Answer:
t_{out} =
t_{in}, t_{out} = 
Explanation:
This in a relative velocity exercise in one dimension,
let's start with the swimmer going downstream
its speed is

The subscripts are s for the swimmer, r for the river and g for the Earth
with the velocity constant we can use the relations of uniform motion
= D / 
D = v_{sg1} t_{out}
now let's analyze when the swimmer turns around and returns to the starting point

= D / 
D = v_{sg 2} t_{in}
with the distance is the same we can equalize

t_{out} = t_{in}
t_{out} =
t_{in}
This must be the answer since the return time is known. If you want to delete this time
t_{in}= D / 
we substitute
t_{out} = \frac{v_s - v_r}{v_s+v_r} ()
t_{out} = 
Answer:
...
<h2>PE=
<em>work done</em></h2><h2>
<em>m</em><em>gh</em><em>=</em><em>2</em><em>0</em><em>×</em><em>1</em><em>0</em><em>×</em><em>2</em><em>0</em><em>.</em><em>.</em></h2>

.
<em>I </em><em>hope</em><em> </em><em>this</em><em> </em><em>helps</em><em> </em><em>you</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>
70-10/70 x 100 percentage change ....
60/70, 6/7 fract change
Answer:
4.25 J
Explanation:
Given that
mass of plastic ball = 11 g
Mass of plastic ball = 0.011 kg
velocity of ball = 29 m/s
We know that from work power energy theorem

We know that kinetic energy of moving mass given as

Now by pitting the values


KE= 4.25 J
So the work done on the ball is 4.25 J
Answer:
F = [MLT⁻²]
Explanation:
Force = ma
m (mass) = [M]
a (acceleration) = [LT⁻²]
F(force) = m x a = [MLT⁻²]