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wariber [46]
3 years ago
14

A small light cylinder and a large heavy cylinder are released at the same time and roll down a ramp without slipping. Which one

reaches the bottom first?
A. Small cylinder
B. Large cylinder
C. Both reach the bottom at the same time.
Physics
1 answer:
forsale [732]3 years ago
6 0

Answer:

C. Both reach the bottom at the same time.

Explanation:

For a rolling object down an inclined plane , the acceleration is given below

a = g sinθ / (1 + k² / r² )

θ is angle of inclination , k is radius of gyration , r is radius of the cylinder

For cylindrical object

k² / r² = 1/2

acceleration =  g sinθ  /( 1 + 1/2 )

= 2 g sinθ / 3  

Since it does not depend upon either mass or radius , acceleration of both the cylinder will be equal . Hence they will reach the bottom simultaneously.

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Answer:

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Explanation:

Let be M and m the masses of the planet and a person standing on the surface of the planet, so that M >> m. The attraction force between the planet and the person is represented by the Newton's Law of Gravitation:

F = G\cdot \frac{M\cdot m}{r^{2}}

Where:

M - Mass of the planet Earth, measured in kilograms.

m - Mass of the person, measured in kilograms.

r - Radius of the Earth, measured in meters.

G - Gravitational constant, measured in \frac{m^{3}}{kg\cdot s^{2}}.

But also, the magnitude of the gravitational field is given by the definition of weight, that is:

F = m \cdot g

Where:

m - Mass of the person, measured in kilograms.

g - Gravity constant, measured in meters per square second.

After comparing this expression with the first one, the following equivalence is found:

g = \frac{G\cdot M}{r^{2}}

Given that G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}, M = 5.972 \times 10^{24}\,kg and r = 6.371 \times 10^{6}\,m, the magnitude of the gravitational field near Earth's surface is:

g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(6.371\times 10^{6}\,m)^{2}}

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The magnitude of the gravitational field strength near Earth's surface is represented by approximately 9.82\,\frac{m}{s^{2}}.

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Answer:

The value is  f  =  0.05 \ Hz

Explanation:

From the question we are told that

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Generally the frequency of the circular motion is  

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