A mass slider m = 0.200 kg rests on a frictionless horizontal air rail connected to a spring with a force constant k = 5.00 N /
m. You pull the slider, stretching the spring 0.100 m, and then release it from rest. The slider returns to its equilibrium position (x = 0). What speed does it have when x = 0.080 m?
1 answer:
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To solve this problem it is necessary to apply the equation related to the Gravitational Force, the equation describes that
Where,
G = Gravitational Universal Constant
M = Mass of Earth (or Bigger star)
m = Mass of Object (or smallest star)
r = Radius
From the statement we know that once the impact is made, the golf ball is subjected to the forces that are exerted in nature. Since the air resistance, which would represent the drag force, is ignored. Only the forces related to gravity remain.
The gravitational force carries 'pushes' or 'attracts' the body towards the earth, while the speed decreases as it reaches its maximum height.
When the ball has reached its maximum height only the force of gravity begins to act on it, generating the attraction to the earth in parabolic motion.
Therefore the correct answer is B.
Answer:
43.41 mm
Explanation:
Given:
thickness of sheet, t = 6 mm
Force exerted by punch, F = 45 KN
Average shearing stress, T = 55 MPa
From average shearing stress T = Force F / Area A
Hence area = force/stress =45000/ 55 =
From area = pi*diameter*thickness
diameter = area/(pi* thickness)
= 818.18/(3.142*6)
= 43.41 mm