Heat
required in a system can be calculated by multiplying the given mass to the
specific heat capacity of the substance and the temperature difference. It is
expressed as follows:<span>
Heat = mC(T2-T1)
Heat = 1 kg (4.18 kJ / kg C)( 1 C)
<span>Heat = 4.18 kJ energy needed</span></span>
The other person who answered this is wrong btw
A. the medium through which the light travels changes.
Explanation:
Light waves will continue to travel in a straight line in all directions from their source unless the medium through which the light travels changes.
A change in medium causes light to exhibit different properties. Also, when light hits an obstacle, they can be diffracted.
- The way light travels on crossing a boundary differs.
- At the boundary between two medium, light can either be reflected back or refracted when they cross the medium
- This will cause the light rays to bend towards or away from the normal depending on the properties of the medium.
Learn more:
Refraction brainly.com/question/12370040
#learnwithBrainly
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
Answer:
Part a)
Part b)
Part c)
Explanation:
Part a)
As we know that there is no external torque on the system of two twins
so here we will use
Part b)
Since angular momentum is conserved here as there is no external torque
so we will have
Part c)
Work done by both of them = change in kinetic energy
so we have
