Answer: (d)
Explanation:
Given
Mass of the first ram 
The velocity of this ram is 
Mass of the second ram 
The velocity of this ram 
They combined after the collision
Conserving the momentum
![\Rightarrow m_1v_1+m_2v_2=(m_1+m_2)v\\\Rightarrow 49\times (-7)+52\times (9)=(52+49)v\\\Rightarrow v=\dfrac{125}{101}\ m/s \quad[\text{east}]](https://tex.z-dn.net/?f=%5CRightarrow%20m_1v_1%2Bm_2v_2%3D%28m_1%2Bm_2%29v%5C%5C%5CRightarrow%2049%5Ctimes%20%28-7%29%2B52%5Ctimes%20%289%29%3D%2852%2B49%29v%5C%5C%5CRightarrow%20v%3D%5Cdfrac%7B125%7D%7B101%7D%5C%20m%2Fs%20%5Cquad%5B%5Ctext%7Beast%7D%5D)
Momentum after the collision will be
Therefore, option (d) is correct
Im pretty sure it’s A eye
Answer:
Centre of mass is the point at which the distribution of mass is equal in all directions, and does not depend on gravitational field. Centre of gravity is the point at which the distribution of weight is equal in all directions, and does depend on gravitational field.
Explanation:
google
In one quadrant there are 90 degrees.
Answer:
The average induced emf in the coil is 0.0286 V
Explanation:
Given;
diameter of the wire, d = 11.2 cm = 0.112 m
initial magnetic field, B₁ = 0.53 T
final magnetic field, B₂ = 0.24 T
time of change in magnetic field, t = 0.1 s
The induced emf in the coil is calculated as;
E = A(dB)/dt
where;
A is area of the coil = πr²
r is the radius of the wire coil = 0.112m / 2 = 0.056 m
A = π(0.056)²
A = 0.00985 m²
E = -0.00985(B₂-B₁)/t
E = 0.00985(B₁-B₂)/t
E = 0.00985(0.53 - 0.24)/0.1
E = 0.00985 (0.29)/ 0.1
E = 0.0286 V
Therefore, the average induced emf in the coil is 0.0286 V
