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aleksklad [387]

3 years ago

15

A train station bell gives off a fundamental tone of 500 Hz as the train approaches the station at a speed of 30.0 m/s. If the s

peed of sound in air is 339 m/s, what will be the apparent frequency of the bell to an observer riding the train?

1 answer:

TiliK225 [7]3 years ago

3 0

Answer:

548.5 Hz

Explanation:

We are given that

Frequency of source=500 Hz

Speed of train=30m/s

Speed of sound in air=339m/s

We have to find the frequency of the bell to an observer riding the train.

The apparent frequency for the listener

$f=\frac{v_0}{v_0-v_s}f_1$

Where $f_1=$ Frequency of source

$v_0$ =Speed of sound

$v_s$ =Speed of source

Using the formula

The apparent frequency of the bell to an observer riding the train= $\frac{339}{339-30}\times 500=548.5 Hz$

Hence, the apparent frequency of the bell to an observer riding the train=548.5 Hz

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A basketball is held over head at a height of 2.4 m. The ball is lobbed to a teammate at 8 m/s at an angle of 40'. If the ball i

cupoosta [38]

Explanation:

since both the teammates are of the same height, their height won't matter. Because now the basketball won't cover any vertical distance.

We have to calculate its range the horizontal distance covered by it when tossed from one teammate to the other.

range can be calculated by the formula :-

$\boxed{\mathfrak{range = \frac{ u {}^{2} \sin 2\theta }{g} }}$

u is the velocity during its take off and $\theta$ is the angle at which its thrown

Given that

u = 8m/ s
$\theta$ = 40°

calculating range using the above formula

$= \frac{ {8}^{2} \sin2(40) }{10}$

$= \frac{64 \times \sin(80) }{10}$

value of sin 80 = 0. 985

$= \frac{64 \times 0.985}{10}$

$= \frac{63.027}{10}$

$= 6.3027$

Hence,

$\mathfrak { \blue{the \: teammate \: is \: \red{\underline{6.3027 \: meters} }\: away } }$

7 0

2 years ago

An experimenter finds that standing waves on a string fixed at both ends occur at 24 Hz and 32 Hz , but at no frequencies in bet

Vera_Pavlovna [14]

Answer:

8 Hz

Explanation:

Given that

Standing wave at one end is 24 Hz

Standing wave at the other end is 32 Hz.

Then the frequency of the standing wave mode of a string having a length, l, is usually given as

f(m) = m(v/2L), where in this case, m could be 1. 2. 3. 4 etc

Also, another formula is given as

f(m) = m.f(1), where f(1) is the fundamental frequency..

Thus, we could say that

f(m+1) - f(m) = (m + 1).f(1) - m.f(1) = f(1)

And as such,

f(1) = 32 - 24

f(1) = 8 Hz

Then, the fundamental frequency needed is 8 Hz

4 0

2 years ago

6. The bat hits a ball. What is the

lana [24]

Answer:

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Explanation:

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2 years ago

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vredina [299]

Answer:

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7 0

2 years ago

g Estimate the number of photons emitted by the Sun in a second. The power output from the Sun is 4 Ã— 1026 W and assume that th

vagabundo [1.1K]

Answer:

The value is $N = 1.107 *10^{45 } \ photons$

Explanation:

From the question we are told that

The power output from the sun is $P_o = 4 * 10^{26} \ W$

The average wavelength of each photon is $\lambda = 550 \ nm = 550 *10^{-9} \ m$

Generally the energy of each photon emitted is mathematically represented as

$E_c = \frac{h * c }{ \lambda }$

Here h is the Plank's constant with value $h = 6.62607015 * 10^{-34} J \cdot s$

c is the speed of light with value $c = 3.0 *10^{8} \ m/s$

So

$E_c = \frac{6.62607015 * 10^{-34} * 3.0 *10^{8} }{ 550 *10^{-9} }$

=> $E_c = 3.614 *10^{-19} \ J$

Generally the number of photons emitted by the Sun in a second is mathematically represented as

$N = \frac{P }{E_c}$

=> $N = \frac{4 * 10^{26} }{3.614 *10^{-19}}$

=> $N = 1.107 *10^{45 } \ photons$

5 0

2 years ago