Consider a compact car that is being driven at 98 km/h.
The acceleration of gravity is 9.8 m/s2.
From what height would the car have to be
dropped to have the same kinetic energy?
Answer in units of m.

The height from which the car would have to be  dropped to have the same kinetic energy is 37.81 meters.

Given the following data:

To find the height from which the car would have to be  dropped to have the same kinetic energy:

First of all, we would convert the value of velocity in km/h to m/s.

Conversion:

Applying the law of conservation of energy:

Substituting the given parameters into the formula, we have;

Height, h = 37.81 meters

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Height h = 37.8 m

Explanation:

Given :

Velocity of car (v) = 98 km / h

Acceleration of gravity = 9.8 m/s²

Computation:

Acceleration of gravity = 9.8 m/s²

Acceleration of gravity = (98)(1,000 m / 3,600 s)

Acceleration of gravity = 27.22 m/s

By using law of conservation of energy ;

(1/2)mv² = mgh

h = v² / 2g

h = 27.22² / 2(9.8)

Height h = 37.8 m

ok

explanation:

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg