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lawyer [7]
3 years ago
8

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

= 1.63 rad/s in the counter clockwise direection when viewed from above. A person with mass m = 73 kg and velocity v = 4.8 m/s runs on a path tangent to the merry-go-round. Once at the merry-go-round the person jumps on and holds on to the rim of the merry-go-round.
What is the magnitude of the initial angular momentum of the merry-go-round?
Physics
1 answer:
RideAnS [48]3 years ago
8 0

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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Answer

given,

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a) satellite's orbital velocity

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b) T= \dfrac{2\pi\ r}{v}

   T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}

   T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}

          T = 14416.92 s

          T = \dfrac{14416.92}{3600}\ hr

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c) gravitational force acting

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In the great shopping cart race, two students push on shopping carts. A having twice the mass of B, with the same force applied
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KB = 1/2 mB x vB^2........................(2)

Diving equation 1 by equation 2, we get -

KA/KB = mA/mB

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dybincka [34]

Heat supplied to the gold will raise the temperature of the gold from 20 degree Celsius to 90 degree Celsius.

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Hence, the heat supplied to the gold to raise the temperature from 20 degree Celsius to 90 degree Celsius = 685.44 Joules

5 0
3 years ago
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