Answer:
e)
Explanation:
In an RC series circuit, at any time, the sum of the voltages through the resistor and the capacitor must be constant and equal to the voltage of the DC voltage source, in order to be compliant with KVL.
At= 0, as the voltage through the capacitor can't change instantaneously, all the voltage appears through the resistor, which means that a current flows, that begins to charge the capacitor, up to a point that the voltage through the capacitor is exactly equal to the DC voltage, so no current flows in the circuit anymore, and the charge in the capacitor reaches to its maximum value.
Answer:
a) -41.1 Joule
b) 108.38 Kelvin
Explanation:
Pressure = P = 290 Pa
Initial volume of gas = V₁ = 0.62 m³
Final volume of gas = V₂ = 0.21 m³
Initial temperature of gas = T₁ = 320 K
Heat loss = Q = -160 J
Work done = PΔV
⇒Work done = 290×(0.21-0.62)
⇒Work done = -118.9 J
a) Change in internal energy = Heat - Work
ΔU = -160 -(-118.9)
⇒ΔU = -41.1 J
∴ Change in internal energy is -41.1 J
b) V₁/V₂ = T₁/T₂
⇒T₂ = T₁V₂/V₁
⇒T₂ = 320×0.21/0.62
⇒T₂ = 108.38 K
∴ Final temperature of the gas is 108.38 Kelvin
Answer:
Speed, mass and acceleration
Explanation:
A scalar quantity is a quantity that has only magnitude but no direction while a vector quantity has both magnitude and direction.
According to the question, the row that has two scalars and one vector is speed, mass and acceleration.
The two scalars in this row are speed and mass while the vector quantity there is the acceleration.
Acceleration has direction since it possess direction. A body accelerating will do so in a particular direction. Speed and mass doesn't possess any direction. Mass only specify the magnitude of the body but no clue as to which direction is the body moving towards.
Speed also only specify the
total distance covered with respect to time but not the direction of the direction.
Answer:
Explanation:
We can assume this problem as two concentric spherical metals with opposite charges.
We have also to take into account the formulas for the electric field and the capacitance. Hence we have
Where k is the Coulomb's constant. Furthermore, by taking into account the expression for the potential and by integrating
![dV=Edr\\\\V=\int_{R_1}^{R_2}Edr=-\int_{R_1}^{R_2}\frac{kQ}{r^2}dr\\\\V=kQ[\frac{1}{R_2}-\frac{1}{R_1}]](https://tex.z-dn.net/?f=dV%3DEdr%5C%5C%5C%5CV%3D%5Cint_%7BR_1%7D%5E%7BR_2%7DEdr%3D-%5Cint_%7BR_1%7D%5E%7BR_2%7D%5Cfrac%7BkQ%7D%7Br%5E2%7Ddr%5C%5C%5C%5CV%3DkQ%5B%5Cfrac%7B1%7D%7BR_2%7D-%5Cfrac%7B1%7D%7BR_1%7D%5D)
Hence, the capacitance is
![C=\frac{1}{k[\frac{1}{R_2}-\frac{1}{R_1}]}](https://tex.z-dn.net/?f=C%3D%5Cfrac%7B1%7D%7Bk%5B%5Cfrac%7B1%7D%7BR_2%7D-%5Cfrac%7B1%7D%7BR_1%7D%5D%7D)
but R1=a and R2=b
HOPE THIS HELPS!!
