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vova2212 [387]
3 years ago
6

Someone pls answer this with steps

Physics
1 answer:
Ivanshal [37]3 years ago
8 0
No worry’s I am here to help
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Help help help help help
mina [271]
Please mark me brainliest

7 0
3 years ago
Is the earth's gravitational force on the sun larger than, smaller than, or equal to the sun's gravitational force on the earth?
Leona [35]

Answer:

The earth's gravitational force on the sun is equal to the sun's gravitational force on the earth

Explanation:

Newton's third law (law of action-reaction) states that:

"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"

In other words, when two objects exert a force on each other, then the magnitude of the two forces is the same (while the directions are opposite).

In this problem, we can call the Sun as "object A" and the Earth as "object B". According to Newton's third law, therefore, we can say that the gravitational force that the Earth exerts on the Sun is equal (in magnitude, and opposite in direction) to the gravitational force that the Sun exerts on the Earth.

6 0
3 years ago
ASAP ONLY CORRECT ANSWERS PLEASEEEEEEEEEEEEEEEEEEEEEEEEEEEE I RLLY NEED HELPPPPPPP
nadya68 [22]

Answer:

attached below

Explanation:

5 0
3 years ago
A 20 kg object in a 2 kg object falls toward earth with a acceleration of 9.8m/sWhat is true about the force of gravity on the t
Vitek1552 [10]
Look it up BYEEEEEEEEEEEEEEEEEEEEEE
5 0
3 years ago
After a fall, a 90 kg rock climber finds himself dangling from the end of a rope that had been 16 m long and 7.8 mm in diameter
Nesterboy [21]

Explanation:

Given that,

Mass of the rock climber, m = 90 kg

Original length of the rock, L = 16 m

Diameter of the rope, d = 7.8 mm

Stretched length of the rope, \Delta L=3.1\ cm

(a) The change in length per unit original length is called strain. So,

\text{strain}=\dfrac{\Delta L}{L}\\\\\text{strain}=\dfrac{3.1\times 10^{-2}}{16}\\\\\text{strain}=0.00193

(b) The force acting per unit area is called stress.

\text{stress}=\dfrac{mg}{A}\\\\\text{stress}=\dfrac{90\times 10}{\pi (3.9\times 10^{-3})^2}\\\\\text{stress}=1.88\times 10^7\ Pa

(c) The ratio of stress to the strain is called Young's modulus. So,

Y=\dfrac{\text{stress}}{\text{strain}}\\\\Y=\dfrac{1.88\times 10^7}{0.00193}\\\\Y=9.74\times 10^9\ N/m^2

Hence, this is the required solution.

8 0
3 years ago
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