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3 years ago

What is the average velocity of the particle from rest to 9 seconds?

1 answer:

7 0

Answer: v = 2[m/s]Explanation:This avarage velocity can be found with the ... B. 2 meters/ second. C. 3 meters/second. D. 4 meters/second. 1.

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

Licemer1 [7]

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

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

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

$\lambda_1 = 0.867 \ m$

4 0

3 years ago

An object is moving in a horizontal direction only. The graph shows its velocity vs. time. (See Graph Below)

zysi [14]

48 M to get the answer add the area of the triangle and rectangle under the line

8 0

3 years ago

Plzzzz help it for civics

Viktor [21]

Answer:

A : the colonists dumped tea in Boston Harbor

6 0

3 years ago

T = mg + ma solve for a

Gnoma [55]

You said T   = mg + ma

Subtract mg
from each side: T - mg =      ma

Divide each side by m : a = (T-mg) / m

      or a = T/m - g

5 0

3 years ago

.prove : s=ut +½ at²

o-na [289]

Explanation:

Let the distance covered by the body be s, initial and final velocities be u and v respectively and time taken be t.

$\therefore Average\: velocity = \frac{u+v}{2} \\\\Now, \:we \:know\: that\\\\Distance \:covered\\ = Average\: velocity \times time\\\\\therefore s= \frac{(u+v) }{2} \times t..... (1)\\\\$

By first equation of motion:

$v = u + at$

Substituting the value of v in equation (1), we find:

$s= \frac{(u+u + at)}{2} \times t\\\\\therefore s= \frac{(2u + at)}{2} \times t\\\\\therefore s= \frac{(2ut + at^2)}{2}\\\\\therefore s= \frac{2ut} {2}+ \frac{at^2}{2}\\\\ \huge \orange {\boxed {\therefore s= ut+ \frac{1}{2}at^2}} \\\\$

Hence proved.

6 0

3 years ago