Answer:
a) When its length is 23 cm, the elastic potential energy of the spring is
0.18 J
b) When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J
Explanation:
Hi there!
a) The elastic potential energy (EPE) is calculated using the following equation:
EPE = 1/2 · k · x²
Where:
k = spring constant.
x = stretched lenght.
Let´s calculate the elastic potential energy of the spring when it is stretched 3 cm (0.03 m).
First, let´s convert the spring constant units into N/m:
4 N/cm · 100 cm/m = 400 N/m
EPE = 1/2 · 400 N/m · (0.03 m)²
EPE = 0.18 J
When its length is 23 cm, the elastic potential energy of the spring is 0.18 J
b) Now let´s calculate the elastic potential energy when the spring is stretched 0.06 m:
EPE = 1/2 · 400 N/m · (0.06 m)²
EPE = 0.72 J
When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J
The given question is incomplete. The complete question is as follows.
A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is
, where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.
Explanation:
We will calculate the work done as follows.
W = 
= 
= ![[14000x + 5000x^{2} - 8666.7x^{3}]^{0.54}_{0}](https://tex.z-dn.net/?f=%5B14000x%20%2B%205000x%5E%7B2%7D%20-%208666.7x%5E%7B3%7D%5D%5E%7B0.54%7D_%7B0%7D)
= 7560 + 1458 - 1364.69
= 7653.31 J
or, = 7.65 kJ (as 1 kJ = 1000 J)
Thus, we can conclude that the work done by the gas on the bullet as the bullet travels the length of the barrel is 7.65 kJ.
10.67m/s²
32N
Explanation:
Given parameters:
Mass of the body = 3kg
velocity of the mass = 4m/s
radius of circle = 0.75m
Unknown:
centripetal acceleration = ?
centripetal force = ?
Solution:
The centripetal force is the force that keeps a radial body in its circular motion. It is directed inward:
Centripetal acceleration = 
v is the velocity of the body
r is the radius of the circle
putting in the parameters:
Centripetal acceleration = 
Centripetal acceleration = 10.67m/s²
Centripetal force = m
m is the mass
Centripetal force = mass x centripetal acceleration
= 3 x 10.67
= 32N
learn more:
Acceleration brainly.com/question/3820012
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Kinetic Energy = 1/2xmassx(velocity)^2
Input values;
K.E=1/2x7kgx(4m/s)^2
K.E.=56J
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..