<h3><u>Answer;</u></h3>
electric potential
<h3><u>Explanation;</u></h3>
Electric potential is the electric potential energy per unit charge.
Mathematically; V =PE/q
Where; PE is the electric potential energy, V is the electric potential and q is the charge.
Electric potential is more commonly known as voltage. If you know the potential at a point, and you then place a charge at that point, the potential energy associated with that charge in that potential is simply the charge multiplied by the potential.
1 kilometre is equal to 1000m
and 4.1 minutes is equal to 246 seconds
thus 1000/246 = 4.065 m/s
and the direction is towards the west
Answer:
I believe its A: Sports biomechanics.
Answer:
the work done by the lawnmower is 236.14 J.
Explanation:
Given;
power exerted by the lawnmower engine, P = 19 hp
time in which the power was exerted, t = 1 minute = 60 s.
1 hp = 745.7 watts
The work done by the lawnmower is calculated as follows;

Therefore, the work done by the lawnmower is 236.14 J.
Answer:
The answer to the question is
The distance d, which locates the point where the light strikes the bottom is 29.345 m from the spotlight.
Explanation:
To solve the question we note that Snell's law states that
The product of the incident index and the sine of the angle of incident is equal to the product of the refractive index and the sine of the angle of refraction
n₁sinθ₁ = n₂sinθ₂
y = 2.2 m and strikes at x = 8.5 m, therefore tanθ₁ = 2.2/8.5 = 0.259 and
θ₁ = 14.511 °
n₁ = 1.0003 = refractive index of air
n₂ = 1.33 = refractive index of water
Therefore sinθ₂ =
=
= 0.1885 and θ₂ = 10.86 °
Since the water depth is 4.0 m we have tanθ₂ =
or x₂ =
=
= 20.845 m
d = x₂ + 8.5 = 20.845 m + 8.5 m = 29.345 m.