Answer:
21.21 m/s
Explanation:
Let KE₁ represent the initial kinetic energy.
Let v₁ represent the initial velocity.
Let KE₂ represent the final kinetic energy.
Let v₂ represent the final velocity.
Next, the data obtained from the question:
Initial velocity (v₁) = 15 m/s
Initial kinetic Energy (KE₁) = E
Final final energy (KE₂) = double the initial kinetic energy = 2E
Final velocity (v₂) =?
Thus, the velocity (v₂) with which the car we travel in order to double it's kinetic energy can be obtained as follow:
KE = ½mv²
NOTE: Mass (m) = constant (since we are considering the same car)
KE₁/v₁² = KE₂/v₂²
E /15² = 2E/v₂²
E/225 = 2E/v₂²
Cross multiply
E × v₂² = 225 × 2E
E × v₂² = 450E
Divide both side by E
v₂² = 450E /E
v₂² = 450
Take the square root of both side.
v₂ = √450
v₂ = 21.21 m/s
Therefore, the car will travel at 21.21 m/s in order to double it's kinetic energy.
There are two forces acting on a rocket at the moment of lift off: Thrust pushes the rocket upwards by pushing gases downwards in the opposite direction.Weight is the force due to gravity pulling the rocket downwards towards the centre of the earth.So I'm thinking the answer is THRUST.
I’m lost at this question, sorry but I would’ve help !
