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4 years ago

Can someone answer these?

Physics

1 answer:

natita [175]4 years ago

4 0

Short Answer

3: C

4: D

Problem Three

Remark

Somewhere we ought to be told that this is the Doppler Effect. I have never done a problem using this formula, so I think I'm doing it correctly, but no guarantees. My guess is that the frequency increases as it comes towards you and decreases as it moves away from you. I think that is correct.

Formula

Givens

f' = observed frequency
f = actual frequency
v = velocity of sound or light waves.
vo = velocity of observer (in both cases 0)
vs = velocity of source.

f' = (v + vo) * f / (v - vs)

Solution

v = 3*10^8 m/s
f' = 1.1 f
f = f
vo = 0 We are standing still while all this is going on.
vs = ???

f'/f = 1.1

1.1 = (3*10^8 + 0 ) / (3*10^8 - vs)

3.3*10^8 - 1.1*vs = 3*10^8

3.3*10^8 - 3*10^8= 1.1 vs

0.3 * 10^8 = 1.1 vs

2.73 * 10^7 = vs

The closest answer is 3.00 * 10^7 which is C

Problem Four

Here what is happening is that you are looking for the frequency resulting from a wave moving towards you at 1/2 the speed of sound. You are not moving.

Givens

v = v
vs = 1/2 v
f ' = ?
f = 1000 hz
vo =0

f' = v/(v - 1/2v) * 1000

f' = v/ (1/2 v) * 1000

f' = 2 * 1000

f' = 2000 which is D

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Complete Question

A commuter train passes a passenger platform at a constant speed of 39.6 m/s. The train horn is sounded at its characteristic frequency of 350 Hz.

(a)

What overall change in frequency is detected by a person on the platform as the train moves from approaching to receding

(b) What wavelength is detected by a person on the platform as the train approaches?

Answer:

$\Delta f = 81.93 \ Hz$

$\lambda_1 = 0.867 \ m$

Explanation:

From the question we are told that

The speed of the train is $v_t = 39.6 m/s$

The frequency of the train horn is $f_t = 350 \ Hz$

Generally the speed of sound has a constant values of $v_s = 343 m/s$

Now according to dopplers equation when the train(source) approaches a person on the platform(observe) then the frequency on the sound observed by the observer can be mathematically represented as

$f_1 = f * \frac{v_s}{v_s - v_t}$

substituting values

$f_1 = 350 * \frac{343 }{343-39.6}$

$f_1 = 395.7 \ Hz$

Now according to dopplers equation when the train(source) moves away from the person on the platform(observe) then the frequency on the sound observed by the observer can be mathematically represented as

$f_2 = f * \frac{v_s}{v_s +v_t}$

substituting values

$f_2 = 350 * \frac{343}{343 + 39.6}$

$f_2 = 313.77 \ Hz$

The overall change in frequency is detected by a person on the platform as the train moves from approaching to receding is mathematically evaluated as

$\Delta f = f_1 - f_2$

$\Delta f = 395.7 - 313.77$

$\Delta f = 81.93 \ Hz$

Generally the wavelength detected by the person as the train approaches is mathematically represented as

$\lambda_1 = \frac{v}{f_1 }$

$\lambda_1 = \frac{343}{395.7 }$

$\lambda_1 = 0.867 \ m$

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4 years ago

2. Two cars A and B are heading forward. The velocity of A and B are 30 m/s and 20 m/s

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Answer:

A.) 27000 kgm/s

18000 kgm/s

B.) Va = 22 m/s

C.) 19800 kgm/s

25200 kgm/s

Explanation: Given that the velocity of A and B are 30 m/s and 20 m/s. And of the same mass M = 9 × 10^5g

M = 9×10^5/1000 = 900 kg

A.) Initial momentum of A

Mu = 900 × 30 = 27000 kgm/s

Initial momentum of B

Mu = 900 × 20 = 18000 kgm/s

B.) if they have an accident and then the velocity of the B is 28 m/s, find out velocity of A.

Momentum before impact = momentum after impact

Given that Vb = 28 m/s

27000 + 18000 = 900Va + 900 × 28

45000 = 900Va + 25200

900Va = 45000 - 25200

900Va = 19800

Va = 19800/900

Va = 22 m/s

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MV = 900 × 22 = 19800 kgm/s

Momentum of B after impact

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3 years ago

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Answer:

a. 0.7 m/s

b. The second runner

c. 4.20 meters

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a. To know the relative speed we can use the following equation:

Relative Speed (Vr) = V1 - V2

The relative speed is used to compared the velocity of one object (V1) to another object (V2). To learn the Relative Velocity from the second runner to the first we do the following:

Relative Speed (Vr) = Second Runner Velocity (Vs) - First Runner Velocity (Vf)

Vr = 4.20 m/s - 3.50 m/s

Vr = 0.7 m/s

This means that the second runner is gaining 0.7 meter from the first runner each second.

b. To know which runner will gain the race, we need to keep in mind, that the seconds runner is 45 meter more away that the first runner from the goal. Knowing this, the winner will be the one that reach the goal on less time.

Because movement is constant, we can use the following equation:

Velocity (V) = Distance (D) / Time (T)

To get the time we do:

T = D / V

Now, for the first runner:

T = 250 meters / (3.50 m/s)

T = 71.42 seconds

For the second runner.

T = (250 meters + 45 meters) / (4.20 m/s)

T = 70,23 seconds

Because the second runner takes less time to reach the goal, he wins the race.

c. Because it is know how much time takes the winner to get to the finish line, we can use that same time to know how far was the first runner from the goal. This is posible we know that how far he is from the goal is as follows.

Distance from goal (Dg) = Total Distance (D) - Distance travelled (Dt)

Where Total Distance (D) is 250 mts. And the distance travelled can be obtained by using the following equation.

Velocity (V) = Distance (D) / Time (T)

Searching for distance:

D = V / T.

D = (3.50 m/s) * 70.23 s

D = 245,80 meters

Soo

Distance from goal (Dg) = Total Distance (D) - Distance travelled (Dt)

Dg = 250 meters - 245,80 meters

Dg = 4,20 meters

The first runner was 4,20 meter behind the second one when the second runner crossed the goal

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