If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be 
.
<h3>What is an escape velocity?</h3>
The ratio of the object's travel distance over a specific period of time is known as its velocity. As a vector quantity, the velocity requires both the magnitude and the direction. the slowest possible speed at which a body can break out of the gravitational pull of a certain planet or another object.
The formula to calculate the escape velocity of earth is given below:-
Given that earth's mass was half its actual value but its radius stayed the same. The escape velocity will be calculated as below:-
.
Therefore, If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be .
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Answer:
Part a)
Part b)
So this speed is independent of the mass of the rider
Explanation:
Part a)
By force equation on the rider at the position of the hump we can say
now we will have
now we have
Part b)
At the top of the loop if the minimum speed is required so that it remains in contact so we will have
at minimum speed
So this speed is independent of the mass of the rider
The gravitational potential energy of the object is 100 J.
Gravitational potential energy stored in an object is the work done in raising the object to a height <em>h</em> against the gravitational force acting on it.
The gravitational force acting on a body is its weight mg, where m is its mass and g, the acceleration due to gravity.
Work done by a force is equal to the product of the force and the displacement made by the point of application of the force.
The weight of the object is given as 20 J and it is raised to a height of 5 m.
The gravitational potential energy of the object is 100 J.
<span>C. switching to cheaper fuel. Physical capital pertains to non-human asset. Under this type were the asset that use to process goods and services like machinery, buildings etc.</span>
Answer:
E_total = 3 N / A
Explanation:
The electric field is a vector magnitude so when adding we must use vectors, in this case as the initial field E = 4N / c goes towards the axis axis and the field created by the fixed charge (E1) is also on the axis x we can add in scalar form.
E_total = E + E₁
the expression for the field of a point charge is
E₁ = k q₁ / r²
for the point x = 2m, they do not say that the total field is zero, so the charge q1 must be negative
E_total = E -k q₁ / r₂
we substitute
0 = E - k q₁ / r²
q₁ = 
let's calculate
q₁ = 
q₁ = 1.78 10⁻⁹ C
now we can calculate the field for position x = 4 m
E_total = 4 - 9 10⁹ 1.78 10⁻⁹ / 4²2
E_total = 3 N / A
