Answer: 0.43 V
Explanation:
L = [μ(0) * N² * A] / l
Where
L = Inductance of the solenoid
N = the number of turns in the solenoid
A = cross sectional area of the solenoid
l = length of the solenoid
7.3*10^-3 = [4π*10^-7 * 450² * A] / 0.24
1.752*10^-3 = 4π*10^-7 * 202500 * A
1.752*10^-3 = 0.255 * A
A = 1.752*10^-3 / 0.255
A = 0.00687 m²
A = 6.87*10^-3 m²
emf = -N(ΔΦ/Δt).........1
L = N(ΔΦ/ΔI) so that,
N*ΔΦ = ΔI*L
Substituting this in eqn 1, we have
emf = - ΔI*L / Δt
emf = - [(0 - 3.2) * 7.3*10^-3] / 55*10^-3
emf = 0.0234 / 0.055
emf = 0.43 V
Answer:
KE = 1.75 J
Explanation:
given,
mass of ball, m₁ = 300 g = 0.3 Kg
mass of ball 2, m₂ = 600 g = 0.6 Kg
length of the rod = 40 cm = 0.4 m
Angular speed = 100 rpm=
=10.47\ rad/s
now, finding the position of center of mass of the system
r₁ + r₂ = 0.4 m.....(1)
equating momentum about center of mass
m₁r₁ = m₂ r₂
0.3 x r₁ = 0.6 r₂
r₁ = 2 r₂
Putting value in equation 1
2 r₂ + r₂ = 0.4
r₂ = 0.4/3
r₁ = 0.8/3
now, calculation of rotational energy
KE = 1.75 J
the rotational kinetic energy is equal to 1.75 J
Answer:
d) -4.0
Explanation:
The magnification of a lens is given by
where
M is the magnification
q is the distance of the image from the lens
p is the distance of the object from the lens
In this problem, we have
p = 50 cm is the distance of the object from the lens
q = 250 cm - 50 cm is the distance of the image from the lens (because the image is 250 cm from the obejct
Also, q is positive since the image is real
So, the magnification is
Answer:
Final velocity of electron,
Explanation:
It is given that,
Electric field, E = 1.55 N/C
Initial velocity at point A,
We need to find the speed of the electron when it reaches point B which is a distance of 0.395 m east of point A. It can be calculated using third equation of motion as :
........(1)
a is the acceleration,
We know that electric force, F = qE
Use above equation in equation (1) as:
v = 647302.09 m/s
or
So, the final velocity of the electron when it reaches point B is . Hence, this is the required solution.