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:
Explanation:
1) acceleration is the change in velocity of a body with respect to time.
acceleration = velocity/time
Given
velocity = 139m/s
time = 20secs
acceleration = 139/20
acceleration = 6.95m/s²
Hence its average acceleration during the first 20 seconds of the launch is 6.95m/s²
2) Speed is the rate of change of distance with respect to time.
average speed = distance/time
Time = distance/speed
Time = 100/7823
Time = 0.013s
3) Using the equation of motion v² = u²+2as
v² = 0²+2(15)(60)
v² = 30*60
v² =1800
v = √1800
v = 42.43m/s
Answer:
Explanation:
Given a parallel plate capacitor of
Area=A
Distance apart =d
Potential difference, =V
If the distance is reduce to d/2
What is p.d
We know that
Q=CV
Then,
V=Q/C
Then this shows that the voltage is inversely proportional to the capacitance
Therefore,
V∝1/C
So, VC=K
Now, the capacitance of a parallel plate capacitor is given as
C= εA/d
When the distance apart is d
Then,
C1=εA/d
When the distance is half d/2
C2= εA/(d/2)
C2= 2εA/d
Then, applying
VC=K
V1 is voltage of the full capacitor V1=V
V2 is the required voltage let say V'
Then,
V1C1=V2C2
V × εA/d=V' × 2εA/d
VεA/d = 2V'εA/d
Then the εA/d cancels on both sides and remains
V=2V'
Then, V'=V/2
The potential difference is half when the distance between the parallel plate capacitor was reduce to d/2
