Question
You have landed on an unknown planet, Newtonia, and want to know what objects will weigh there. You find that when a certain tool is pushed on a frictionless horizontal surface by a 12.2 N force, it moves 16.3 m in the first 2.30 s, starting from rest. You next observe that if you release this tool from rest at 10.6 m above the ground, it takes 2.68 s to reach the ground.
What does the tool weigh on Newtonia?
What would it weigh on Earth?
The average velocity after pushing 16.3 m is
V = 2*(16.3m/2.3s) = 14.17 m/s
F X = (1/2) M V^2
M = 2 F X/V^2
Solve for M. Then get the acceleration of gravity, g', on the planet from the second measurement
(1/2) g' t^2 = 10.6 m
Solve that for g'. It will be less than the value on Earth
The weight is M g' on Newtonia and M g on Earth
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What does the tool weigh on Newtonia?
What would it weigh on Earth?
The average velocity after pushing 16.3 m is
V = 2*(16.3m/2.3s) = 14.17 m/s
F X = (1/2) M V^2
M = 2 F X/V^2
Solve for M. Then get the acceleration of gravity, g', on the planet from the second measurement
(1/2) g' t^2 = 10.6 m
Solve that for g'. It will be less than the value on Earth
The weight is M g' on Newtonia and M g on Earth
afdadf
Answers
Why is The average velocity after pushing V = 2*(16.3m/2.3s)
Where did u get the 2 from?
Where did u get the 2 from?
When x is known and here it is x = 16.3m we use the equation of motion for constant acceleration
Vx^2 = Vix^2 + 2Ax(x-xi) where
Vx is our velocity
Vix is our initial velocity
A is our acceleration or distance/time
x is our final point
and xi is our starting point
so to answer your question, the 2 is just part of the equation.
Vx^2 = Vix^2 + 2Ax(x-xi) where
Vx is our velocity
Vix is our initial velocity
A is our acceleration or distance/time
x is our final point
and xi is our starting point
so to answer your question, the 2 is just part of the equation.
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