Answer:
a) The velocity of rock at 1 second, v = 9.8 m/s
b) The velocity of rock at 3 second, v = 29.4 m/s
c) The velocity of rock at 5.5 second, v = 53.9 m/s
Explanation:
Given data,
The rock is dropped from a bridge.
The initial velocity of the rock, u = 0
a) The velocity of rock at 1 second,
Using the first equation of motion
v = u + gt
v = 0 + 9.8 x 1
v = 9.8 m/s
b) The velocity of rock at 3 second,
v = u + gt
v = 0 + 9.8 x 3
v = 29.4 m/s
c) The velocity of rock at 5.5 second,
v = u + gt
v = 0 + 9.8 x 5.5
v = 53.9 m/s
Answer:w=mxg
2x10 =20 N
Explanation:force acting downwards is mg mass into gravitional feild
Answer:
(a) Force must be grater than 283.87 N
(B) Force will be equal to 193.945 N
Explanation:
We have given mass of the crate m = 49.6 kg
Acceleration due to gravity 
Coefficient of static friction 
Coefficient of kinetic friction 
(a) Static friction force is given by 
So to just start the crate moving we have to apply more force than 283.87 N
(B) This force will be equal to kinetic friction force
We know that kinetic friction force is given by 
Answer:
B) 1.2 N, toward the center of the circle
Explanation:
The circumference of the circle is:
C = 2πr
C = 2π (0.70 m)
C = 4.40 m
So the velocity of the ball is:
v = C/t
v = 4.40 m / 0.60 s
v = 7.33 m/s
Sum of the forces in the radial direction:
∑F = ma
T = m v² / r
T = (0.015 kg) (7.33 m/s)² / (0.70 m)
T = 1.2 N
The tension force is 1.2 N towards the center of the circle.
