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Kay [80]
3 years ago
8

A person having a mass of 85 kg sits on the edge of a horizontal rotating platform, 2.2 m from the center of the platform, and h

as a tangential speed of 6.6 m/s. calculate the angular momentum of the person
Physics
1 answer:
gulaghasi [49]3 years ago
3 0
The angular momentum of the person is given by:
L=mvr
where m=85 Kg is the mass of the person, v=6.6 m/s is the tangential velocity, and r=2.2 m is the distance of the person from the center of rotation. Using these data, we find
L=(85 Kg)(6.6 m/s)(2.2 m)=1234.2 Kg \cdot m^2/s
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A 0.600 kg block is attached to a spring with spring constant 15 N/m. While the block is sitting at rest, a student hits it with
gulaghasi [49]

Answer:

8.8 cm

31.422 cm/s

Explanation:

m = Mass of block = 0.6 kg

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x = Compression of spring

v = Velocity of block

A = Amplitude

As the energy of the system is conserved we have

\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{0.6\times 0.44^2}{15}}\\\Rightarrow A=0.088\ m\\\Rightarrow A=8.8\ cm

Amplitude of the oscillations is 8.8 cm

At x = 0.7 A

Again, as the energy of the system is conserved we have

\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\dfrac{15(0.088^2-(0.7\times 0.088)^2)}{0.6}}\\\Rightarrow v=0.31422\ m/s

The block's speed is 31.422 cm/s

4 0
3 years ago
In this experiment, you need to examine the idea of thermal energy transfer. Using a controlled experiment, what might a good qu
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We'll look at two properties:

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By taking an example;

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Conclusion of experiment;

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Answer:

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