Do you remember this formula for the distance traveled while accelerated ?
<u>Distance = (initial speed) x (t) plus (1/2) x (acceleration) x (t²)</u>
I think this is exactly what we need for this problem.
initial speed = 20 m/s down
acceleration = 9.81 m/s² down
t = 3.0 seconds
Distance down = (20) x (3) plus (1/2) x (9.81) x (3)²
Distance = (60) plus (4.905) x (9)
Distance = (60) plus (44.145) = 104.145 meters
Choice <em>D)</em> is the closest one.
Accourding to newtons second law of motion:
Force = mass * acceleration
F = ma
a = F/m
Answer:
a. 17 μC b. 14.45 J
Explanation:
Here is the complete question
A pair of 10μF capacitors in a high-power laser are charged to 1.7 kV.
a. What charge is stored in each capacitor?
b. How much energy is stored in each capacitor?
a. The charge Q stored in a capacitor is Q = CV where C = capacitance and V = voltage. Here, C = 10μF = 10 × 10⁻⁶ F and V = 1.7 kV = 1.7 × 10³ V.
So, Q = CV = 10 × 10⁻⁶ × 1.7 × 10³ = 1.7 × 10⁻² C = 17 × 10⁻³ C = 17 μC
b. The energy stored in a capacitor is given by W = 1/2CV² = 1/2 × 10 × 10⁻⁶ F × (1.7 × 10³ V)² = 1/2 × 10 × 10⁻⁶ F × 2.89 × 10⁶ V² = 1/2 × 10 × 2.89 × 10⁶ × 10⁻⁶ J = 14.45 J