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Andrei [34K]
3 years ago
9

A swimmer, capable of swimming at a speed of 1.0 m/s in still water (i.e., the swimmer can swim with a speed of 1.0 m/s relative

to the water), starts to swim directly across a 3.0-km-wide river. However, the current is 0.91 m/s, and it carries the swimmer downstream.
(a) How long does it take the swimmer to cross the river?
(b) How far downstream will the swimmer be upon reaching the other side of the river?
Physics
1 answer:
Vesnalui [34]3 years ago
5 0
If the current takes him downstream we must find the resultant vector of the velocities: V res= \sqrt{1^{2}+0.91^{2}  } = \sqrt{1.8281}= 1.3520747 Then if the river is 3000 m-wide the swimmer will have to pass:
 1.3520747 · 300 = 4056.14 m                t = 4056.14 m : 1 m/s
a ) It takes 4056.15 seconds ( 1 hour 7 minutes and 36 seconds ) to cross the river.  
b ) 0.91 · 3000 = 2730 m
He will be 2730 m downstream.
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a_r=389\ rev.s^{-2}

n=58350 rev

Explanation:

Given:

time of constant deceleration, t=10\ s

A.

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<u>Using equation of motion:</u>

N_f=N_i+a_r.t

0=3890+a_r\times 10

a_r=389\ rev.s^{-2}

B.

Using eq. of motion for no. of revolutions, we have:

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Answer:

The initial velocity is 0.5114 m/s or 511.4 mm/s

Explanation:

Let the initial velocity be 'v'.

Given:

Mass of the ball (m) = 130 g = 0.130 kg   [ 1 g = 0.001 kg]

Initial height of the ball (h₁) = 1.4 mm = 0.0014 m   [ 1 mm = 0.001 m]

Final height of the ball (h₂) = 15 mm = 0.015 m

Now, from conservation of energy principle, energy can neither be created nor be destroyed but converted from one form to another.

Here, the kinetic energy of the ball is converted to gravitational potential energy of the ball after reaching the final height.

Change in kinetic energy is given as:

\Delta KE=\frac{1}{2}m(v_f^2-v_i^2)\\Where\ v_f\to Final\ velocity\\v_i\to Initial\ velocity

As it just touches the 15 mm high roof, the final velocity will be zero. So,

v_f=0\ m/s.

Now, the change in kinetic energy is equal to:

\Delta KE = \frac{1}{2}\times 0.130\times v^2\\\\\Delta KE = 0.065v^2

Change in gravitational potential energy = Final PE - Initial PE

So,

\Delta U=mg(h_f-h_i)\\\\\Delta U=0.130\times 9.8\times (0.015-0.0014)\\\\\Delta U=0.017\ J                    [ g = 9.8 m/s²]

Now, Change in KE = Change in PE

0.065v^2=0.017\\\\v=\sqrt{\frac{0.017}{0.065}}\\\\v=0.5114\ m/s\\\\1\ m=1000\ mm\\\\So,0.5114\ m=511.4\ mm\\\\\therefore v=511.4\ mm/s

Therefore, the initial velocity is 0.5114 m/s or 511.4 mm/s

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3 years ago
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Answer:

5.88×10⁸ W

Explanation:

Power = energy / time

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P = (m/t) gh

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4

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