269915
A boat moves perpendicular to the bank with a velocity of\(7.2 \mathrm{~km} / \mathrm{h}\). The current carries it \(150 \mathrm{~m}\) downstream, find the velocity of the current(The width of the river is \(0.5 \mathrm{~km}\) ).
1 \(0.4 \mathrm{~ms}^{-1}\)
2 \(1.2 \mathrm{~ms}^{-1}\)
3 \(0.5 \mathrm{~ms}^{-1}\)
4 \(0.6 \mathrm{~ms}^{-1}\)
Explanation:
\(x=V_{W} \cdot \frac{d}{V_{B}} \quad\)
Motion in Plane
269916
A swimmer is capable of swimming\(1.65 \mathrm{~ms}^{-1}\) in still water. If she swims directly across a \(180 \mathrm{~m}\) wide river whose current is \(0.85 \mathrm{~ms}^{-1}\), how far downstream(from a point opposite her starting point) will she reach
1 \(92.7 \mathrm{~m}\)
2 \(40 \mathrm{~m}\)
3 \(48 \mathrm{~m}\)
4 \(20 \mathrm{~m}\)
Explanation:
\(x=V_{W} \cdot \frac{d}{V_{B}} \quad\)
Motion in Plane
269917
A person swims at\(135^{\circ}\) to current of river, to meet target on reaching opposite point. The ratio of person's velocity to river water velocity is
1 \(\sqrt{3}: 1\)
2 \(\sqrt{2}: 1\)
3 \(1: \sqrt{2}\)
4 \(1: \sqrt{3}\)
Explanation:
\(\cdot \sin \theta=\frac{V_{w}}{V_{b}}\)
Motion in Plane
269954
A boat takes 2 hours to travel \(8 \mathrm{~km}\) and back in still water lake. With water velocity of 4 kmph, the time taken for going upstream of \(\mathbf{8 k m}\) and coming back is
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Motion in Plane
269915
A boat moves perpendicular to the bank with a velocity of\(7.2 \mathrm{~km} / \mathrm{h}\). The current carries it \(150 \mathrm{~m}\) downstream, find the velocity of the current(The width of the river is \(0.5 \mathrm{~km}\) ).
1 \(0.4 \mathrm{~ms}^{-1}\)
2 \(1.2 \mathrm{~ms}^{-1}\)
3 \(0.5 \mathrm{~ms}^{-1}\)
4 \(0.6 \mathrm{~ms}^{-1}\)
Explanation:
\(x=V_{W} \cdot \frac{d}{V_{B}} \quad\)
Motion in Plane
269916
A swimmer is capable of swimming\(1.65 \mathrm{~ms}^{-1}\) in still water. If she swims directly across a \(180 \mathrm{~m}\) wide river whose current is \(0.85 \mathrm{~ms}^{-1}\), how far downstream(from a point opposite her starting point) will she reach
1 \(92.7 \mathrm{~m}\)
2 \(40 \mathrm{~m}\)
3 \(48 \mathrm{~m}\)
4 \(20 \mathrm{~m}\)
Explanation:
\(x=V_{W} \cdot \frac{d}{V_{B}} \quad\)
Motion in Plane
269917
A person swims at\(135^{\circ}\) to current of river, to meet target on reaching opposite point. The ratio of person's velocity to river water velocity is
1 \(\sqrt{3}: 1\)
2 \(\sqrt{2}: 1\)
3 \(1: \sqrt{2}\)
4 \(1: \sqrt{3}\)
Explanation:
\(\cdot \sin \theta=\frac{V_{w}}{V_{b}}\)
Motion in Plane
269954
A boat takes 2 hours to travel \(8 \mathrm{~km}\) and back in still water lake. With water velocity of 4 kmph, the time taken for going upstream of \(\mathbf{8 k m}\) and coming back is
269915
A boat moves perpendicular to the bank with a velocity of\(7.2 \mathrm{~km} / \mathrm{h}\). The current carries it \(150 \mathrm{~m}\) downstream, find the velocity of the current(The width of the river is \(0.5 \mathrm{~km}\) ).
1 \(0.4 \mathrm{~ms}^{-1}\)
2 \(1.2 \mathrm{~ms}^{-1}\)
3 \(0.5 \mathrm{~ms}^{-1}\)
4 \(0.6 \mathrm{~ms}^{-1}\)
Explanation:
\(x=V_{W} \cdot \frac{d}{V_{B}} \quad\)
Motion in Plane
269916
A swimmer is capable of swimming\(1.65 \mathrm{~ms}^{-1}\) in still water. If she swims directly across a \(180 \mathrm{~m}\) wide river whose current is \(0.85 \mathrm{~ms}^{-1}\), how far downstream(from a point opposite her starting point) will she reach
1 \(92.7 \mathrm{~m}\)
2 \(40 \mathrm{~m}\)
3 \(48 \mathrm{~m}\)
4 \(20 \mathrm{~m}\)
Explanation:
\(x=V_{W} \cdot \frac{d}{V_{B}} \quad\)
Motion in Plane
269917
A person swims at\(135^{\circ}\) to current of river, to meet target on reaching opposite point. The ratio of person's velocity to river water velocity is
1 \(\sqrt{3}: 1\)
2 \(\sqrt{2}: 1\)
3 \(1: \sqrt{2}\)
4 \(1: \sqrt{3}\)
Explanation:
\(\cdot \sin \theta=\frac{V_{w}}{V_{b}}\)
Motion in Plane
269954
A boat takes 2 hours to travel \(8 \mathrm{~km}\) and back in still water lake. With water velocity of 4 kmph, the time taken for going upstream of \(\mathbf{8 k m}\) and coming back is
269915
A boat moves perpendicular to the bank with a velocity of\(7.2 \mathrm{~km} / \mathrm{h}\). The current carries it \(150 \mathrm{~m}\) downstream, find the velocity of the current(The width of the river is \(0.5 \mathrm{~km}\) ).
1 \(0.4 \mathrm{~ms}^{-1}\)
2 \(1.2 \mathrm{~ms}^{-1}\)
3 \(0.5 \mathrm{~ms}^{-1}\)
4 \(0.6 \mathrm{~ms}^{-1}\)
Explanation:
\(x=V_{W} \cdot \frac{d}{V_{B}} \quad\)
Motion in Plane
269916
A swimmer is capable of swimming\(1.65 \mathrm{~ms}^{-1}\) in still water. If she swims directly across a \(180 \mathrm{~m}\) wide river whose current is \(0.85 \mathrm{~ms}^{-1}\), how far downstream(from a point opposite her starting point) will she reach
1 \(92.7 \mathrm{~m}\)
2 \(40 \mathrm{~m}\)
3 \(48 \mathrm{~m}\)
4 \(20 \mathrm{~m}\)
Explanation:
\(x=V_{W} \cdot \frac{d}{V_{B}} \quad\)
Motion in Plane
269917
A person swims at\(135^{\circ}\) to current of river, to meet target on reaching opposite point. The ratio of person's velocity to river water velocity is
1 \(\sqrt{3}: 1\)
2 \(\sqrt{2}: 1\)
3 \(1: \sqrt{2}\)
4 \(1: \sqrt{3}\)
Explanation:
\(\cdot \sin \theta=\frac{V_{w}}{V_{b}}\)
Motion in Plane
269954
A boat takes 2 hours to travel \(8 \mathrm{~km}\) and back in still water lake. With water velocity of 4 kmph, the time taken for going upstream of \(\mathbf{8 k m}\) and coming back is