Answer:
Approximately 
. (Assuming that the drag on this ball is negligible, and that 
.)
Explanation:
Assume that the drag (air friction) on this ball is negligible. Motion of this ball during the descent:
- Horizontal: no acceleration, velocity is constant (at
is constant throughout the descent.) - Vertical: constant downward acceleration at , starting at
.
The horizontal velocity of this ball is constant during the descent. The horizontal distance that the ball has travelled during the descent is also given: 
. Combine these two quantities to find the duration of this descent:
.
In other words, the ball in this question start at a vertical velocity of 
, accelerated downwards at , and reached the ground after 
.
Apply the SUVAT equation 
to find the vertical displacement of this ball.
![\begin{aligned}& x(\text{vertical}) \\[0.5em] &= -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t\\[0.5em] &= - \frac{1}{2} \times 10\; \rm m \cdot s^{-2} \times (7.5\; \rm s)^{2} \\ & \quad \quad + 0\; \rm m \cdot s^{-1} \times 7.5\; s \\[0.5em] &= -281.25\; \rm m\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26%20x%28%5Ctext%7Bvertical%7D%29%20%5C%5C%5B0.5em%5D%20%26%3D%20-%5Cfrac%7B1%7D%7B2%7D%5C%2C%20g%20%5Ccdot%20t%5E%7B2%7D%20%2B%20v_0%5Ccdot%20t%5C%5C%5B0.5em%5D%20%26%3D%20-%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%2010%5C%3B%20%5Crm%20m%20%5Ccdot%20s%5E%7B-2%7D%20%5Ctimes%20%287.5%5C%3B%20%5Crm%20s%29%5E%7B2%7D%20%5C%5C%20%26%20%5Cquad%20%5Cquad%20%2B%200%5C%3B%20%5Crm%20m%20%5Ccdot%20s%5E%7B-1%7D%20%5Ctimes%207.5%5C%3B%20s%20%5C%5C%5B0.5em%5D%20%26%3D%20-281.25%5C%3B%20%5Crm%20m%5Cend%7Baligned%7D)
.
In other words, the ball is below where it was before the descent (hence the negative sign in front of the number.) The height of this cliff would be 
.
Answer:
The correct option is;
(b) The end A of the solenoid behaves like a north pole
Explanation:
According to Lenz's law we have that the induced emf direction in the solenoid due to the rapid introduction of the bar magnet will be such that the electric current induced will have a resultant magnet field that will oppose to the movement of the north pole of the bar magnet that resulted in the magnetic field
Therefore, the opposing magnetic pole to the north pole of a magnet is a north pole and the solenoid end A will act like the north pole.
Answer:
0.08kg
Explanation:
K.E = 1/2 mv^2
v = 970m/s
K.E = 3.9x 10^3J= 3900J
K.E = 1/2 mv^2
3900 = 1/2 m x 970x 970
3900 = 1/2 ×940900m
3900 = 470450m
m = 3900/470450 = 0.00828993516 = 0.008kg
