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Sonbull [250]
3 years ago
6

If a 8.0 kg mass is hung on the end of a spring, it is stretched 0.78 meters as a result. What is the force constant of the spri

ng?
Physics
2 answers:
Alborosie3 years ago
3 0

Answer:

K=100.62N/m

Explanation:

Given:

             m= 8.0 Kg\\x=0.78 m

We know from Hook's law

           F= k x

where

           F= m g

So equating we get

             mg=kx\\k=\frac{mg}{x} \\k=\frac{8*9.8}{0.78}\\k=100.62 N/m

Hatshy [7]3 years ago
3 0

Answer:

k=100.5 n/m

Explanation:

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A music fan at a swimming pool is listening to a radio on a diving platform. The radio is playing a constant- frequency tone whe
joja [24]

Answer:

The Doppler Effect is given by the following relation;

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:

7 0
3 years ago
A merry-go-round with a a radius of R = 1.9 m and moment of inertia I = 209 kg-m2 is spinning with an initial angular speed of ω
RideAnS [48]

Answer:

340.67 kgm²/s

Explanation:

R = Radius of merry-go-round = 1.9 m

I = Moment of inertia = 209 kgm²

\omega_i = Initial angular velocity = 1.63 rad/s

m = Mass of person = 73 kg

v = Velocity = 4.8 m/s

Initial angular momentum is given by

L=I\omega_i\\\Rightarrow L=209\times 1.63\\\Rightarrow L=340.67\ kgm^2/s

The initial angular momentum of the merry-go-round is 340.67 kgm²/s

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3 years ago
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Answer:

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Explanation:

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What would happen if you removed the jaguar from this food web?
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If it was removed it would not be a predator
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Which planet has the lowest surface temperature of any planet and an atmosphere comprised mostly of nitrogen?
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It is the answer C. Neptune
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