Answer:

Explanation:
From frequency of oscillation

Initially with the suspended string, the above equation is correct for the relation, hence

where k is force constant and m is the mass
When the spring is cut into half, by physics, the force constant will be doubled as they are inversely proportional

Employing f2/ f1, we have

Answer:

Explanation:
We know that
Acceleration due to gravity g is given by the formula

G= gravitational constant
M= mass of the Earth
r= radius of the earth
Let acc. due to gravity after radius is 20% greater be g_2
then

=> g1/g2 = (r_2/r_1)^2 => g2 = 9.81/1.2^2 = 6.8125
Answer: The unpolarized light's intensity is reduced by the factor of two when it passes through the polaroid and becomes linearly polarized in the plane of the Polaroid. When the polarized light passes through the polaroid with the plane of polarization at an angle
with respect to the polarization plane of the incoming light, the light's intensity is reduced by the factor of
(this is the Law of Malus).
Explanation: Let us say we have a beam of unpolarized light of intensity
that passes through two parallel Polaroid discs with the angle of
between their planes of polarization. We are asked to find
such that the intensity of the outgoing beam is
. To solve this we follow the steps below:
Step 1. It is known that when the unpolarized light passes through a polaroid its intensity is reduced by the factor of two, meaning that the intensity of the beam passing through the first polaroid is

This beam also becomes polarized in the plane of the first polaroid.
Step 2. Now the polarized beam hits the surface of the second polaroid whose polarization plane is at an angle
with respect to the plane of the polarization of the beam. After passing through the polaroid, the beam remains polarized but in the plane of the second polaroid and its intensity is reduced, according to the Law of Malus, by the factor of
This yields
. Substituting from the previous step we get

yielding

and finally,

Answer:
18.1 × 10⁻⁶ A = 18.1 μA
Explanation:
The current I in the wire is I = ∫∫J(r)rdrdθ
Since J(r) = Br, in the cylindrical wire. With width of 10.0 μm, dr = 10.0 μm. r = 1.20 mm. We have a differential current dI. We integrate first by integrating dθ from θ = 0 to θ = 2π.
So, dI = J(r)rdrdθ
dI/dr = ∫J(r)rdθ = ∫Br²dθ = Br²∫dθ = 2πBr²
Now I = (dI/dr)dr at r = 1.20 mm = 1.20 × 10⁻³ m and dr = 10.0 μm = 0.010 mm = 0.010 × 10⁻³ m
I = (2πBr²)dr = 2π × 2.00 × 10⁵ A/m³ × (1.20 × 10⁻³ m)² × 0.010 × 10⁻³ m = 0.181 × 10⁻⁴ A = 18.1 × 10⁻⁶ A = 18.1 μA
It's difficult to measure that because it's hard to make sure it is only a uniform layer of gas in whatever you're measuring it in