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3 years ago

8

With what tension must a rope with length 2.20 m and mass 0.150 kg be stretched for transverse waves of frequency 43.0 hz to hav

e a wavelength of 0.800 m ?

1 answer:

Blababa [14]3 years ago

5 0

Solution:

$v=sqrt(\frac{t}{u})\\v=40\times0.80=32m/s\\\frac{t}{u}=900\\t=900\times\frac{0.150}{2.20}\\=61.36N$

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A person of mass 70 kg stands at the center of a rotating merry-go-round platform of radius 2.9 m and moment of inertia 900 kg⋅m

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Explanation:

It is given that,

Mass of person, m = 70 kg

Radius of merry go round, r = 2.9 m

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Part A,

Let $\omega_2$ is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :

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Here, $I_2=I_1+mr^2$

$I_1\omega_1=(I_1+mr^2)\omega_2$

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Solving the above equation, we get the value as :

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Part B,

The initial rotational kinetic energy is given by :

$k_i=\dfrac{1}{2}I_1\omega_1^2$

$k_i=\dfrac{1}{2}\times 900\times (0.95)^2$

$k_i=406.12\ rad/s$

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$k_f=\dfrac{1}{2}(I_1+mr^2)\omega_1^2$

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Answer:

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$f' = \left (\dfrac{v + v_0}{v - v_s} \right) \times f$

Where;

f' = The frequency the observer hears

f = Actual frequency of the wave

v = The velocity of the sound wave

$v_o$ = The velocity of the observer

$v_s$ = The velocity of the source

Where the observer is stationary, we have;

(i) When the source is moving in the direction of the observer

$f' = \left (\dfrac{v }{v - v_s} \right) \times f$

(ii) When the source is receding from the observer, we have;

$f' = \left (\dfrac{v }{v + v_s} \right) \times f$

Therefore;

(a) A person left behind on the platform

For a person left behind on the platform, we have that the radio source is receding, therefore, we have;

$f' = \left (\dfrac{v }{v + v_s} \right) \times f$

(1) Given that (v + $v_s$ ) > v, therefore, v < (v + $v_s$ ), f' < f, the frequency heard by the person left on the platform, f', is smaller (lower) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

(3) During the fall, the speed of the source continuously increases under the effect of gravitational attraction and therefore the frequency heard by the person on the platform becomes progressively smaller

(b) A person down below floating on a rubber raft

For the the person down below on the rubber raft, the radio source is advancing

Therefore, the radio source is moving towards the person at rest down on the rubber raft, therefore, we have;

$f' = \left (\dfrac{v }{v - v_s} \right) \times f$

(1) Given that (v - $v_s$ ) < v, therefore, f' > f, the frequency heard by the person down below floating on the rubber raft, f', is greater (higher) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

(3) During the fall, the speed of the source continuously increases under the effect of gravitational attraction and therefore the frequency heard by the person on the platform becomes progressively greater (higher)

Explanation:

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3 years ago