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bazaltina [42]
2 years ago
15

A 20-cm-long stick of m = 0.600 kg is lifted by a rope tied 5.0 cm from the upper end. The other end touches a smooth floor. The

stick makes an angle θ with the floor. Find the magnitude of the normal reaction from the floor on the stick.
Physics
1 answer:
Liula [17]2 years ago
3 0

Answer:

https://www.chegg.com/homework-help/questions-and-answers/part-20-cm-long-stick-m-0400-kg-lifted-rope-tied-70-cm-upper-end-end-touches-smooth-floor--q81268572

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A ball is attached to a string of length 3 m to make a pendulum. The pendulum is placed at a location that is away from the Eart
Musya8 [376]

1) 0.61 m/s^2

2) 13.9 s

Explanation:

1)

The acceleration due to gravity is the acceleration that an object in free fall (acted upon the force of gravity only) would have.

It can be calculated using the equation:

g=\frac{GM}{r^2} (1)

where

G is the gravitational constant

M=5.98\cdot 10^{24} kg is the Earth's mass

r is the distance of the object from the Earth's center

The pendulum in the problem is at an altitude of 3 times the radius of the Earth (R), so its distance from the Earth's center is

r=4R

where

R=6.37\cdot 10^6 m is the Earth's radius

Therefore, we can calculate the acceleration due to gravity at that height using eq.(1):

g=\frac{GM}{(4R)^2}=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})0.}{(4\cdot 6.37\cdot 10^6)^2}=0.61 m/s^2

2)

The period of a simple pendulum is the time the pendulum takes to complete one oscillation. It is given by the formula

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration due to gravity at the location of the pendulum

Note that the period of a pendulum does not depend on its mass.

For the pendulum in this problem, we have:

L = 3 m is its length

g=0.61 m/s^2 is the acceleration due to gravity (calculated in part 1)

Therefore, the period of the pendulum is:

T=2\pi \sqrt{\frac{3}{0.61}}=13.9 s

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Answer:

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Explanation:

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Answer:

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Explanation:

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Answer:

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