The velocity is 60 because you divide your distance by your time (240÷4=60)
Answer:
Option B. 5 nC
Explanation:
From the question given above, the following data were obtained:
Capicitance (C) = 100 pF
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:
1 pF = 1×10¯¹² F
Therefore,
100 pF = 100 pF × 1×10¯¹² F / 1 pF
100 pF = 1×10¯¹⁰ F
Next, we shall determine the quantity of charge. This can be obtained as follow:
Capicitance (C) = 1×10¯¹⁰ F
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Q = CV
Q = 1×10¯¹⁰ × 50
Q = 5×10¯⁹ C
Finally, we shall convert 5×10¯⁹ C to nano coulomb (nC). This can be obtained as follow:
1 C = 1×10⁹ nC
Therefore,
5×10¯⁹ C = 5×10¯⁹ C × 1×10⁹ nC / 1 C
5×10¯⁹ C = 5 nC
Thus, the quantity of charge is 5 nC
38*10=380 N
To be more exact, 38 should be multiplied by 9.8 instead of 10.
Answer:
0.0002 C.
Explanation:
Charge: This can be defined as the ratio of current to time flowing in a circuit. The S.I unit of charge is Coulombs (C)
Mathematically, charge can be expressed as
Q = CV ................................. Equation 1
Where Q = amount of charge, C = capacitance of the capacitor, V = potential difference across the plates.
Given: C = 2.0-μF = 2×10⁻⁶ F, V = 100 V.
Substitute into equation 1
Q = 2×10⁻⁶× 100
Q = 2×10⁻⁴ C
Q = 0.0002 C.
The amount of charge accumulated = 0.0002 C
