Work = force x distance
200 Newtons x 20 meters
= 4,000 Joules
Answer:
true
Explanation:
the wavelength of AM radio is 10 m whereas FM has 1 m
Answer:
W = 1250 J = 1.25 KJ
Explanation:
The work done by the boy is due to the change in the position of the cement vertically. Hence, the work done in this case will be equal to the potential energy of the cement:

where,
W = Work done = ?
mg = W = weight of cement = 500 N
h = height covered = 2.5 m
Therefore,

<u>W = 1250 J = 1.25 KJ</u>
the independent variable is what you're testing or changing in an experiment, so the answer is the temperature of the ball when its dropped.
i hope that helped <3
Answer:
Loss, 
Explanation:
Given that,
Mass of particle 1, 
Mass of particle 2, 
Speed of particle 1, 
Speed of particle 2, 
To find,
The magnitude of the loss in kinetic energy after the collision.
Solve,
Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.
Applying the conservation of momentum to find the speed of two particles after the collision.



V = 6.71 m/s
Initial kinetic energy before the collision,



Final kinetic energy after the collision,



Lost in kinetic energy,



Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.