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Softa [21]
3 years ago
13

You have landed on an unknown planet, newtonia, and want to know what objects will weigh there. you find that when a certain too

l is pushed on a frictionless horizontal surface by a 12.7 n force, it moves 16.1 m in the first 2.50 s , starting from rest. you next observe that if you release this tool from rest at 10.3 m above the ground, it takes 2.88 s to reach the ground.

Physics
1 answer:
AleksandrR [38]3 years ago
3 0
Refer to the diagram shown.

Because the surface is frictionless, the resistive for, R, is zero.

Let m = the mass of the object.
Let a =  acceleration due to the applied force.
Therefore
12.7 N = (m kg)*(a m/s²)
a = 12.7/m  m/s²

The object travels  16.1 m in 2.5 s, starting from rest. Therefore
16.1 N = (1/2)*(12.7/m m/s²)*(2.5 s)² = 39.6875/m  N
m = 16.1/39.6875 = 0.4057 kg

For freefall, let g =  acceleration due to gravity.
The time to fall from 10.3 m is 2.88 s, therefore
10.3 m = (1/2)*(g m/s²)*(2.88 s)² = 4.1472g m
g = 10.3/4.1472 = 2.484 m/s²

Answer:
The gravitational acceleration on the planet is 2.5 m/s² (nearest tenth)


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Answer:

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The latent heat of fusion is 481.76 J/mol

Melting point is given by

T_m=\dfrac{L_f}{\Delta S}\\\Rightarrow T_m=\dfrac{8\times 10^{-22}\times 6.022\times 10^{23}}{6\times 10^{-24}\times 6.022\times 10^{23}}\\\Rightarrow T_f=133.33\ K

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3 years ago
near the surface of the earth, objects in free fall (but not terminal velocity) experience a. constant distance. b. constant acc
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<h3><u>Answer;</u></h3>

B. constant acceleration.

<h3><u>Explanation</u>;</h3>
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The speed of the plate is 3.2048179721\times 10^{-9}\ m/s

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