a) Remember Newton's second Law:
F = m*a
This means that the forces that is being exerted over an object is equal to the mass of the object times the acceleration that it has.
In this case, in order to hold the mass you need a force with the same magnitude but opposite direction to the gravitational force. The magnitude of this force would be the mass of the object times the gravitactional acceleration:
F = 2kg*9.80665m/s² = 19.6133 N
b) Using again Newton's second Law, we can issolate mass from that equation:
m = F/a
Then, the mass that a force of 1N can support is equal to:
m = 1N/9.80665m/s² = 0.102 kg
Your question: The strong nuclear force felt by a single proton in a large nucleus _______________________.
Answer: is about the same as that felt by a single proton in a small nucleus.
If the student (which I’m naming student “x”) starts off at running 200m/30secs, then accelerated to 300m/30secs. If it was one minute later, I naturally assume she increases 100m/20secs.
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
b
Explanation:
From the question we are told that
The maximum height is H = 2.70
The Range is R = 12.9 m
Generally from projectile motion we have that
Generally from trigonometric identity
So
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Also the maximum height is
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Dividing equation 2 by (1)
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So
From equation 1
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Generally the vertical component of the initial velocity is mathematically evaluated as
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Generally the time taken is mathematically represented as
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