Answer:
Explanation:
Given
mass of lead piece
mass of water in calorimeter
Initial temperature of water
Initial temperature of lead piece
we know heat capacity of lead and water are and
respectively
Let us take be the final temperature of the system
Conserving energy
heat lost by lead=heat gained by water
Explanation:
Let us assume that the separation of plate be equal to d and the area of plates is . As the capacitance of capacitor is given as follows.
C =
It is known that the dielectric strength of air is as follows.
E =
Expression for maximum potential difference is that the capacitor can with stand is as follows.
dV = E × d
And, maximum charge that can be placed on the capacitor is as follows.
Q = CV
=
=
=
=
or, = 10.62 nC
Thus, we can conclude that charge on capacitor is 10.62 nC.
Answer:
The magnitude of the induced electric field at a point 2.5 cm from the axis of the solenoid is 8.8 x 10⁻⁵ V/m
Explanation:
given information:
radius, r = 2.0 cm
N = 700 turns/m
decreasing rate, dI/dt = 9.0 A/s
the magnitude of the induced electric field at a point 2.5 cm (r = 2.5 cm = 0.025 m) from the axis of the solenoid?
the magnetic field at the center of solenoid
B = μ₀nI
where
B = magnetic field (T)
μ₀ = permeability (1.26× 10⁻⁶ T.m/A)
n = the number turn per unit length (turn/m)
I = current (A)
dB/dt = μ₀n dI/dt (1)
now we calculate the induced electric field by using
E =
= 2E/r (2)
where
E = the induced electric field (V/m)
we substitute the firs and second equation, thus
dB/dt = μ₀n dI/dt
2E/r = μ₀n dI/dt
E = (1/2) r μ₀n dI/dt
= (1/2) (0.025) (1.26× 10⁻⁶) (700) (8)
= 8.8 x 10⁻⁵ V/m