The correct option is C. i.e. <span>Using more coils of wire
</span>
Explanation:
The strength of an electromagnet can be increased in the following ways:
1. By increasing the number of coils.
2. By increasing the current/voltage
3. Using an iron core, increases the strength of the electromagnet.
The diffusion coefficient of the gas is proportional to the average rate of thermal motion of the molecules.
the average velocity is inversely proportional to the square root of the molar mass
so
The gas diffusion rate is inversely proportional to the square root of its molecular weight.
Sounds like the shingle/ball is thrown from the roof horizontally, so that the distance it travels <em>x</em> after time <em>t</em> horizontally is
<em>x</em> = (7.2 m/s) <em>t</em>
The object's height <em>y</em> at time <em>t</em> is
<em>y</em> = 9.4 m - 1/2 <em>gt</em>²
where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity, and its vertical velocity is
<em>v</em> = -<em>gt</em>
(a) The object hits the ground when <em>y</em> = 0:
0 = 9.4 m - 1/2 <em>gt</em>²
<em>t</em>² = 2 * (9.4 m) / (9.80 m/s²)
<em>t</em> ≈ 1.92 s
at which time the object's vertical velocity is
<em>v</em> = -<em>g</em> (1.92 s) = -18.8 m/s ≈ -19 m/s
(b) See part (a); it takes the object about 1.9 s to reach the ground.
(c) The object travels a horizontal distance of
<em>x</em> = (7.2 m/s) * (1.92 s) ≈ 13.8 m ≈ 14 m
Answer:
Explanation:
As we know that melting point of silver is
T = 961.8 degree C
Latent heat of fusion of silver is given as
L = 111 kJ/kg
specific heat capacity of silver is given as
now we will have
now from above equation
Answer:
a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center
Explanation:
Let n₁ and n₂ be no of lines per unit length of grating A and B respectively.
λ₁ and λ₂ be wave lengths of green and red respectively , D be distance of screen and d₁ and d₂ be distance between two slits of grating A and B ,
Distance of first maxima for green light
= λ₁ D/ d₁
Distance of first maxima for red light
= λ₂ D/ d₂
Given that
λ₁ D/ d₁ = λ₂ D/ d₂
λ₁ / d₁ = λ₂ / d₂
λ₁ / λ₂ = d₁ / d₂
But
λ₁ < λ₂
d₁ < d₂
Therefore no of lines per unit length of grating A will be more because
no of lines per unit length ∝ 1 / d
If grating B is illuminated with green light first maxima will be at distance
λ₁ D/ d₂
As λ₁ < λ₂
λ₁ D/ d₂ < λ₂ D/ d₂
λ₁ D/ d₂ < 1 m
In this case position of first maxima will be less than 1 meter.
Option a is correct .
