Answer:
Explanation:
Force on q due to Q
= k Qq / ( a² + d²)
x component
= k Qq / ( a² + d²) x d /√ ( a² + d²)
F = kQq d/ ( a² + d²)³/²
differentiating F with respect to d
dF / Dd = kQq [ d. -3/2 ( a² + d²)⁻⁵/² 2d + ( a² + d²)⁻³/²]=0 for maximum F
- 3d² / ( a² + d²) + 1 = 0
a² + d² = 3 d²
a² = 2d²
d = a / √2
Gravitational Force is F=G*m1*m2/r^2
Given Information:
Resistance = R = 1.9 k
Ω
Capacitance = C = 30 uF
Initial charge = 30 uC
Final charge = 5 uC
Required Information:
Time taken to reduce the capacitor's charge to 5.0 μC = ?
Answer:
t = 0.101 seconds
Explanation:
The voltage across the capacitor is given by
V = V₀e^(–t/τ)
Where τ is time constant, V₀ is the initial voltage, V is the voltage after some time t
τ = RC
τ = 1900*30x10⁻⁶
τ = 0.057 sec
The initial voltage across the capacitor was
V₀ = Q/C
V₀ = 30/30
V₀ = 1 V
Voltage to reduce the charge to 5 uF
V = 5/30
V = 0.167 V
V = V₀e^(–t/τ)
0.167 = 1*e^(–t/0.057)
take ln on both sides
ln(0.167) = ln(e^(–t/0.057))
-1.789 = -t/0.057
t = 1.789*0.057
t = 0.101 seconds
This is <span>what happens to the system after heat is added : </span><span>Kinetic energy within the system increases and is transformed into mechanical energy as the piston moves upward.
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