Yes, the above-given statement is true 
<u>Explanation:</u>
- The product of the mass x the velocity will be the same for both. Momentum is the action of a body with a particular mass through space and there is the conservation of momentum.
- Momentum is described as the mass of the object multiplied by its velocity.
- <u>Momentum (p) = Mass (M) * Velocity (v)</u>
- Therefore for two objects with many masses to have a similar momentum, then the lighter one has to be moving quicker than the heavier object.
 
 
 
        
             
        
        
        
Answer:
emf induced in the loop, at the instant when 9.0s have passed = 1.576 * 10 ⁻² V.
Direction is counter clockwise.
Explanation:
See attached pictures. 
 
        
             
        
        
        
<h2>When two object P and Q are supplied with the same quantity of heat, the temperature change in P is observed to be twice that of Q. The mass of P is half that of Q. The ratio of the specific heat capacity of P to Q</h2>
Explanation:
Specific heat capacity 
It is defined as amount of heat required to raise the temperature of a substance by one degree celsius .
It is given as :
Heat absorbed = mass of substance x specific heat capacity x rise in temperature 
or , 
Q= m x c x t 
In above question , it is given :
For Q 
mass of Q = m 
Temperature changed =T₂/2 
Heat supplied = x 
Q= mc t 
or 
X=m x C₁ X T₁ 
or, X =m x C₁ x T₂/2 
or, C₁=X x 2 /m x T₂                                 (equation 1 )
For another quantity : P 
mass of P =m/2 
Temperature= T₂ 
Heat supplied is same that is : X 
so, X= m/2 x C₂ x T₂                             
or, C₂=2X/m. T₂                                   (equation 2 )
Now taking ratio of C₂ to c₁, We have 
C₂/C₁= 2X /m.T₂  /2X  /m.T₂
so, C₂/C₁= 1/1 
so, the ratio is 1: 1 
 
        
             
        
        
        
The answer is C.
The Kinetic energy which was exerted and experience pulling the string of a bow is kept as a potential energy at the end of the arrow in contact with the string. Once release from aim at stationary position the potential energy is again transformed.