<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:
U = (ε0AV^2) / 2d
Explanation:
Where C= capacitance of the capacitor
ε0= permittivity of free space
A= cross sectional area of plates
d= distance between the plates
V= potential difference
First, the capacitance of a capacitor is obtained by:
C = ε0A/d.
Starting at the formula , U= (CV^2)/2. Formula for energy stored in a capacitor
Substitute in for C:
U = (ε0A/d) * V^2 / 2
Hence:
U = (ε0AV^2) / 2d
Answer:
.409 N
Explanation:
For this to balance, the moments around the fulcrum must sum to zero.
On the left you have .21 ( is that down? I will assume it is)
Counterclockwise moments :
.21 * 40 + 1.0 * 20
Clockwise moments :
.5 * 20 + F * 45
these moments must equal each other
.21*40 + 1 *20 = .5 * 20 + F * 45
F = .409 N
Answer:
71.19 C
Explanation:
25C = 25 + 273 = 298 K
Applying the ideal gas equation we have
where P, V and T are the pressure, volume and temperature of the gas at 1st and 2nd stage, respectively. We can solve for the temperature and the 2nd stage: