<span>when it returns to its original level after encountering air resistance, its kinetic energy is
decreased.
In fact, part of the energy has been dissipated due to the air resistance.
The mechanical energy of the ball as it starts the motion is:
</span>

<span>where K is the kinetic energy, and where there is no potential energy since we use the initial height of the ball as reference level.
If there is no air resistance, this total energy is conserved, therefore when the ball returns to its original height, the kinetic energy will still be 100 J. However, because of the presence of the air resistance, the total mechanical energy is not conserved, and part of the total energy of the ball has been dissipated through the air. Therefore, when the ball returns to its original level, the kinetic energy will be less than 100 J.</span>
Answer:
4km
Explanation:
15 minutes is 1/4 of an hour.
1/4 of 16 is 4.
Complete question:
An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0 V battery. Find the value of its capacitance.
Answer:
The value of its capacitance is 1.027 x 10⁻¹² F
Explanation:
Given;
area of the plate, A = 2.9 cm² = 2.9 x 10⁻⁴ m²
separation distance of the plates, d = 2.5 mm = 2.5 x 10⁻³ m
voltage of the battery, V = 18 V
The value of its capacitance is calculated as;

Therefore, the value of its capacitance is 1.027 x 10⁻¹² F
Answer:
See explanation below
Explanation:
The equation to use for this is the following:
dU = q + w
As the heat is being release, this value is negative, and same here happens with the work done, because it's in the surroundings.
Therefore the change in the energy would be:
dU = -2.59x10^4 - 6.46^4
dU = -9.05x10^4 kJ