Answer:
The resistance will be twice the original resistance
Explanation:
This is a fairly simple question, the formular for the resistance of a wire is given as
R = rho *length / area
Where R = resistance
Rho = resistivity
L = length
A = area
Since the density and area are constant I.e they do not change
R/ length = rho/ area
The initial length is given as L, when this length is stretched to twice its original length ,it becomes 2×L = 2L
Let x represent the resistance when the length is doubled
R/ L = x / 2L
x = 2LR / L ; dividing by L
We have that x = 2R ; twice the resistance
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
b
c
Explanation:
From the question we are told that
From the question we are told to find 
when t=0 equals the time constant (
)
That is to obtain .This is mathematically represented as
Substituting 12 V for 
and 150Ω for R
From the question we are told to find 
when t=0 equals the 3 times the time constant ()
That is to obtain 
.This is mathematically represented as
As tends to infinity 
So 
would be mathematically evaluated as
Answer:
True
Explanation:
When no net force is applied to a moving object, it still comes to rest because of its inertia.
Answer: The reason a light bulb glows is that electricity is forced through tungsten, which is a resistor. The energy is released as light and heat. A conductor is the opposite of a resistor. Electricity travels easily and efficiently through a conductor, with almost no other energy released as it passes.
Explanation:
Answer:
1.696 nm
Explanation:
For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,
dsinθ = mλ = (1)λ = λ
dsinθ = λ
sinθ = λ/d.
Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m
From trig ratios 1 + cot²θ = cosec²θ
1 + (1/tan²θ) = 1/(sin²θ)
substituting the values of sinθ and tanθ we have
1 + (D/w)² = (d/λ)²
(D/w)² = (d/λ)² - 1
(w/D)² = 1/[(d/λ)² - 1]
(w/D) = 1/√[(d/λ)² - 1]
w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹ = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.
w is also the distance from the center to the other principal maximum on the other side.
So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm
So, the minimum width of the screen must be 1.696 nm
