the Heat
    12·1 answer
  • Examples of impact printers​
    6·1 answer
  • A student thinks that any real vibration must be damped. Is the student correct? If so, give convincing reasoning. If not, give
    8·1 answer
  • An astronomy class is so excited by the discovery of planets around other stars that they decide to do a library exhibit on the
    12·1 answer
  • Air is being pumped into a spherical balloon so that its volume increases at a rate of 150 cm3/s. How fast is the radius of the
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!