Answer:
a) Deceleration = 201.76 m/s^2
b) Distance traveled = 197.68 m
Explanation:
Initial Speed = 632 mi/h
Initial Speed (in meters/second) = (632 * 1.609 * 1000) / (60 * 60) = 282.4 m/s
Time to decelerate = 1.4 seconds
a) Change in speed = Acceleration * time
-282.4 = Acceleration * 1.4
Acceleration = -201.76 m/s^2
Deceleration = 201.76 m/s^2
b) Distance traveled = average speed * time
average speed = 282.4 / 2 = 141.2 m/s
Distance traveled = 141.2 * 1.4
Distance traveled = 197.68 m
Force = mass • acceleration
1) Find acceleration
-) 8kg ball
a= v/t
a= .2/ 10 = .02 m/s^2
Force = 8kg • .02 = .16N
-) 4kg ball
a= v/t
a= 1.0/ 10 = .1 m/s^2
Force = 4kg • .1 = .4N
Answer: The 8kg ball would need more force because it requires .16N, while the 4kg only need .4N .
Because the speed is constant, then that means it's not accelerating. If it's not accelerating, the force is balanced, so the net force is always zero. The frictional force is the same as the force applied. Which is 85N.
Answer:
film is at distance of 3.07 cm from lens
Explanation:
Given data
focal length = 3.06 cm
distance = 10.4 m = 1040 cm
to find out
How far must the lens
solution
we apply here lens formula that is
1/f = 1/p + 1/q
here f = 3.06 and p = 1040 so we find q
1/f = 1/p + 1/q
1/3.06 = 1/1040 + 1/q
1/ q = 0.3258
q = 3.0690 cm
so film is at distance of 3.07 cm from lens