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Snezhnost [94]
2 years ago
7

A 1-kilogram stone is being dropped from a bridge 100 meters above a gorge.

Physics
2 answers:
ipn [44]2 years ago
6 0
<h2><u>ANSWER</u><u>:</u></h2>

<h3><u>C</u><u>.</u><u> </u><u>1</u><u>0</u><u>0</u><u> </u><u>NEWTONS</u></h3>

<h2><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u></h2>

<u>CARRY</u><u> </u><u>ON</u><u> LEARNING</u>

<u>CAN</u><u> </u><u>Y</u><u>OU</u><u> BRAINLEST</u><u> ME</u>

ziro4ka [17]2 years ago
4 0

Answer:

C. 100 NEWTONS

Explanation:

just my brain being ly = brainly is the name so my brian is being ly

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Due to the earth's rotation, a person would be traveling faster at the equator than they would at a position halfway from the eq
Kisachek [45]

Answer:

False

Explanation:

It is actually the exact opposite.

8 0
3 years ago
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A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g
natali 33 [55]

(a) 5.65 times the Earth's radius

The escape velocity for a projectile on Earth is

v_e=\sqrt{\frac{2GM}{R}}

where

G is the gravitational constant

M is the Earth's mass

R is the Earth's radius

If the projectile has an initial speed of 0.421 escape speed,

v=0.421 v_e

So its initial kinetic energy will be

K=\frac{1}{2}m(0.421 v)^2=0.089 m(\sqrt{\frac{2GM}{R}})^2=0.177 \frac{GMm}{R}

where m is the mass of the projectile

At the point of maximum altitude, all this energy is converted into gravitational potential energy:

K=U\\0.177 \frac{GMm}{R}=\frac{GMm}{r}

where r is the distance from the Earth's centre reached by the projectile. We can write r as a multiple of R, the Earth's radius:0.177 \frac{GMm}{R}=\frac{GMm}{nR}

And solving the equation we find

n=\frac{1}{0.177}=5.65

So, the projectile reaches a radial distance of 5.65 times the Earth's radius.

b) 2.36 times the Earth's radius

The kinetic energy needed to escape is:

K=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}

This time, the projectile has 0.421 times this energy:

K=0.421 \frac{GMm}{R}

Again, at the point of maximum altitude, all this energy will be converted into potential energy:

0.421 \frac{GMm}{R}=\frac{GMm}{nR}

and by solving for n we find

n=\frac{1}{0.421}=2.36

So, the projectile reaches a radial distance of 2.36 times the Earth's radius.

c) E=U=\frac{GMm}{R}

The least initial mechanical energy needed for the projectile to escape Earth is equal to the gravitational potential energy of the projectile at the Earth's surface:

E=U=\frac{GMm}{R}

Indeed, the kinetic energy of the projectile must be equal to this value. In fact, if we use the formula of the escape velocity inside the formula of the kinetic energy, we find

K_e=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}

6 0
3 years ago
Which of the following is a form of potential energy
Zepler [3.9K]

the ancwer is B sound hope it helps

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How much space is taken up by a package with a length of 45 cm, a width of 10 cm, and a height of 8 cm?
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The answer is C. 3,600cm cubed

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Which statement best describes the adiabatic process? A.the temperature remains constant B.the temperature increases at a consta
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For an adiabatic process Heat is not allowed to cross the boundary of the system.

So:

E.no heat flows into or out of the system

 

7 0
3 years ago
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