Ask question

Ask question

All categories

2 years ago

11

A train is approaching a signal tower at a speed of 40m/s. The train engineer sounds the 1000-Hz whistle, while a switchman in t

he tower responds by sounding 1200-Hz siren. THe air still and the speed is 340m/s. What is the frequency of the train whistle tone that is heard by the switchman?
300/340x1000Hz
380/340x1000Hz
340/300x1000Hz
1000Hz
340/380x1000Hz

1 answer:

stealth61 [152]2 years ago

3 0

v = speed of the source of sound or the train towards the listener or switchman = 40 m/s

V = actual speed of sound = 340 m/s

f = actual frequency of sound as emitted from source or the train = 1000 Hz

f' = frequency as observed by the listener or by switchman = ?

Using Doppler's law , frequency observed by a listener from a source moving towards it is given as

f' = V f /(V - v)

inserting the values

f' = 340 x 1000 /(340 - 40)

f' = 340 x 1000/300

You might be interested in

A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo

-BARSIC- [3]

Answer:

It would take $\tau(\ln 9 - \ln 8)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{8}{9} \, q_0$ .

It would take $\tau(\ln 9 - \ln 7)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{7}{9}\, q_0$ .

Note that $\ln 9 = 2\,\ln 3$ , and that $\ln 8 = 3\, \ln 2$ .

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is $\tau$ , and the initial charge of the capacitor be $q_0$ . Then at time $t$ , the charge stored in the capacitor would be:

$\displaystyle q(t) = q_0 \, e^{-t / \tau}$ .

<h3>a)</h3>

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{8}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{8}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{8}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{8}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 8 - \ln 9$ .

$t = - \tau \, \left(\ln 8 - \ln 9\right) = \tau(\ln 9 - \ln 8)$ .

<h3>b)</h3>

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{7}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{7}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{7}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 7 - \ln 9$ .

$t = - \tau \, \left(\ln 7 - \ln 9\right) = \tau(\ln 9 - \ln 7)$ .

7 0

3 years ago

Which gas giant has a rotation axis so tilted that the planet rotated like a bowling ball as it orbits the sun?

Anestetic [448]

The answer to your question is OPTION B

3 0

3 years ago

Two blocks, A and B, are connected by a rope. A second rope is connected to block B and a steady, horizontal tension force of T

Oksanka [162]

Answer:

40 N

Explanation:

We are given that

Speed of system is constant

Therefore, acceleration=a=0

Tension applied on block B=T=50 N

Friction force=f=10 N

We have to find the friction force acting on block A.

Let T' be the tension in string connecting block A and block B and friction force on block A be f'.

For Block B

$T-f-T'=m_Ba$

Where $m_B$ =Mass of block B

Substitute the values

$50-10-T'=m_B\times 0=0$

$T'==40 N$

For block A

$T'-f'=m_Aa$

Where $m_A=$ Mass of block A

Substitute the values

$40-f'=m_A\times 0=0$

$f'=40 N$

Hence, the friction force acting on block A=40 N

3 0

3 years ago

A plane travels at a speed of 205mph in still air. Flying with a tailwind, the plane is clocked over a distance of 1000 miles. F

vaieri [72.5K]

While plane is moving under tailwind condition it took time "t"

so here we will have

$t = \frac{d}{v_{net}}$

here net speed of the plane will be given as

$v_{net} = v + v_w$

$t = \frac{1000}{205 + v_w}$

similarly when it moves under the condition of headwind its net speed is given as

$v_{net} = v - v_w$

now time taken to cover the distance is 2 hours more

$t + 2 = \frac{1000}{205 - v_w}$

now solving two equations

$\frac{1000}{205 + v_w} + 2 = \frac{1000}{205 - v_w}$

solving above for v_w we got

$v_w = 40.4 mph$

6 0

3 years ago

Question 3 of 10

Mazyrski [523]

Answer:

B . energy cannot be created or destroyed

3 0

2 years ago