Answer
given,
mass of glider = 0.23 Kg
spring constant = k = 4.50 N/m
spring stretched to 0.130 m
The springs potential energy =
U = 0.038 J
at x = 0,the only energy will be kinetic .
v² = 0.3304
v = 0.575 m/s
displacement of the glider
using conservation of energy
x = 0.678 m
2 someone's glasses could fall of because of the inertia
roller coaster is in motion but your glasses are opposing and resisting that motion due to inertia....
1 there is a possibility that electrons move from wheels to the rail and vice versa which also means that it is producing some kind of electrical energy
3 question can't answer
Answer:
a)15 N
b)12.6 N
Explanation:
Given that
Weight of block (wt)= 21 N
μs = 0.80 and μk = 0.60
We know that
Maximum value of static friction given as
Frs = μs m g = μs .wt
by putting the values
Frs= 0.8 x 21 = 16.8 N
Value of kinetic friction
Frk= μk m g = μk .wt
By putting the values
Frk= 0.6 x 21 = 12.6 N
a)
When T = 15 N
Static friction Frs= 16.8 N
Here the value of static friction is more than tension T .It means that block will not move and the value of friction force will be equal to the tension force.
Friction force = 15 N
b)
When T= 35 N
Here value of tension force is more than maximum value of static friction that is why block will move .We know that when body is in motion then kinetic friction will act on the body.so the value of friction force in this case will be 12.6 N
Friction force = 12.6 N
Answer:
remains the same
Explanation:
Momentum refers to the quantity of motion of a body. When any body of mass moves, it possess momentum. Numerically,
Momentum = mass x velocity
i.e. momentum is the product of the mass x velocity
Momentum of a body is always conserved.
In the context, the skateboard has certain momentum before Freddy lands on it. After Freddy lands, the momentum of skateboard remains the same, there is no change in the momentum.
This is because, here the momentum is conserved. After Freddy lands on the skateboard, the total mass on the skateboard increases and so the velocity decreases making the momentum same before the landing.
