Answer:
3.536*10^-6 C
Explanation:
The magnitude of the charge is expresses as Q = CV
C is the capacitance of the capacitor
V is the voltage across the capacitor
Get the capacitance
C = ε0A/d
ε0 is the permittivity of the dielectric = 8.84 x 10-12 F/m
A is the area = 0.2m²
d is the plate separation = 0.1mm = 0.0001m
Substitute
C = 8.84 x 10-12 * 0.2/0.0001
C = 1.768 x 10-8 F
Get the potential difference V
Using the formula for Electric field intensity
E = V/d
2.0 × 10^6 = V/0.0001
V = 2.0 × 10^6 * 0.0001
V = 2.0 × 10^2V
Get the charge on each plate.
Q = CV
Q = 1.768 x 10-8 * 2.0 × 10^2
Q = 3.536*10^-6 C
Hence the magnitude of the charge on each plate should be 3.536*10^-6 C
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Answer:
The time constant is 1.049.
Explanation:
Given that,
Charge 
We need to calculate the time constant
Using expression for charging in a RC circuit
![q(t)=q_{0}[1-e^{-(\dfrac{t}{RC})}]](https://tex.z-dn.net/?f=q%28t%29%3Dq_%7B0%7D%5B1-e%5E%7B-%28%5Cdfrac%7Bt%7D%7BRC%7D%29%7D%5D)
Where, 
= time constant
Put the value into the formula
![0.65q_{0}=q_{0}[1-e^{-(\dfrac{t}{RC})}]](https://tex.z-dn.net/?f=0.65q_%7B0%7D%3Dq_%7B0%7D%5B1-e%5E%7B-%28%5Cdfrac%7Bt%7D%7BRC%7D%29%7D%5D)
Hence, The time constant is 1.049.
Answer:
1.551×10^-8 Ωm
Explanation:
Resistivity of a material is expressed as shown;.
Resistivity = RA/l
R is the resistance of the material
A is the cross sectional area
l is the length of the wire.
Given;
R = 0.0310 Ω
A = πd²/4
A = π(2.05×10^-3)²/4
A = 0.000013204255/4
A = 0.00000330106375
A = 3.30×10^-6m
l = 6.60m
Substituting this values into the formula for calculating resistivity.
rho = 0.0310× 3.30×10^-6/6.60
rho = 1.023×10^-7/6.60
rho = 1.551×10^-8 Ωm
Hence the resistivity of the material is 1.551×10^-8 Ωm
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