If a charge of 12 C flows past any point along a circuit in _____ seconds, the current at that point would be 3 A.
1 answer:
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Gravitational potential energy can be calculated using the formula:
Where:
PEgrav = Gravitational potential energy
m= mass
g = acceleration due to gravity
h = height
On Earth acceleration due to gravity is a constant 9.8 but since the scenario is on Mars, the pull of gravity is different. In this case, it is 3.7, so we will use that for g.
So put in what you know and solve for what you don't know.
m = 10kg
g = 3.7m/s^2
h = 1m
So we put that in and solve it.
Where:
PEgrav = Gravitational potential energy
m= mass
g = acceleration due to gravity
h = height
On Earth acceleration due to gravity is a constant 9.8 but since the scenario is on Mars, the pull of gravity is different. In this case, it is 3.7, so we will use that for g.
So put in what you know and solve for what you don't know.
m = 10kg
g = 3.7m/s^2
h = 1m
So we put that in and solve it.
Answer:
0.3619sec
Explanation:
Given that
Mass,m=148 g
Length,L=13 cm
Velocity,u'(0)=10 cm/s
We have to find the position u of the mass at any time t
We know that
Where
u(0)=0
Substitute the value
Substitute u'(0)=10
Substitute the values
Period =T = 2π/8.68
After half period
π/8.68 it returns to equilibruim
π/8.68 = 0.3619sec
Answer:
(a) Charge of 4.25 μF capacitor is 35.46 μC.
(b) Charge of 1.13 μF capacitor is 10.05 μC.
(c) Charge of 2.85 μF capacitor is 25.36 μC.
Explanation:
Let C₁ , C₂ and C₃ are the capacitor which are connected to the battery having voltage V. According to the problem, C₁ and C₂ are connected in parallel. There equivalent capacitance is:
C₄ = C₁ + C₂
Substitute 1.13 μF for C₁ and 2.85 μF for C₂ in the above equation.
C₄ = ( 1.13 + 2.85 ) μF = 3.98 μF
Since, C₄ and C₃ are connected in series, there equivalent capacitance is:
C₅ =
Substitute 4.25 μF for C₃ and 3.98 μF for C₄ in the above equation.
C₅ =
C₅ = 2.05 μF
The charge on the equivalent capacitance is determine by the relation :
Q = C₅ V
Substitute 2.05 μF for C₅ and 17.3 volts for V in the above equation.
Q = 2.05 μF x 17.3 = 35.46 μC
Since, the capacitors C₃ and C₄ are connected in series, so the charge on these capacitors are equal to the charge on the equivalent capacitor C₅.
Charge on the capacitor, C₃ = 35.46 μC
Charge on the capacitor, C₄ = 35.46 μC
Voltage on the capacitor C₄ = =
= 8.90 volts
Since, C₁ and C₂ are connected in parallel, the voltage drop on both the capacitors are same, that is equal to 8.90 volts.
Charge on the capacitor, C₁ = C₁ V = 1.13 μF x 8.90 = 10.05 μC
Charge on the capacitor, C₂ = C₂ V = 2.85 μF x 8.90 = 25.36 μC