By using the equation speed = distance/time we can solve for distance. The speed is 4 m/s and the time is 12 seconds. We need to rearrange the equation to Speed * Time = distance. 4(12) = 48; 48 = distance. The cliff is 48 meters high.
Answer:
(a)2.7 m/s
(b) 5.52 m/s
Explanation:
The total of the system would be conserved as no external force is acting on it.
Initial momentum = final momentum
⇒(4.30 g × 943 m/s) + (730 g × 0) = (4.30 g × 484 m/s) + (730 g × v)
⇒ 730 ×v = (4054.9 - 2081.2) =1973.7
⇒v=2.7 m/s
Thus, the resulting speed of the block is 2.7 m/s.
(b) since, the momentum is conserved, the speed of the bullet-block center of mass would be constant.

Thus, the speed of the bullet-block center of mass is 5.52 m/s.
Answer:
0.5 Amperes
Explanation:
Information that we have:
Power: 
Voltage: 
and we use the equation that relates this two quantities to the Current (I):

where I is the curent, P is the Power and V is the Voltage.
we substitute the values given for the Power and the Voltage to find the current:

the Current running into the bulb is 0.5 Amperes
Given:
Mass of the rail road car, m = 2 kg
velocity of the three cars coupled system, v' = 1.20 m/s
velocity of first car,
= 3 m/s
Solution:
a) Momentum of a body of mass 'm' and velocity 'v' is given by:
p = mv
Now for the coupled system according to law of conservation of momentum, total momentum of a system before and after collision remain conserved:
(1)
where,
= velocity of the first car
= velocity of the 2 coupled cars after collision
Now, from eqn (1)


v' = 1.80 m/s
Therefore, the velocity of the combined car system after collision is 1.80 m/s
Answer: 
Explanation:
Given
Cross-sectional area 
Dielectric constant 
Dielectric strength 
Distance between capacitors 
Maximum charge that can be stored before dielectric breakdown is given by
![\Rightarrow Q=CV\\\\\Rightarrow Q=\dfrac{k\epsilon_oA}{d}\cdot (Ed)\quad\quad [V=E\cdot d]\\\\\Rightarrow Q=k\epsilon_oAE\\\\\Rightarrow Q=4\times 8.85\times 10^{-12}\times 0.4\times 10^{-4}\times 2\times 10^8\\\\\Rightarrow Q=28.32\times 10^{-8}\\\\\Rightarrow Q=283.2\times 10^{-9}\ nC](https://tex.z-dn.net/?f=%5CRightarrow%20Q%3DCV%5C%5C%5C%5C%5CRightarrow%20Q%3D%5Cdfrac%7Bk%5Cepsilon_oA%7D%7Bd%7D%5Ccdot%20%28Ed%29%5Cquad%5Cquad%20%5BV%3DE%5Ccdot%20d%5D%5C%5C%5C%5C%5CRightarrow%20Q%3Dk%5Cepsilon_oAE%5C%5C%5C%5C%5CRightarrow%20Q%3D4%5Ctimes%208.85%5Ctimes%2010%5E%7B-12%7D%5Ctimes%200.4%5Ctimes%2010%5E%7B-4%7D%5Ctimes%202%5Ctimes%2010%5E8%5C%5C%5C%5C%5CRightarrow%20Q%3D28.32%5Ctimes%2010%5E%7B-8%7D%5C%5C%5C%5C%5CRightarrow%20Q%3D283.2%5Ctimes%2010%5E%7B-9%7D%5C%20nC)