The Potential energy stored in the system is 1 J
<u>Explanation:</u>
Given-
Mass, m = 4 kg
Spring constant, k = 800 N/m
Distance, x = 5cm = 0.05m
Potential energy, U = ?
We know,
Change in potential energy is equal to the work done.
So,

By plugging in the values we get,

Therefore, Potential energy stored in the system is 1 J
Answer:
Explanation:
a ) speed of passenger = circumference / time
= 2π R / Time
= 2 x 3.14 x 50 / 60
= 5.23 m /s
b )
centrifugal force = m v² /R
= (882 /9.8 ) x 5.23² / 50
= 77.47 N
Apparent weight at the highest point
real weight - centrifugal force
= 882 - 77.47
= 804.53 N
Apparent weight at the lowest point
real weight + centrifugal force
= 882 +77.47
= 959.47 N
c ) if the passenger’s apparent weight at the highest point were zero
centrifugal force = weight
mv² /R = mg
v² = gR
= 9.8 X 50
v = 22.13 m /s
d )
apparent weight
mg - mv² / R
= 882 - (882 / 9.8 )x 22.13²/50
= 882 + 882
= 1764 N
=
The conservation of momentum states that the total momentum in a system is constant if there is no external force acting on the system. The total momentum in the gun bullet system is 0 so it must stay that way.
The momentum of the bullet is mv = 0.015*500=7.5
The momentum of the gun must be the same to keep the total momentum of the system equal to zero, so we know that p = 7.5 for the gun.
Substituting this in we get:
7.5=3.1x
x=7.5/3.1
x=2.42
So the speed of the gun is 2.4m/s.
Answer:
The acceleration of a point on the wheel is 11.43 m/s² acting radially inward.
Explanation:
The centripetal acceleration acts on a body when it is performing a circular motion.
Here, a point on the bicycle is performing circular motion as the rotation of the wheel produces a circular motion.
The centripetal acceleration of a point moving with a velocity
and at a distance of
from the axis of rotation is given as:

Here, 
∴ 
Therefore, the acceleration of a point on the wheel is 11.43 m/s² acting radially inward.
Answer:
I tend to slap thicc trees a#ss
Explanation: