When a force acts (pushes or pulls) on an object, it changes the object's speed or direction (in other words, makes it accelerate). The bigger the force, the more the object accelerates. When a force acts on an object, there's an equal force (called a reaction) acting in the opposite direction.
The power the swimmer must provide to overcome that drag force is 50 W.
<h3>What is work done?</h3>
Work done is equal to product of force applied and distance moved.
Work = Force x Distance
Given is an athlete, swimming at a constant speed, covers a distance of 203 m in a time period of 5 minutes and 50 seconds = 350 seconds. The drag force exerted by the water on the swimmer is 86.0 N.
Work done by drag force is W= F x d
So, W = 86 N x 203 m
W = 17.458 kJ
Power , a swimmer is provided P = Work done/ Time
P = 17458 / 350
P = 50 W
Thus, the power, the swimmer must provide to overcome that drag force is 50 W.
Learn more about Work done
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Answer:
I believe it has 4 legs as well
Explanation:
It looks like 4 legs, it never gets on 2 legs so it should be 4.
Answer:
13330.86 J
Explanation:
mass of volume of 30.1 cc of water = density x volume
= 1 x 30.1 = 30.1 gm
= 30.1 x 10⁻³ kg
Heat to be withdrawn to cool water from 22.4 degree to 0 degree
= mass x specific heat x fall of temperature
= 30.1 x 10⁻³ x 4186 x 22.4
= 2822.36 J
Heat to be withdrawn to cool water 0 degree ice
mass x latent heat of freezing
30.1 x 10⁻³ x 334000
= 10053.4 J
Heat to be withdrawn to cool ice from 0 degree to -7.2 degree
mass x specific heat of ice x fall of temperature
= 30.1 x 10⁻³ x 2100 x 7.2
= 455.1 J
Total heat to be withdrawn
= 2822.36 + 10053.4 + 455.1
= 13330.86 J
Answer:
The average force exerted on the pier is 
N
Explanation:
Given :
Mass of ship 
Kg
Distance 
m
Velocity 
Now find the average velocity,
Now find the time taken by ship to travel 4.44 m.
sec
Now calculating average force,
Where 
= 0.652
N
Therefore, the average force exerted on the pier is N
