Answer:
Explanation:
Let the velocity after first collision be v₁ and v₂ of car A and B . car A will bounce back .
velocity of approach = 1.5 - 0 = 1.5
velocity of separation = v₁ + v₂
coefficient of restitution = velocity of separation / velocity of approach
.8 = v₁ + v₂ / 1.5
v₁ + v₂ = 1.2
applying law of conservation of momentum
m x 1.5 + 0 = mv₂ - mv₁
1.5 = v₂ - v₁
adding two equation
2 v ₂= 2.7
v₂ = 1.35 m /s
v₁ = - .15 m / s
During second collision , B will collide with stationary A . Same process will apply in this case also. Let velocity of B and A after collision be v₃ and v₄.
For second collision ,
coefficient of restitution = velocity of separation / velocity of approach
.5 = v₃ + v₄ / 1.35
v₃ + v₄ = .675
applying law of conservation of momentum
m x 1.35 + 0 = mv₄ - mv₃
1.35 = v₄ - v₃
adding two equation
2 v ₄= 2.025
v₄ = 1.0125 m /s
v₃ = - 0 .3375 m / s
<h2>Question</h2>
why freezer is made in the upper part of refrigrator
<h2>✒ Answer</h2>
the cold air produced from it is denser than the warmer air in the bottom
<h3>Explaination</h3>
Freezer is normally provided at the top of the refrigerator, because density of the cold air is high compared to the hot air. In a refrigerator the air contacts with the cooling coil and gets cooling.Because of the high density the cold air gets down and the warm air/hot air moves upward and gets cooling from the cooling coil/evaporator coil. This process is repeated. If the Freezer is provided at the bottom place of the refrigerator, the cold air can't to move full area of the refrigerator. So the freezer is normally provided at the top at the refrigerator
The displacement of a moving object is the straight-line distance between the place it starts from and the place where it stops.
The displacement of anything moving along a circular track depends on how far around it goes before it stops. The greatest displacement it can possibly have is the diameter of the track ... 100m on this particular one ... because that's as far apart as two places on a circle can ever be.
The most interesting case is when the object goes around the circle exactly once. Then it stops at the same place it started from, the distance between the starting point and ending point is zero, and after all that motion, the displacement is zero.
Factors affecting friction
The intensity of friction depends on following factors: i) The area involved in friction. ii) The pressure applied on the surfaces. Force = Pressure ´ Area Frictional force will increase, if the area of contact will increase or if pressure applied on the surface increased.
Methods to reduce friction
i) Polish the contact surface. ii) Put oil or grease so that it fills in the small gaps of the flat parts. iii) Use ball bearings to reduce area of contact between rotating parts.
Lubrication
Following methods can be used to reduce friction: Oil is either thin or viscous. It depends upon SAE No. of oil. (SAE means Society of Automotive Engineers). If we use very viscous oil, it does not reach all the parts. Very thin oil will flows away easily and gets wasted. Grease is used in such cases. It is generally used around ball-bearing. Normal grease or oil is never used where there is high pressure, high temperature and high speed. Special lubricants are used in such cases. In cold season the oil becomes thick and in hot season it becomes thin. Therefore selection of lubrication also depends on the season. It is always advisable to refer operating manual of the equipment before selecting the lubricant.
