Since the track is friction less, block’s kinetic energy at the bottom of the track is equal to its potential energy at the top of the track.
PE = 81 * 9.8 * 3.8 = 3016.44 J
Work = 1/2 * 1888 * d^2
PE = Kinetic energy at the base.
1/2 * 1888 * d^2 = 3016.44
d = 1.78 approx 1.8
F = Ke = 1888 * 1.8 = 3398.4N
Answer:
98%
Explanation:
Given parameters
Mass of motor = 10kg
Height = 2m
Time = 2s
Power input = 100w
Unknown
Efficiency = ?
Solution
Efficiency is the percentage of the power output to the power input.
Power is the rate at which work is done.
Power output = mass x g x height / time
g is the acceleration due to gravity
Power output = 10x 2 x 9.8 / 2 = 98W
Efficiency = power output/ power input x 100
Efficiency = 98/100 x 100 = 98%
Answer:
The speed of Susan is 2.37 m/s
Explanation:
To visualize better this problem, we need to draw a free body diagram.
the work is defined as:
here we have the work done by Paul and the friction force, so:
Now the change of energy is:
Report this clown who put the first answer he’s trying to get your ip
Answer:
1.566 x 10^2
Move the decimal to where the number being multiplied by 10^x is greater than 1 but less than 10. Then multiply it by 10^x
X is the number of times you moved the decimal, so in this case it would be 10^2
