Answer:
It is explained in the explanation section
Explanation:
When the lift starts going downwards, it will start accelerating downwards. After a while, it will start moving with a constant velocity.
Constant velocity means that acceleration is zero and so the man will not feel any weight loss.
Now, Once the lift achieves constant velocity the acceleration is zero hence he will not experience any weight loss.
However, when the lift is in uniform motion, the lift and the man will fall down with an acceleration(a) that is less than that due to gravity(g) . Thus, the man will feel an apparent weight F which is not equal to zero.
Answer:
The force of gravity exerts a downward force. The floor exerts an upward force. Since these two forces are of equal magnitude and in opposite directions, they balance each other.
Explanation:
Assuming the wall is frictionless, there are four forces acting on the ladder.
Weight pulling down at the center of the ladder (mg).
Reaction force pushing to the left at the wall (Rw).
Reaction force pushing up at the foot of the ladder (Rf).
Friction force pushing to the right at the foot of the ladder (Ff).
(a) Calculate the reaction force at the wall.
Take the sum of the moments about the foot of the ladder.
∑τ = Iα
Rw (3.0 sin 60°) − mg (1.5 cos 60°) = 0
Rw (3.0 sin 60°) = mg (1.5 cos 60°)
Rw = mg / (2 tan 60°)
Rw = (10 kg) (9.8 m/s²) / (2√3)
Rw = 28 N
(b) State the friction at the foot of the ladder.
Take the sum of the forces in the x direction.
∑F = ma
Ff − Rw = 0
Ff = Rw
Ff = 28 N
(c) State the reaction at the foot of the ladder.
Take the sum of the forces in the y direction.
∑F = ma
Rf − mg = 0
Rf = mg
Rf = 98 N
Answer:
40 Hz
Explanation:
f = 1/T = 1 / 0.025 = 40 Hz
Answer:
The maximum height the pebble reaches is approximately;
A. 6.4 m
Explanation:
The question is with regards to projectile motion of an object
The given parameters are;
The initial velocity of the pebble, u = 19 m/s
The angle the projectile path of the pebble makes with the horizontal, θ = 36°
The maximum height of a projectile, 
, is given by the following equation;
Therefore, substituting the known values for the pebble, we have;
Therefore, the maximum height of the pebble projectile, ≈ 6.4 m.
