Explanation:
It is given that,
Mass of the football player, m = 92 kg
Velocity of player, v = 5 m/s
Time taken, t = 10 s
(1) We need to find the original kinetic energy of the player. It is given by :


k = 1150 J
In two significant figure, 
(2) We know that work done is equal to the change in kinetic energy. Work done per unit time is called power of the player. We need to find the average power required to stop him. So, his final velocity v = 0
i.e. 

P = 115 watts
In two significant figures, 
Hence, this is the required solution.
<span>Self-monitoring would be the best way to </span><span>determine your own correct intensity level. I hope this helps! <3
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Answer:
The final velocity of the runner at the end of the given time is 2.7 m/s.
Explanation:
Given;
initial velocity of the runner, u = 1.1 m/s
constant acceleration, a = 0.8 m/s²
time of motion, t = 2.0 s
The velocity of the runner at the end of the given time is calculate as;

where;
v is the final velocity of the runner at the end of the given time;
v = 1.1 + (0.8)(2)
v = 2.7 m/s
Therefore, the final velocity of the runner at the end of the given time is 2.7 m/s.
Answer:
Explanation:
Let the angle between the first polariser and the second polariser axis is θ.
By using of law of Malus
(a)
Let the intensity of light coming out from the first polariser is I'
.... (1)
Now the angle between the transmission axis of the second and the third polariser is 90 - θ. Let the intensity of light coming out from the third polariser is I''.
By the law of Malus

So,



(b)
Now differentiate with respect to θ.


Answer:
The magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.
Explanation:
Given;
distance half way between the parallel wires, r = ¹/₂ (40 cm) = 20 cm = 0.2 m
current carried in opposite direction, I₁ and I₂ = 10 A and 20 A respectively
The magnitude of the magnetic field halfway between the wires can be calculated as;

where;
B is magnitude of the magnetic field halfway between the wires
I₁ is current in the first wire
I₂ is current the second wire
μ₀ is permeability of free space
r is distance half way between the wires

Therefore, the magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.