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antoniya [11.8K]
4 years ago
14

A power source of 2.0 V is attached to the ends of a capacitor. The capacitance is 4.0 μF. What is the amount of charge stored i

n this capacitor?
Physics
1 answer:
ioda4 years ago
8 0
C = 4 \ \mu F = 4 \cdot 10^{-6} \ F. \newline
q = Cu = 4 \cdot 10^{-6} \cdot 2 = 8 \cdot 10^{-6} = 0.000008 \ C.
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Explanation:

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maxonik [38]

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Explanation:

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5 0
2 years ago
Find the amount of force required to move an object of 1200 kg at a velocity of 54 km/hr?​​
Mkey [24]

Answer:

0 Newtons

Explanation:

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4 0
3 years ago
What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of −25.0 nc? g?
Daniel [21]

To calculate the force between two negative charges, we use the formula which is given by the Coulomb`s Law as

F=\frac{kq_{1}q_{2}  }{r^{2} }

Here, q_{1} andq_{2} are the charges on the pith balls, r is the separation between the charges and k is constant and its value is  8.99\times 10^9 N m^2/C^2.

Given  q_{1} =q_{2} =-25 nC=-25\times 10^{-9}C and  r=8 cm=8\times10^{-2} m.

Substituting these values in above formula we get,

F= 8.99\times 10^9 N m^2/C^2\frac{(-25\times 10^{-9}C)(-25\times 10^{-9}C)}{(8\times10^{-2} m)^2} \\\\\ F= 877929.7\times 10^{-9}  N\\\\F=8.8\times10^{-4}N

Thus, the repulsive force between two pith balls is 8.8\times10^{-4}N.

7 0
3 years ago
Read 3 more answers
PLS HELP ME
NeTakaya
The answer is true...
4 0
3 years ago
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