Answer:
189 m/s
Explanation:
The pilot will experience weightlessness when the centrifugal force, F equals his weight, W.
So, F = W
mv²/r = mg
v² = gr
v = √gr where v = velocity, g = acceleration due to gravity = 9.8 m/s² and r = radius of loop = 3.63 × 10³ m
So, v = √gr
v = √(9.8 m/s² × 3.63 × 10³ m)
v = √(35.574 × 10³ m²/s²)
v = √(3.5574 × 10⁴ m²/s²)
v = 1.89 × 10² m/s
v = 189 m/s
Answer:
The size of the force that pushes the wall is 12,250 N.
Explanation:
Given;
mass of the wrecking ball, m = 1500 kg
speed of the wrecking ball, v = 3.5 m/s
distance the ball moved the wall, d = 75 cm = 0.75 m
Apply the principle of work-energy theorem;
Kinetic energy of the wrecking ball = work done by the ball on the wall
¹/₂mv² = F x d
where;
F is the size of the force that pushes the wall
¹/₂mv² = F x d
¹/₂ x 1500 x 3.5² = F x 0.75
9187.5 = 0.75F
F = 9187.5 / 0.75
F = 12,250 N
Therefore, the size of the force that pushes the wall is 12,250 N.
Answer
Nature of the surfaces.
Explanation
Friction is the the force that opposes motion between two surfaces that are in relative motion. If two objects are in contact, one of it or both must be in motion for friction to exist.
Friction is affected by a number of factors.
One is the weight of the object. The more the weight of the object the higher the friction between it and the surface.
The other factor is the nature of the surfaces. Rough surfaces contribute to high friction while smooth surfaces reduces the friction between surfaces.
Answer:
Explanation:
Relative velocity is defined as the velocity of an object B in the rest frame of another object A.
