A lead bullet of mass 0.05kg is fired with a velocity of ...
A lead bullet of mass 0.05kg is fired with a velocity of 200ms−1 into a block of mass 0.95kg. Given that the lead block can move freely, the final kinetic energy after impact is?
50J
100J
150J
200J
Correct answer is A
Given
m1 = 0.05kg, u1 = 200ms−1, m2 = 0.95kg
K.E = 12mrv2
m1u1 = v(m1 + m2) [law of conversation of momentum]
v = 0.05×2000.05+95 = 10ms−1
K.E = 12(1)102 = 50J
The effect of closing the key K in the circuit shown above would be to ...
The maximum density of water occurs at a temperature of ...
Short sightedness can be corrected by what type of lens? ...
Which of the following pairs is NOT part of the electromagnetic spectrum? i. Radio waves. ii. Be...
Calculate the amount of heat generated in an external load of resistance 8Ω if an a...
Using the data on the diagram above, calculate the potential difference across the 20- Ω resis...