Some of these frictions depend on the Pressure, temperature of atmosphere.
Static Friction: This is the friction force when two objects in contact are not moving relative to each other. This friction is higher than kinetic friction.
Kinetic or Dynamic friction: this the friction force opposing the motion of objects, when two objects in contact are in motion relative to each other. It is less than the static friction. The two surfaces are rubbing against each other as they move.
Rolling friction: This is the friction when two objects are in contact and one object is rolling over the other - like a wheel on a road. The point of contact appears as stationary. The rolling friction is very less compared to static friction & dynamic friction.
Lubricated friction: this is the friction between two solid surfaces in contact with a layer of lubricant fluid flowing in between them. This friction is the least.
Fluid friction - viscosity : this is friction between two adjacent layers that are moving relative to each other at different speeds in a fluid. This is not high.
Internal friction: when an object is compressed and forced to deform, like in a piece of rubber, there is friction between the layers, that opposes this deformation.
Skin friction is the friction that opposes movement of a fluid across a solid surface. This is also called drag. When a coin is dropped in water, there is a friction called drag on the coin. Same is the case when a ball is thrown, a drag is experienced by the ball due to the drag of air.
Explanation:
Given that,
Force with which a child hits a ball is 350 N
Time of contact is 0.12 s
We need to find the impulse received by the ball. The impulse delivered is given by :
So, the impulse is 42 N-m..
We know that he change in momentum is also equal to the impulse delivered.
So, impulse = 42 N-m and change in momentum =42 N-m.
Answer:
Answer:
Speed of the wave in the string will be 3.2 m/sec
Explanation:
We have given frequency in the string fixed at both ends is 80 Hz
Distance between adjacent antipodes is 20 cm
We know that distance between two adjacent anti nodes is equal to half of the wavelength
So \frac{\lambda }{2}=20cm
2
λ
=20cm
\lambda =40cmλ=40cm
We have to find the speed of the wave in the string
Speed is equal to v=\lambda f=0.04\times 80=3.2m/secv=λf=0.04×80=3.2m/sec
So speed of the wave in the string will be 3.2 m/sec
