I'm not sure if a figure or some choices go along with this, but the closer to the sea floor the diver is, the lower the potential energy
By using your bones and the ligaments are being used
Answer:
Answer:
The velocity of the ball when it reaches its highest point is 0
Explanation:
The velocity of the ball when it reaches its highest point is 0
Once the ball is tossed into the air, as it goes up, the initial velocity with which it was thrown, reduces, as the motion of the ball is hindered by several forces such as gravity and air resistance. This slows down the velocity of the ball, up until it reaches a point, where the upwards velocity of the ball becomes zero. at this point, the ball begins to fall back to the ground.
Answer:
Explanation:
Given that,
First Capacitor is 10 µF
C_1 = 10 µF
Potential difference is
V_1 = 10 V.
The charge on the plate is
q_1 = C_1 × V_1 = 10 × 10^-6 × 10 = 100µC
q_1 = 100 µC
A second capacitor is 5 µF
C_2 = 5 µF
Potential difference is
V_2 = 5V.
Then, the charge on the capacitor 2 is.
q_2 = C_2 × V_2
q_2 = 5µF × 5 = 25 µC
Then, the average capacitance is
q = (q_1 + q_2) / 2
q = (25 + 100) / 2
q = 62.5µC
B. The two capacitor are connected together, then the equivalent capacitance is
Ceq = C_1 + C_2.
Ceq = 10 µF + 5 µF.
Ceq = 15 µF.
The average voltage is
V = (V_1 + V_2) / 2
V = (10 + 5)/2
V = 15 / 2 = 7.5V
Energy dissipated is
U = ½Ceq•V²
U = ½ × 15 × 10^-6 × 7.5²
U = 4.22 × 10^-4 J
U = 422 × 10^-6
U = 422 µJ
