When the breaks in a car are applied, the frictional force on the tyres is?
a disadvantage because it is in the direction of the motion of the car
a disadvantage because it is in the opposite direction of the motion of the car
an advantage because it is in the direction of the motion of the car
an advantage because it is in the opposite direction of the motion of the car
Correct answer is D
Frictional force opposes motion
A car of mass 800kg attains a speed of 25m/s in 20secs. The power developed in the engine is?
1.25 x 10\(^{4}\)W
2.50x 10\(^{4}\)W
1.25 x 10\(^{6}\)W
2.50 x 10\(^{6}\)W
Correct answer is B
P = \(\frac{mv^{2}}{t} = \frac{800 \times 25 \times 25}{20}\)
2.5 x 10\(^{4}\)W
5J
10J
15J
20J
Correct answer is C
m = 0.1kg
u = 10ms\(^{-1}\)
h = 10m
g = 10ms\(^{-2}\)
At h\(_{max}\); v = 0
Using
\(\not v\) = \(\not u\) = \(\not gt\)
\(\not 0\) = \(\not w\) - \(\not 10t\)
v\(^{2}\) = u\(^{2}\) - 2gh
0\(^{2}\) = 10\(^{2}\) - 2 x 10h
20h = 100
h = 5m
Total height = 10 + 5
= 15m
Total energy = mghr = 0.1 x 10 x1 5
= 15J
50J
100J
150J
200J
Correct answer is A
Given
m\(_{1}\) = 0.05kg, u\(_{1}\) = 200ms\(^{-1}\), m\(_{2}\) = 0.95kg
K.E = \(\frac{1}{2}\)m\(_{r}\)v\(^{2}\)
m\(_{1}\)u\(_{1}\) = v(m\(_{1}\) + m\(_{2}\)) [law of conversation of momentum]
v = \(\frac{0.05 \times 200}{0.05 + 95}\) = 10ms\(^{-1}\)
K.E = \(\frac{1}{2}\)(1)10\(^{2}\) = 50J
2N
20N
200N
400N
Correct answer is B
Given
m = 10\(^{-2}\)
w = 100rads\(^{-1}\)
F = mw\(^{2}\)r
r = 0.2m
F = ?
= 10\(^{-2}\) x (100)\(^{2}\) x 0.2
= 20N