Answer:
E) True. Ball B will go four times as high as ball A because it had four times the initial kinetic energ
Explanation:
To answer the final statements, let's pose the solution of the exercise
Energy is conserved
Initial
Em₀ = K
Em₀ = ½ m v²
Final
Emf = U = mg h
Em₀ = emf
½ m v² = mgh
h = v² / 2g
For ball A
h_A = v² / 2g
For ball B
h_B = (2v)² / 2g
h_B = 4 (v² / 2g) = 4 h_A
Let's review the claims
A) False. The neck acceleration is zero, it has the value of the acceleration of gravity
B) False. Ball B goes higher
C) False has 4 times the gravitational potential energy than ball A
D) False. It goes 4 times higher
E) True.
As per the third law of Newton, the force exerted by the boat over the student is equal in magnitude to the force that the student exerted on the boat.
So, calculate the force on the student using the second law of Newton, Force = mass * acceleration.
Force on the student = 60 kg * 2.0 m/s^2 = 120 N.
=> horizontal force exerted by the student on the boat = 120 N
Answer: option d. 120 N. toward the back of the boat.
Of course it is toward the back because that is where the student jumped from..
Answer:
Spring constant, k = 0.3 N/m
Explanation:
It is given that,
Force acting on DNA molecule, 
The molecule got stretched by 5 nm, 
Let k is the spring constant of that DNA molecule. It can be calculated using the Hooke's law. It says that the force acting on the spring is directly proportional to the distance as :



k = 0.3 N/m
So, the spring constant of the DNA molecule is 0.3 N/m. Hence, this is the required solution.
Answer:
80 J
Explanation:
Ep = mgh
Ep = (4 kg) (10 m/s²) (2 m)
Ep = 80 J
To solve this problem we will apply the first law of thermodynamics which details the relationship of energy conservation and the states that the system's energy has. Energy can be transformed but cannot be created or destroyed.
Accordingly, the rate of work done in one cycle and the heat transferred can be expressed under the function,

Substitute 1W for
and 1.5 W for 


Now calculcate the rate of specific internal energy increase,



The rate of specific internal energy increase is 1.6667W/kg