It's called cellular differentiation. I think.
Answer:
B.
It will be greater than 10 J.
Explanation:
The total mechanical energy of an object is the sum of its potential energy (PE) and its kinetic energy (KE):
E = PE + KE
According to the law of conservation of energy, when there are no frictional forces on an object, its mechanical energy is conserved.
The potential energy PE is the energy due to the position of the object: the highest the object above the ground, the highest its PE.
The kinetic energy KE is the energy due to the motion of the object: the highest its speed, the largest its KE.
Here at the beginning, when it is at the top of the roof, the baseball has:
PE = 120 J
KE = 10 J
So the total energy is
E = 120 + 10 = 130 J
As the ball falls down, its potential energy decreases, since its height decreases; as a result, since the total energy must remain constant, its kinetic energy increases (as its speed increases).
Therefore, when the ball reaches the ground, its kinetic energy must be greater than 10 J.
Friction stole the book's kinetic energy, and turned it into heat energy ... which blew away in the breeze.
Answer:
0 J
Explanation:
= Work done on the satellite in circular orbit about earth by earth
= Force on gravity on satellite by earth
= displacement of the satellite
= Angle between the force on gravity and displacement = 90
We know that, Work done is given as

Answer:
A
B

C

D

Explanation:
Considering the first question
From the question we are told that
The spring constant is 
The potential energy is 
Generally the potential energy stored in spring is mathematically represented as 
=>
=>
=>
Considering the second question
From the question we are told that
The mass of the dart is m = 0.050 kg
Generally from the law of energy conservation

=> 
=> 
Considering the third question
The height at which the dart was fired horizontally is 
Generally from the law of energy conservation

Here KE is kinetic energy of the dart which is mathematical represented as

=> 
=> 
=> 
Considering the fourth question
Generally the total time of flight of the dart is mathematically represented as

=> 
=> 
Generally the horizontal distance from the equilibrium position to the ground is mathematically represented as

=> 
=> 