270222
In the above problem its displacement after 3 seconds is
1 \(78.4 \mathrm{~m}\)
2 \(44.15 \mathrm{~m}\)
3 \(10.57 \mathrm{~m}\)
4 Zero
Explanation:
\(S=u t+\frac{1}{2} a t^{2}, a=g\left(\sin \theta-\mu_{k} \cos \theta\right)\)
Laws of Motion
270223
A block sliding down on a rough \(45^{\circ}\) inclined plane has half the velocity it would have been, the inclined plane is smooth.The coefficient of sliding friction between the block and the inclined plane is
1 \(\frac{1}{4}\)
2 \(\frac{3}{4}\)
3 \(\frac{1}{2 \sqrt{2}}\)
4 \(\frac{1}{\sqrt{2}}\)
Explanation:
\(v^{2}-u^{2}=2 a s\)
\(V_{R}=\sqrt{2 g l\left(\sin \theta-\mu_{k} \cos \theta\right)}, V_{S}=\sqrt{2 g l \sin \theta}\)
Laws of Motion
270282
The horizontal acceleration that should be given to a smooth inclined plane of angle sin\(^{-1}\) A to keep an object stationary on the plane, relative to the inclined plane is
1 \(\frac{g}{\sqrt{l^{2}-1}}\)
2 \(g \sqrt{l^{2}-1}\)
3 \(\frac{\sqrt{l^{2}-1}}{g} 4\)
4 \(\frac{g}{\sqrt{l^{2}+1}}\)
Explanation:
Resolve\(m g\) and pseudo force into components At equilibrium \(m g \sin \theta=m a \cos \theta\)
270222
In the above problem its displacement after 3 seconds is
1 \(78.4 \mathrm{~m}\)
2 \(44.15 \mathrm{~m}\)
3 \(10.57 \mathrm{~m}\)
4 Zero
Explanation:
\(S=u t+\frac{1}{2} a t^{2}, a=g\left(\sin \theta-\mu_{k} \cos \theta\right)\)
Laws of Motion
270223
A block sliding down on a rough \(45^{\circ}\) inclined plane has half the velocity it would have been, the inclined plane is smooth.The coefficient of sliding friction between the block and the inclined plane is
1 \(\frac{1}{4}\)
2 \(\frac{3}{4}\)
3 \(\frac{1}{2 \sqrt{2}}\)
4 \(\frac{1}{\sqrt{2}}\)
Explanation:
\(v^{2}-u^{2}=2 a s\)
\(V_{R}=\sqrt{2 g l\left(\sin \theta-\mu_{k} \cos \theta\right)}, V_{S}=\sqrt{2 g l \sin \theta}\)
Laws of Motion
270282
The horizontal acceleration that should be given to a smooth inclined plane of angle sin\(^{-1}\) A to keep an object stationary on the plane, relative to the inclined plane is
1 \(\frac{g}{\sqrt{l^{2}-1}}\)
2 \(g \sqrt{l^{2}-1}\)
3 \(\frac{\sqrt{l^{2}-1}}{g} 4\)
4 \(\frac{g}{\sqrt{l^{2}+1}}\)
Explanation:
Resolve\(m g\) and pseudo force into components At equilibrium \(m g \sin \theta=m a \cos \theta\)
270222
In the above problem its displacement after 3 seconds is
1 \(78.4 \mathrm{~m}\)
2 \(44.15 \mathrm{~m}\)
3 \(10.57 \mathrm{~m}\)
4 Zero
Explanation:
\(S=u t+\frac{1}{2} a t^{2}, a=g\left(\sin \theta-\mu_{k} \cos \theta\right)\)
Laws of Motion
270223
A block sliding down on a rough \(45^{\circ}\) inclined plane has half the velocity it would have been, the inclined plane is smooth.The coefficient of sliding friction between the block and the inclined plane is
1 \(\frac{1}{4}\)
2 \(\frac{3}{4}\)
3 \(\frac{1}{2 \sqrt{2}}\)
4 \(\frac{1}{\sqrt{2}}\)
Explanation:
\(v^{2}-u^{2}=2 a s\)
\(V_{R}=\sqrt{2 g l\left(\sin \theta-\mu_{k} \cos \theta\right)}, V_{S}=\sqrt{2 g l \sin \theta}\)
Laws of Motion
270282
The horizontal acceleration that should be given to a smooth inclined plane of angle sin\(^{-1}\) A to keep an object stationary on the plane, relative to the inclined plane is
1 \(\frac{g}{\sqrt{l^{2}-1}}\)
2 \(g \sqrt{l^{2}-1}\)
3 \(\frac{\sqrt{l^{2}-1}}{g} 4\)
4 \(\frac{g}{\sqrt{l^{2}+1}}\)
Explanation:
Resolve\(m g\) and pseudo force into components At equilibrium \(m g \sin \theta=m a \cos \theta\)
270222
In the above problem its displacement after 3 seconds is
1 \(78.4 \mathrm{~m}\)
2 \(44.15 \mathrm{~m}\)
3 \(10.57 \mathrm{~m}\)
4 Zero
Explanation:
\(S=u t+\frac{1}{2} a t^{2}, a=g\left(\sin \theta-\mu_{k} \cos \theta\right)\)
Laws of Motion
270223
A block sliding down on a rough \(45^{\circ}\) inclined plane has half the velocity it would have been, the inclined plane is smooth.The coefficient of sliding friction between the block and the inclined plane is
1 \(\frac{1}{4}\)
2 \(\frac{3}{4}\)
3 \(\frac{1}{2 \sqrt{2}}\)
4 \(\frac{1}{\sqrt{2}}\)
Explanation:
\(v^{2}-u^{2}=2 a s\)
\(V_{R}=\sqrt{2 g l\left(\sin \theta-\mu_{k} \cos \theta\right)}, V_{S}=\sqrt{2 g l \sin \theta}\)
Laws of Motion
270282
The horizontal acceleration that should be given to a smooth inclined plane of angle sin\(^{-1}\) A to keep an object stationary on the plane, relative to the inclined plane is
1 \(\frac{g}{\sqrt{l^{2}-1}}\)
2 \(g \sqrt{l^{2}-1}\)
3 \(\frac{\sqrt{l^{2}-1}}{g} 4\)
4 \(\frac{g}{\sqrt{l^{2}+1}}\)
Explanation:
Resolve\(m g\) and pseudo force into components At equilibrium \(m g \sin \theta=m a \cos \theta\)
