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Pie
3 years ago
5

If the 50 kg boy were in a spacecraft 5.0r from the center of the earth, what would his weight be?

Physics
1 answer:
Lady_Fox [76]3 years ago
7 0
·The acceleration of gravity is proportional to

                       1 / (the square of the distance from the center) .

When we're on the surface, we're 1 radius from the center of the Earth,
and the acceleration of gravity is  9.8 m/s² .

The boy's weight = (mass) · (gravity) = (50kg) · (9.8 m/s²)

                                                             =      490 newtons .

At the distance of 5 radii from the center (4 radii altitude from the surface),
the acceleration of gravity is

                 (9.8 m/s²) · (1/5)²  =  0.39 m/s² .

The boy's weight is    (mass) · (gravity)  =  (50kg) · (0.39 m/s²)

                                                                  =     19.6 newtons .

Just as we expected, his weight at that distance is

         (19.6 / 490) = 0.04  =  1/25  =  1/5²  of his weight on the surface.
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A golf club rotates 215 degrees and has a length (radius) equal to 29 inches. The time it took to swing the club was 0.8 seconds
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Answer:

The average linear velocity (inches/second) of the golf club is 136.01 inches/second

Explanation:

Given;

length of the club, L = 29 inches

rotation angle, θ = 215⁰

time of motion, t = 0.8 s

The angular speed of the club is calculated as follows;

\omega = (\frac{\theta}{360} \times 2\pi, \ rad) \times \frac{1}{t} \\\\\omega =  (\frac{215}{360} \times 2\pi, \ rad) \times \frac{1}{0.8 \ s} \\\\\omega = 4.69 \ rad/s

The average linear velocity (inches/second) of the golf club is calculated as;

v = ωr

v = 4.69 rad/s  x  29 inches

v = 136.01 inches/second

Therefore, the average linear velocity (inches/second) of the golf club is 136.01 inches/second

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3 years ago
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A massless string connects a 10.00 kg mass to a 13.00 kg cart which is resting on a frictionless horizontal surface. The mass ha
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The cart's acceleration to the right after the mass is released  is determined as 7.54 m/s².

<h3>Acceleration of the cart</h3>

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Thus, the cart's acceleration to the right after the mass is released  is determined as 7.54 m/s².

Learn more about acceleration here: brainly.com/question/14344386

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