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
Which of the following is NOT an effect that can be produced by an electric current? ...
Two 50μF parallel plate capacitors are connected in series. The combined capacitor is then con...
I. Jet propelled aircraft II. Rocket propulsion III. The recoil of a gun IV. A person walking Wh...
A stroboscope can be used to make the wave appear …………. ...
Myopic defects in the human eye can be corrected through the use of a ...
If the threshold frequency for tungsten is 1.3 x 1015Hz, what is its work function? [H = 6.6 x ...