Answer:
(a) 
(b) 
Explanation:
Given data
Area A=4.0 cm²=0.0004 m²
Electric permittivity ε=8.854×10⁻¹² farads/meter
Charge q=0.708 nC
To find
(a) Potential at distance=1.00mm
(b) Potential at distance=2.00mm
Solution
For (a) Potential at distance=1.00mm
First we need to find the capacitance
So

As we know that

So Electric potential is given as:

For (b) Potential at distance=2.00mm
First we need to find the capacitance
So

As we know that

So Electric potential is given as:

Answer:
The major difference between a primary cell and the secondary cell is that primary cells are the ones that cannot be charged but secondary cells are the ones that are rechargeable
Explanation:
Answer:
(a) 31.75 mph
(b) 31.77 mph
Explanation:
(a) For t = 10 min = 10 / 60 = 1 / 6 hour
According to the question,
dx /dt = 15 mph
dy / dt = 28 mph
After 10 minutes
x = 1/6 x 15 = 2.5 miles
y = 1/6 x 28 = 4.67 miles

According to diagram

Differentiate both sides with respect to t.
2D dD/dt = 2 x dx/dt + 2y dy/dt
D dD/dt = x dx/dt + y dy/dt
5.3 dD/dt = 2.5 (15) + 4.67 (28)
dD/dt = 31.75 miles/hour
(b) For t = 55 minutes = 55 / 60 hours
After 55 / 60 hours
x = 55 / 60 (15) = 13.75 miles
y = 55 / 60 (28) = 25.67 miles

According to diagram

Differentiate both sides with respect to t.
2D dD/dt = 2 x dx/dt + 2y dy/dt
D dD/dt = x dx/dt + y dy/dt
29.12 dD/dt = 13.75 (15) + 25.67 (28)
dD/dt = 31.77 miles/hour
Answer:

Explanation:
<u>Given Data:</u>
Cycles = c = 3 cycles
Time = t = 2 seconds
<u>Required:</u>
Time period = T = ?
<u>Formula:</u>
1) f = c/t
2) T = 1/f
<u>Solution:</u>
<u>Finding frequency first:</u>
f = c / t
f = 3/2
f = 1.5 Hz
<u>Now, finding period:</u>
T = 1/f
T = 1 / 1.5
T = 0.67 seconds
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