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3 years ago

The second hand on a clock is 3.00 cm long. What is the speed of the outermost tip of that second hand

1 answer:

4 0

In 60 minutes or 3600 seconds, the tip of the minute hand traverses the circumference of a circle with radius 3.00 cm, so it moves with a tangential speed of

(3.00 cm)/(3600 s) ≈ 0.00083 cm/s = 8.3 μm/s

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Nicolaus Copernicus was considered: a heretic because his theory disagreed with the church a hero because his theory agreed with

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Nicolaus Copernicus was considered the <span>"Father of Astronomy" because he believed the earth was the center of the solar system. The rest of the choices do not answer the question above.</span>

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4 years ago

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A distance-time graph indicates that an object moves 20 m in 10 s and then remains at rest for 20s. What is the average speed of

boyakko [2]

Answer:

v = 2 m/s

Explanation:

The object moves 20 m in 10 s initially. Then it remains at rest for 20 s. It is required to find the average speed of the object.

Fro 10 s it covers 20 m. For that time, speed is :

$v=\dfrac{d}{t}\\\\v=\dfrac{20\ m}{10\ s}\\\\v=2\ m/s$

For 20 s, it was at rest means distance covered is 0. So, speed is 0.

The average speed of the object is equal to the sum of speed for 10 s and for 20 s i.e.

v = 2 m/s + 0 = 2 m/s

So, the average speed of the object is 2 m/s.

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3 years ago

It usually takes more force to start an object sliding than it does to keep an object sliding because static friction is usually

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The answer is already in the blank for, its was greater

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4 years ago

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:

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):

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

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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3 years ago

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