270166
Brakes are applied to a car moving with disengaged engine, bringing it to a halt after 2s. Its velocity at the moment when the breaks are applied if the coefficient of friction between the road and the tyres is 0.4 is
1 \(3.92 \mathrm{~ms}^{-1}\)
2 \(7.84 \mathrm{~ms}^{-1}\)
3 \(11.2 \mathrm{~ms}^{-1}\)
4 \(19.6 \mathrm{~ms}^{-1}\)
Explanation:
\(v=u+a t, a=\mu g\)
Laws of Motion
270167
A book of weight \(20 \mathrm{~N}\) is pressed between two hands and each hand exerts a force of 40N. If the book just starts to slide down. Coefficient of friction is
1 0.25
2 0.2
3 0.5
4 0.1
Explanation:
\(f_{s}=\mu_{s} N, \mu=\frac{f_{s}}{N}=\frac{m g}{2 F}\)
Laws of Motion
270168
A car running with a velocity \(72 \mathrm{kmph}\) on a level road, is stopped after travelling a distance of \(30 \mathrm{~m}\) after disengaging its engine \(\left(g=10 \mathrm{~ms}^{-2}\right)\). The coefficient of friction between the road and the tyres is
1 0.33
2 4.5
3 0.67
4 0.8
Explanation:
\(v^{2}-u^{2}=2 a s, a=\mu g\)
Laws of Motion
270169
In the above problem car got a stopping distance of \(80 \mathrm{~m}\) on cement road then \(\mu_{\mathrm{k}}\) is \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{sec}^{2}\right)\)
1 0.2
2 0.25
3 0.3
4 0.35
Explanation:
\(v^{2}-u^{2}=2 a s, a=\mu g\)
Laws of Motion
270170
A \(10 \mathrm{~kg}\) mass is resting on a horizontal surface and horizontal force of \(80 \mathrm{~N}\) is applied. If \(\mu=0.2\), the ratio of acceleration without and with friction is \(\left(g=10 \mathrm{~ms}^{-2}\right)\)
1 \(3 / 4\)
2 \(4 / 3\)
3 \(1 / 2\)
4 2
Explanation:
\(F_{n e t}=m a, a_{1}=\frac{F}{m}, a_{2}=\frac{F-\mu_{k} m g}{m}\)
270166
Brakes are applied to a car moving with disengaged engine, bringing it to a halt after 2s. Its velocity at the moment when the breaks are applied if the coefficient of friction between the road and the tyres is 0.4 is
1 \(3.92 \mathrm{~ms}^{-1}\)
2 \(7.84 \mathrm{~ms}^{-1}\)
3 \(11.2 \mathrm{~ms}^{-1}\)
4 \(19.6 \mathrm{~ms}^{-1}\)
Explanation:
\(v=u+a t, a=\mu g\)
Laws of Motion
270167
A book of weight \(20 \mathrm{~N}\) is pressed between two hands and each hand exerts a force of 40N. If the book just starts to slide down. Coefficient of friction is
1 0.25
2 0.2
3 0.5
4 0.1
Explanation:
\(f_{s}=\mu_{s} N, \mu=\frac{f_{s}}{N}=\frac{m g}{2 F}\)
Laws of Motion
270168
A car running with a velocity \(72 \mathrm{kmph}\) on a level road, is stopped after travelling a distance of \(30 \mathrm{~m}\) after disengaging its engine \(\left(g=10 \mathrm{~ms}^{-2}\right)\). The coefficient of friction between the road and the tyres is
1 0.33
2 4.5
3 0.67
4 0.8
Explanation:
\(v^{2}-u^{2}=2 a s, a=\mu g\)
Laws of Motion
270169
In the above problem car got a stopping distance of \(80 \mathrm{~m}\) on cement road then \(\mu_{\mathrm{k}}\) is \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{sec}^{2}\right)\)
1 0.2
2 0.25
3 0.3
4 0.35
Explanation:
\(v^{2}-u^{2}=2 a s, a=\mu g\)
Laws of Motion
270170
A \(10 \mathrm{~kg}\) mass is resting on a horizontal surface and horizontal force of \(80 \mathrm{~N}\) is applied. If \(\mu=0.2\), the ratio of acceleration without and with friction is \(\left(g=10 \mathrm{~ms}^{-2}\right)\)
1 \(3 / 4\)
2 \(4 / 3\)
3 \(1 / 2\)
4 2
Explanation:
\(F_{n e t}=m a, a_{1}=\frac{F}{m}, a_{2}=\frac{F-\mu_{k} m g}{m}\)
270166
Brakes are applied to a car moving with disengaged engine, bringing it to a halt after 2s. Its velocity at the moment when the breaks are applied if the coefficient of friction between the road and the tyres is 0.4 is
1 \(3.92 \mathrm{~ms}^{-1}\)
2 \(7.84 \mathrm{~ms}^{-1}\)
3 \(11.2 \mathrm{~ms}^{-1}\)
4 \(19.6 \mathrm{~ms}^{-1}\)
Explanation:
\(v=u+a t, a=\mu g\)
Laws of Motion
270167
A book of weight \(20 \mathrm{~N}\) is pressed between two hands and each hand exerts a force of 40N. If the book just starts to slide down. Coefficient of friction is
1 0.25
2 0.2
3 0.5
4 0.1
Explanation:
\(f_{s}=\mu_{s} N, \mu=\frac{f_{s}}{N}=\frac{m g}{2 F}\)
Laws of Motion
270168
A car running with a velocity \(72 \mathrm{kmph}\) on a level road, is stopped after travelling a distance of \(30 \mathrm{~m}\) after disengaging its engine \(\left(g=10 \mathrm{~ms}^{-2}\right)\). The coefficient of friction between the road and the tyres is
1 0.33
2 4.5
3 0.67
4 0.8
Explanation:
\(v^{2}-u^{2}=2 a s, a=\mu g\)
Laws of Motion
270169
In the above problem car got a stopping distance of \(80 \mathrm{~m}\) on cement road then \(\mu_{\mathrm{k}}\) is \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{sec}^{2}\right)\)
1 0.2
2 0.25
3 0.3
4 0.35
Explanation:
\(v^{2}-u^{2}=2 a s, a=\mu g\)
Laws of Motion
270170
A \(10 \mathrm{~kg}\) mass is resting on a horizontal surface and horizontal force of \(80 \mathrm{~N}\) is applied. If \(\mu=0.2\), the ratio of acceleration without and with friction is \(\left(g=10 \mathrm{~ms}^{-2}\right)\)
1 \(3 / 4\)
2 \(4 / 3\)
3 \(1 / 2\)
4 2
Explanation:
\(F_{n e t}=m a, a_{1}=\frac{F}{m}, a_{2}=\frac{F-\mu_{k} m g}{m}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Laws of Motion
270166
Brakes are applied to a car moving with disengaged engine, bringing it to a halt after 2s. Its velocity at the moment when the breaks are applied if the coefficient of friction between the road and the tyres is 0.4 is
1 \(3.92 \mathrm{~ms}^{-1}\)
2 \(7.84 \mathrm{~ms}^{-1}\)
3 \(11.2 \mathrm{~ms}^{-1}\)
4 \(19.6 \mathrm{~ms}^{-1}\)
Explanation:
\(v=u+a t, a=\mu g\)
Laws of Motion
270167
A book of weight \(20 \mathrm{~N}\) is pressed between two hands and each hand exerts a force of 40N. If the book just starts to slide down. Coefficient of friction is
1 0.25
2 0.2
3 0.5
4 0.1
Explanation:
\(f_{s}=\mu_{s} N, \mu=\frac{f_{s}}{N}=\frac{m g}{2 F}\)
Laws of Motion
270168
A car running with a velocity \(72 \mathrm{kmph}\) on a level road, is stopped after travelling a distance of \(30 \mathrm{~m}\) after disengaging its engine \(\left(g=10 \mathrm{~ms}^{-2}\right)\). The coefficient of friction between the road and the tyres is
1 0.33
2 4.5
3 0.67
4 0.8
Explanation:
\(v^{2}-u^{2}=2 a s, a=\mu g\)
Laws of Motion
270169
In the above problem car got a stopping distance of \(80 \mathrm{~m}\) on cement road then \(\mu_{\mathrm{k}}\) is \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{sec}^{2}\right)\)
1 0.2
2 0.25
3 0.3
4 0.35
Explanation:
\(v^{2}-u^{2}=2 a s, a=\mu g\)
Laws of Motion
270170
A \(10 \mathrm{~kg}\) mass is resting on a horizontal surface and horizontal force of \(80 \mathrm{~N}\) is applied. If \(\mu=0.2\), the ratio of acceleration without and with friction is \(\left(g=10 \mathrm{~ms}^{-2}\right)\)
1 \(3 / 4\)
2 \(4 / 3\)
3 \(1 / 2\)
4 2
Explanation:
\(F_{n e t}=m a, a_{1}=\frac{F}{m}, a_{2}=\frac{F-\mu_{k} m g}{m}\)
270166
Brakes are applied to a car moving with disengaged engine, bringing it to a halt after 2s. Its velocity at the moment when the breaks are applied if the coefficient of friction between the road and the tyres is 0.4 is
1 \(3.92 \mathrm{~ms}^{-1}\)
2 \(7.84 \mathrm{~ms}^{-1}\)
3 \(11.2 \mathrm{~ms}^{-1}\)
4 \(19.6 \mathrm{~ms}^{-1}\)
Explanation:
\(v=u+a t, a=\mu g\)
Laws of Motion
270167
A book of weight \(20 \mathrm{~N}\) is pressed between two hands and each hand exerts a force of 40N. If the book just starts to slide down. Coefficient of friction is
1 0.25
2 0.2
3 0.5
4 0.1
Explanation:
\(f_{s}=\mu_{s} N, \mu=\frac{f_{s}}{N}=\frac{m g}{2 F}\)
Laws of Motion
270168
A car running with a velocity \(72 \mathrm{kmph}\) on a level road, is stopped after travelling a distance of \(30 \mathrm{~m}\) after disengaging its engine \(\left(g=10 \mathrm{~ms}^{-2}\right)\). The coefficient of friction between the road and the tyres is
1 0.33
2 4.5
3 0.67
4 0.8
Explanation:
\(v^{2}-u^{2}=2 a s, a=\mu g\)
Laws of Motion
270169
In the above problem car got a stopping distance of \(80 \mathrm{~m}\) on cement road then \(\mu_{\mathrm{k}}\) is \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{sec}^{2}\right)\)
1 0.2
2 0.25
3 0.3
4 0.35
Explanation:
\(v^{2}-u^{2}=2 a s, a=\mu g\)
Laws of Motion
270170
A \(10 \mathrm{~kg}\) mass is resting on a horizontal surface and horizontal force of \(80 \mathrm{~N}\) is applied. If \(\mu=0.2\), the ratio of acceleration without and with friction is \(\left(g=10 \mathrm{~ms}^{-2}\right)\)
1 \(3 / 4\)
2 \(4 / 3\)
3 \(1 / 2\)
4 2
Explanation:
\(F_{n e t}=m a, a_{1}=\frac{F}{m}, a_{2}=\frac{F-\mu_{k} m g}{m}\)