The resistance R of a piece of wire is given by

where

is the resistivity of the material, L is the length of the wire and A is its cross-sectional area.
Using this formula, and labeling with A the aluminum and with T the tungsten wire, we can write the ratio between

(the resistance of the tungsten wire) and

(the resistance of the aluminum wire):

the two wires are identical, so L and A are the same for the two wires and simplify in the ratio, and we get:

By using the resistivity of the aluminum:

and the resistivity of the tungsten:

m we can get the resistance of the tungsten wire:
When the mass of the spring changed from 0.2kg to 0.1kg, the time period changed from 1 sec to 0.5 seconds
<u>Explanation:</u>
Given-
Mass, m1 = 0.2kg
Time period, T1 = 1s
m2 = 0.1 kg
T2 = ?
We know,

where,
T = Time period
m = mass
k = spring constant
From the equation, we can see that T is directly proportion to the square root of mass, m
T ∝ √m
So,
If m1 = 0.2kg , T1 = 1s and m2 = 0.1kg
The T2 would be:


Therefore, when the mass of the spring changed from 0.2kg to 0.1kg, the time period changed from 1 sec to 0.5 seconds
Answer:
a. push
Explanation:
Voltage is the force or pressure that is responsible for pushing the charge or electrons to flow in a closed-looped electrical circuit. This flow of electrons (charge) is called the electric current. It is also defined as the difference in electric potential per unit charge between two points in an electric field.
Answer:
I₂ = 25.4 W
Explanation:
Polarization problems can be solved with the malus law
I = I₀ cos² θ
Let's apply this formula to find the intendant intensity (Gone)
Second and third polarizer, at an angle between them is
θ₂ = 68.0-22.2 = 45.8º
I = I₂ cos² θ₂
I₂ = I / cos₂ θ₂
I₂ = 75.5 / cos² 45.8
I₂ = 155.3 W
We repeat for First and second polarizer
I₂ = I₁ cos² θ₁
I₁ = I₂ / cos² θ₁
I₁ = 155.3 / cos² 22.2
I₁ = 181.2 W
Now we analyze the first polarizer with the incident light is not polarized only half of the light for the first polarized
I₁ = I₀ / 2
I₀ = 2 I₁
I₀ = 2 181.2
I₀ = 362.4 W
Now we remove the second polarizer the intensity that reaches the third polarizer is
I₁ = 181.2 W
The intensity at the exit is
I₂ = I₁ cos² θ₂
I₂ = 181.2 cos² 68.0
I₂ = 25.4 W
Answer: Option d.)
R_eq = (1/R_1 + 1/R_2 + 1/R_3)^-1
Explanation:
Since, there are three resistors connected in parallel, the reciprocal of the total resistance of the resistor combination (R_eq) is obtained by adding the reciprocal of each resistance.
i.e 1/R_eq = (1/R_1 + 1/R_2 + 1/R_3)
So, R_eq = (1/R_1 + 1/R_2 + 1/R_3)^-1
Thus, the total resistance (R_eq) is equal to the inverse of the sum of the reciprocal of each resistance.