Answer: 2940 J
Explanation: solution attached:
PE= mgh
Substitute the values:
PE= 10kg x 9.8 m/s² x 30 m
= 2940 J
Answer:
Magnitude of the average force exerted on the wall by the ball is 800N
Explanation:
Given
Contact Time = t = 0.05 seconds
Mass (of ball) = 0.80kg
Initial Velocity = u = 25m/s
Final Velocity = 25m/s
Magnitude of the average force exerted on the wall by the ball is given by;
F = ma
Where m = 0.8kg
a = Average Acceleration
a = (u + v)/t
a = (25 + 25)/0.05
a = 50/0.05
a = 1000m/s²
Average Force = Mass * Average Acceleration
Average Force = 0.8kg * 1000m/s²
Average Force = 800kgm/s²
Average Force = 800N
Hence, the magnitude of the average force exerted on the wall by the ball is 800N
The correct answer is:
Work is negative, the environment did work on the object, and the energy of the system decreases.
In fact, the work-energy theorem states that the work done by the system is equal to its variation of kinetic energy:
In this problem, the variation of kinetic energy 
is negative (because the final velocity is less than the initial velocity), so the work is negative, and this means that the environment did work on the object, and its energy decreased.
Answer:
phle follow karo yrr tab hee bat u ga
Answer:
35m/s[57o].
X = 35*Cos57 =
Y = 35*sin7 =Explanation:
learn man but there u go
