Answer:
Explanation:
Given that,
The distance between two spheres, r = 25 cm = 0.25 m
The capacitance, C = 26 pF = 26×10⁻¹² F
Charge, Q = 12 nC = 12 × 10⁻⁹ C
We need to find the work done in moving the charge. We know that, work done is given by :
Put all the values,
So, the work done is 
.
The correct answer is, A) Straight line motion
I took the quiz
This is a power problem which requires the rearranging of a formula. The lamps energy used is 5 N, and the TV’s usage is 116.7 N (rounded from 116.6666repeating). Here my work:
Explanation:
acceleration is 2 m/s^2
v-u/t
distance travelled is 30.25 meter
(v^2-u^2)/2a
11*11/2*2
121/4
30.25 m
Answer:
Change in Q = 2.1x 10^-3 C
Explanation:
We are given that
The Initialcapacitance C1 = 6.0μF
Initial charge oncapacitor
Q1 = C1 V
= 6.00 x 10^-6 x 100
= 6.00 x 10^-4 C
So the Final capacitance C2 will be
= K x C1 = 4.5 x 6.00 x 10^-6
= 2.7 x 10^ -5 F
So to get Finalcharge
We use Q2 = C2 x V
= 2.7 x 10^ - 5 x 100
= 27 x 10^ -4 C
So Charge flown in thecapacitor is change in Q
Which is = Q2 - Q1
= 27 x 10^-4 - 6.0 x 10^ -4
Change in Q = 2.1x 10^-3 C
