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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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Answer:

The new speed of the ball is 176.43 m/s

Explanation:

Given;

mass of the ball, m = 7 kg

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Apply Newton's second law of motion;

F = ma = \frac{m(v-u)}{t}\\\\m(v-u) = Ft\\\\v-u = \frac{Ft}{m}\\\\v =  \frac{Ft}{m} + u

where;

v is new speed of the ball

v =  \frac{Ft}{m} + u\\\\v =\frac{300*4}{7} + 5\\\\v = 176.43 \ m/s

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Answer:c

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You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.
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(a) 2.7\cdot 10^{25} kg

The acceleration due to gravity on the surface of the planet is given by

g=\frac{GM}{R^2} (1)

where

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R is the radius of the planet

Here we know:

g=22.4 m/s^2

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R=\frac{d}{2}=\frac{1.8\cdot 10^7 m}{2}=9\cdot 10^6 m

So we can re-arrange eq.(1) to find M, the mass of the planet:

M=\frac{gR^2}{G}=\frac{(22.4 m/s^2)(9\cdot 10^6 m)^2}{6.67\cdot 10^{-11}}=2.7\cdot 10^{25} kg

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The planet is orbiting the star, so the centripetal force is equal to the gravitational attraction between the planet and the star:

m\frac{v^2}{r}=\frac{GMm}{r^2} (1)

where

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I could only tell you if i knew what the figure looked like
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