Answer:
C2, C1, C4, C5 and C6 are in parallel. Therefore, we use the formula Cp = C1 + C2 + ....
Cp = C2 + C1 + C4 + C5 + C6 = ( 7 * 10 ^-3) + (18 * 10^-6) + (0.8F) + (200 * 10^-3 F) + (750 * 10^-6) = 1.008F
Now, Cp will become one capacitor and it will be aligned with C3, therefore it will now become a circuit in series.
We use the formula: 1/Cs = 1/C1 + 1/C2 + .... + ....1/Cn
Thus,
1/Cs = 1/C3 + 1/Cp
1/Cs = 1/(14 * 10^-3 F) + 1/(1.008F)
Cs = 1.4 * 10 ^-2 or if we do not round too much it will give exactly 0.0138 F
So the answer should be a)
Answer:
-1m/s
Explanation:
We can calculate the speed of block A after collision
According to collision theory:
MaVa+MbVb = MaVa+MbVb (after collision)
Substitute the given values
5(3)+10(0) = 5Va+10(2)
15+0 = 5Va + 20
5Va = 15-20
5Va = -5
Va = -5/5
Va = -1m/s
Hence the velocity of ball A after collision is -1m/s
Note that the velocity of block B is zero before collision since it is stationary
Answer:
Rs. 480.00
Explanation:
1kW = 1000W
therefore 500W = 0.5kW
20 × 24hrs = 480hrs in total.
0.5kW × 480hrs = 240kWh
if rs. 2 for 1kWh
then, 240kWh × 2 = Rs. 480.
Because they perform specific tasks repeatedly throughout your program, as needed
Answer:
V = 576 V
Explanation:
Given:
- The area of the two plates A = 0.070 m^2
- The space between the two plates d = 6.3 mm
- Te energy density u = 0.037 J /m^3
Find:
- What must the potential difference between the plates V?
Solution:
- The energy density of the capacitor with capacitance C and potential difference V is given as:
u = 0.5*ε*E^2
- Where the Electric field strength E between capacitor plates is given by:
E = V / d
Hence,
u = 0.5*ε*(V/d)^2
Where, ε = 8.854 * 10^-12
V^2 = 2*u*d^2 / ε
V = d*sqrt ( 2*u / ε )
Plug in values:
V = 0.0063*sqrt ( 2 * 0.037 / (8.854 * 10^-12) )
V = 576 V