A 100 Kg man is diving off a 50 meter cliff. What is his kinetic energy when he is 20 meters from the water?
1 answer:
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Answer:
The force exerted on the is
Explanation:
From the question we are told that
The area is
The magnitude of charge placed on them is
The charge placed between the plate is
The electric field generated around the plate is mathematically represented as
Substituting values
The force exerted the charge is mathematically represented as
Substituting values
<h2>Answer:</h2>
Torque = <em>2.05 x 10²⁸ Nm</em>
Energy = <em>3.54 x 10³³ J</em>
Average power = <em>1.02 x 10²⁸ W</em>
<h2>Explanation:</h2>(a) Torque (τ) is the rotational effect of a given force.
It is given by
τ = I x α -------------(i)
Where;
I = rotational inertia of the object
α = angular acceleration of the object.
In this case, the object is the Earth. Therefore,
I = 9.71 x 10³⁷ kg m²
α = ω / t
Where;
ω = angular velocity of earth = 2π rad / day
<em>Since </em>
<em>1 day = 24 hours and 1 hour = 3600seconds</em>
<em>1 day = 24 x 3600 seconds = 86400seconds</em>
<em>=> ω = 2π rad / 86400seconds</em>
<em>=> ω = 7.29 × 10⁻⁵ rad/s</em>
<em />
t = 4 days = 4 x 24 x 3600 seconds = 345600 seconds
=> α = ω / t
=> α = 7.29 × 10⁻⁵ / 345600
=> α = (7.29 × 10⁻⁵) / (3.456 x 10⁵)
=> α = (7.29 × 10⁻⁵⁻⁵) / (3.456)
=> α = (7.29 × 10⁻¹⁰) / (3.456)
=> α = 2.11 × 10⁻¹⁰ rad/s²
Now substitute the values of I and α into equation (i)
τ = 9.71 x 10³⁷ x 2.11 × 10⁻¹⁰
τ = 9.71 x 10²⁷ x 2.11
τ = 20.5 x 10²⁷ Nm
τ = 2.05 x 10²⁸ Nm
(ii) The energy (rotational energy) E is given by;
E = x I x ω
E = x 9.71 x 10³⁷ x 7.29 × 10⁻⁵
E = 35.4 x 10³² J
E = 3.54 x 10³³ J
(iii) The average power P, is given by;
P = E / t
P = 3.54 x 10³³ / 345600
P = 1.02 x 10²⁸ W
Answer:
4v/3
Explanation:
Assume elastic collision by the law of momentum conservation:
where v is the original speed of car 1, v1 is the final speed of car 1 and v2 is final speed of car 2. m1 and m2 are masses of car 1 and car 2, respectively
Substitute
Divide both side by , then multiply by 6 we have
So the final speed of the second car is 4/3 of the first car original speed