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lions [1.4K]
3 years ago
8

If you swim with the current in a river, your speed is increased by the speed of the water; if you swim against the current, you

r speed is decreased by the water's speed. The current in a river flows at 0.52 m/s. In still water you can swim at 1.73 m/s. If you swim downstream a certain distance, then back again upstream, how much longer, in percent, does it take compared to the same trip in still water?
Physics
1 answer:
Blababa [14]3 years ago
5 0

Answer:

11.23%

Explanation:

Lets take

Speed of man in still water =u= 1.73 m/s

Speed of flow of water = v=0.52 m/s

When swims in downward direction then speed of man = u + v

When swims in upward direction then speed of man = u - v

Lets time taken by man when he swims in downward direction is t_1 and when he swims in downward direction is t_2

Lets distance is d and it will be remain constant in both the case

d=(u+v)t_1

d=(u-v)t_2

(1.73+0.52)t_1=(1.73-0.52)t_2

t_2=1.85t_1

Time taken in still water

2 d= t x 1.73

t=1.15 x d sec

t_1=0.44d\ sec

t_2=0.82d\ sec

total time in current = 0.82 +0.44 d=1.26 d sec

So the percentage time

percentage\ time =\dfrac{1.28-1.15}{1.15}

 Percentage time =11.32%

So it will take 11.32% more time as compare to still current.

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A rope of total mass m hnd length L is suspended vertically with an object of mass M suspended from the lower end. Find an expre
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Answer:

Part a)

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Part b)

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Explanation:

Part a)

Tension in the rope at a distance x from the lower end is given as

T = \frac{m}{L}xg + Mg

so the speed of the wave at that position is given as

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here we know that

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now we have

v = \sqrt{\frac{ \frac{m}{L}xg + Mg}{m/L}

v = \sqrt{xg + \frac{MLg}{m}}

Part b)

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t = \int \frac{dx}{\sqrt{xg + \frac{MLg}{m}}}

t = \frac{1}{g}(2\sqrt{xg + \frac{MLg}{m}})

t = \frac{2}{9.8}(\sqrt{(39.2\times 9.8) + \frac{8(39.2)(9.8)}{1}})

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3 years ago
Suppose the electric field in problems 2 was caused by a point charge. The test charge is moved to a distance twice as far from
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Answer:

it is reduced four times.

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a bird flies 25.0 m in the direction 55° east of south to its nest. the bird then flies 75.0 m in the direction 55° west of nort
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The northward components of the resultant displacement is 40.96 m and the westward components of the resultant displacement of the bird from its nest is 28.68 m.

<h3>Displacement of the bird</h3>

The displacement of the bird is the change in the position of the bird.

<h3>Vertical component of the bird's displacement </h3>

Vy₁ = -25 m  x   sin(55)

Vy₁ = -20.48 m

Vy₂ = 75 m   x    sin(55)

Vy₂ = 61.44 m

Total vertical displacement = 61.44 m - 20.48 m = 40.96 m

<h3>Horizontal component of the bird's displacement </h3>

Vx₁ = -25 m  x   cos(55)

Vx₁ = -14.34 m

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Vx₂ = 43.02 m

Total horizontal displacement = 43.02 m - 14.34 m = 28.68 m

Learn more about displacement here: brainly.com/question/2109763

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