PHYSICS COLLEGE

Part 1) A cop car traveling at 25 m/s has a siren producing a frequency of 700 Hz. A felon jumps on his motorcycle and speed off in the opposite direction of 15 m/s. What frequency does the felon hear as he sped away (speed of sound is 343 m/s)?Part 2) The cop does a U-turn and speeds towards the felon at 30 m/s, while the felon speeds up to 20 m/s. What frequency does the felon hear as he sped away (speed of sound is 343 m/s)?
Part 3) What if the felon then sped up to 30 m/s and all other conditions remained the same?

Answers

Answer 1

Answer:

1) 621.8 Hz

2) 719.3 Hz

3) 700 Hz

Explanation:

The Doppler effect occurs when there is a source of a wave in relative motion with respect to an observer.

When this happens, the frequency of the wave appears shifted to the observer, according to the equation:

$f'=(v\pm v_o)/(v \pm v_s)f$

where

f is the real frequency of the sound

f' is the apparent frequency of the sound

v is the speed of the sound wave

$v_o$ is the velocity of the observer, which is negative if the observer is moving away from the source, positive if the observer is moving towards the source

$v_s$ is the velocity of the source, which is negative if the source is moving towards the observer, positive if the source is moving away

In this problem we have:

f = 700 Hz is the frequency of the siren

v = 343 m/s is the speed of sound

$v_s=-25 m/s$ is the velocity of the car with the siren

$v_o = +15 m/s$ is the velocity of the felon (he's moving away from the siren)

So, the frequency heard by the felon is

$f=(343-25)/(343+15)(700)=621.8 Hz$

In this case, the cop does a U-turn and speeds towards the felon at 30 m/s.

This means that now the siren is moving towards the observer (so, $v_s$ becomes positive), while the sign of $v_o$ still remains positive.

So we have:

f = 700 Hz is the frequency of the siren

v = 343 m/s is the speed of sound

$v_s=+30 m/s$ is the velocity of the car with the siren

$v_o = +20 m/s$ is the velocity of the felon

So, the frequency heard by the felon is

$f=(343+30)/(343+20)(700)=719.3 Hz$

In this case, the felon speeds up to 30 m/s.

This means that now the felon and the siren are moving with the same relative velocity: so, it's like they are not moving relative to each other, so the frequency will not change.

In fact we have:

f = 700 Hz is the frequency of the siren

v = 343 m/s is the speed of sound

$v_s=+30 m/s$ is the velocity of the car with the siren

$v_o = +30 m/s$ is the velocity of the felon

So, the frequency heard by the felon is

$f=(343+30)/(343+30)(700)=700 Hz$

So, the frequency will not change.

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As you take the stoppered part of the tube up the staircase you begin to see the water level drop around the 4th floor. As you continue up it does not continue up with you but stays at a constant level. What does that mean?a. The pressure in the tubing is equal to the barometric pressure.
b. The tubing was unable to supply any more water to the tube for use.
c. The pressure outside the tube is higher that the water pressure inside the tube.

Answers

Answer:

a. The pressure in the tubing is equal to the barometric pressure.

Explanation:

Since in the question it is mentioned that the if you take the stoppert part of the tube than the level of warer would be fall approx 4th floor and if it is continued than it wont be continue but remains constant.

Now here first we do that the tube i.e. connected to the bucket should be taken up. In the first instance, the bucket supplies the water to the tube but it would not increased far away to the level of the barometric pressure

Hence, the correct option is a.

A block slides from rest with negligible friction down the track above, descending a vertical height of 5.0 m to point P at the bottom. It then slides on the horizontal surface. The coefficient of friction between the block and the horizontal surface is 0.20. How far does the block slide on the horizontal surface before it comes to rest?

Answers

The block slide on the horizontal surface is "24.99 m" far.

According to the question,

Vertical height = 5.0 m
Coefficient of friction = 0.20

Let,

The time taken be "t".

Now,

→ $s = ut+ (1)/(2) at^2$

By substituting the values, we get

$5 = (1)/(2)* 9.8* t^2$

$t = 1.01 \ sec$

The final velocity will be:

→ $v_1 = gt$

$= 9.8* 1.01$

$= 9.899 \ m/s$

Now,

→ $t = (u)/(a)$

$= (9.899)/(0.2* 9.8)$

$= 5.05 \ seconds$

hence,

The distance will be:

→ $s = ut+0.5* at^2$

$= 9.899(5.05)-0.5* (0.2* 9.8* 5.05^2)$

$= 24.99 \ m$

Thus the above approach is right.

Learn more about friction here:

brainly.com/question/18851133

Answer:

The block slides on the horizontal surface 25 m before coming to rest.

Explanation:

Hi there!

For this problem, we have to use the energy-conservation theorem. Initially, the block has only gravitational potential energy (PE) that can be calculated as follows:

PE = m · g · h

Where:

m = mass of the block.

g = acceleration due to gravity.

h = height at which the block is located.

As the block starts to slide down the track, its height diminishes as well as its potential energy. Due to the conservation of energy, energy can´t disappear, so the loss of potential energy is compensated by an increase of kinetic energy (KE). In other words, as the block slides, the potential energy is converted into kinetic energy. The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

m = mass of the block.

v = speed of the block.

Then, at the bottom of the ramp, the kinetic energy of the block will be equal to the potential energy that the block had at the top of the ramp.

Initial PE = KE at the bottom

When the block starts sliding horizontally, friction force does work to stop the block. According to the energy-work theorem, the change in the kinetic energy of an object is equal to the net work done on that object. In other words, the amount of work needed to stop the block is equal to its kinetic energy. Then, the work done by friction will be equal to the kinetic energy of the block at the bottom, that is equal to the potential energy of the block at the top of the track:

initial PE = KE at the bottom = work done by friction

The work done by friction is calculated as follows:

W = Fr · Δx

Where:

W = work

Fr = friction force.

Δx = traveled distance.

And the friction force is calculated as follows:

Fr = μ · N

Where:

μ = coefficient of friction.

N = normal force.

Since the block is not accelerated in the vertical direction, in this case, the normal force is equal to the weight (w) of the block:

Sum of vertical forces = ∑Fy = N - w = 0 ⇒N = w

And the weight is calculated as follows:

w = m · g

Where m is the mass of the block and g the acceleration due to gravity.

Then, the work done by friction can be expressed as follows:

W = μ · m · g · Δx

Using the equation:

intial PE = work done by friction

m · g · h = μ · m · g · Δx

Solving for Δx

h/μ = Δx

5.0 m / 0.20 = Δx

Δx = 25 m

The block slides on the horizontal surface 25 m before coming to rest.

Which of the following statements is true about the variation of pressure in function of the depth? O Pressure decreases exponentially with the depth O Pressure increases exponentially with the depth O Pressure decreases linearly with the depth o Pressure increases linearly with the depth O None of the above

Answers

Answer:

The answer is: Pressure increases linearly with the depth

Explanation:

In this case, the definition of pressure is:

$P = (F)/(A)$

where F = mg is the weight of the fluid over the body, and A is the area of the surface to which the force is exerted. If we consider $\rho = m/V$ , then

$P = (mg)/(A) = (\rho Vg)/(A)$ .

Volume can be expressed as V = A*h, where A is the cross section of the column of the fluid over the body and h is the height of the column, in other words, the depth.

$P = (A\rho gh)/(A)= \rho g h$ ,

which means that pressure increases linearly with the depth in a factor of $\rho g$ .

You have two identical pure silver ingots. You place one of them in a glass of water and observe it to sink to the bottom. You place the other in a container full of mercury and observe that it floats. Comparing the buoyant forces in the two cases you conclude that a.) the buoyant force in water is smaller than in mercury

b.) the buoyant force in the water is larger than that in mercury

c.) the buoyant force in the water is zero and that in mercury is non - zero

d.) the buoyant force in the water is equal to that in mercury

e.) no conclusion can be made about the respective values of the buoyant forces

Answers

Answer: a)

Explanation:

The buoyant force, as stated by Archimedes’ principle, is equal to the weight of the liquid that occupies the same volumen as the submerged object, as follows:

Fb = δ.V.g

If this force is larger than the weight of the object (that means that the fluid is denser than the solid), the object floats, which is the case for silver and mercury.

Instead, silver density is larger than water density, which explains why the pure silver ingot sinks.

Finally, as mercury is denser than water, we conclude that for a same object, the buoyant force in mercury is larger than in water (exactly 13.6 times greater).

A large convex lens stands on the floor. The lens is 180 cm tall, so the principal axis is 90 cm above the floor. A student holds a flashlight 120 cm off the ground, shining straight ahead (parallel to the floor) and passing through the lens. The light is bent and intersects the principal axis 60 cm behind the lens. Then the student moves the flashlight 30 cm higher (now 150 cm off the ground), also shining straight ahead through the lens. How far away from the lens will the light intersect the principal axis now?A. 30 cm
B. 60 cm
C. 75 cm
D. 90 cm

Answers

B. 60 cm

All parallel light rays are bent through the focal point of a convex lens, so the rays from the flashlight 150 cm above the floor must go through the same point on the principal axis as the rays from the flashlight 120 cm above the floor. The location of the focal point does not change when the position of the object is moved either vertically or horizontally.

Write down the DE of motion of a particle moving under the influence of gravity and experiencing a resistive force. .