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:
21.3 V, 1.2 A
Explanation:
1.
These resistors are in series, so the net resistance is:
R = R₁ + R₂ + R₃
R = 20 + 30 + 45
R = 95
So the current is:
V = IR
45 = I (95)
I = 9/19
So the voltage drop across R₃ is:
V = IR
V = (9/19) (45)
V ≈ 21.3 V
2.
First, we need to find the equivalent resistance of R₂ and R₃, which are in parallel:
1/R₂₃ = 1/R₂ + 1/R₃
1/R₂₃ = 1/10 + 1/10
R₂₃ = 5
Now we find the overall resistance by adding the resistors in series:
R = R₁ + R₂₃ + R₄
R = 10 + 5 + 10
R = 25
So the current through R₁ is:
V = IR
30 = I (25)
I = 1.2 A
2) Unbalanced. Mike will push the box with a force of 20 N. The forces would be balanced if the box responded with 30 N.
3) Balanced. Both boys are pulling with the same force. Neither is winning.
4) Unbalanced. The rope will move with 10 N to the west. The teachers are winning.
5) Unbalanced. The kids are pulling 220 N to the east. The kids are winning.
6) Balanced. You and the dog are pulling with the same force.
Answer:
It looks like its moving north.
Explanation:
