Answer:
a) y = 2.4 x 10⁻³ m = 0.24 cm
b) y = 3.2 x 10⁻³ m = 0.32 cm
Explanation:
The formula of Young's Double Slit experiment will be used here:

where,
y = distance between dark spots = ?
λ = wavelength
L = distance of screen = 2 m
d = slit width = 4 x 10⁻⁴ m
a) FOR λ = 480 nm = 4.8 x 10⁻⁷ m:

<u>y = 2.4 x 10⁻³ m = 0.24 cm</u>
<u></u>
a) FOR λ = 640 nm = 6.4 x 10⁻⁷ m:

<u>y = 3.2 x 10⁻³ m = 0.32 cm</u>
Answer:
W = 1.06 MJ
Explanation:
- We will use differential calculus to solve this problem.
- Make a differential volume of water in the tank with thickness dx. We see as we traverse up or down the differential volume of water the side length is always constant, hence, its always 8.
- As for the width of the part w we see that it varies as we move up and down the differential element. We will draw a rectangle whose base axis is x and vertical axis is y. we will find the equation of the slant line that comes out to be y = 0.5*x. And the width spans towards both of the sides its going to be 2*y = x.
- Now develop and expression of Force required:
F = p*V*g
F = 1000*(2*0.5*x*8*dx)*g
F = 78480*x*dx
- Now, the work done is given by:
W = F.s
- Where, s is the distance from top of hose to the differential volume:
s = (5 - x)
- We have the work as follows:
dW = 78400*x*(5-x)dx
- Now integrate the following express from 0 to 3 till the tank is empty:
W = 78400*(2.5*x^2 - (1/3)*x^3)
W = 78400*(2.5*3^2 - (1/3)*3^3)
W = 78400*13.5 = 1058400 J
They have some but not very much, the particles in the ice are still vibrating just not as much as in water. the only time a substance would have 0 kinetic energy is when that substance is at 0 degrees kelvin(absolute zero) so far no place in the universe has been recorded at absolute zero though
Answer:
r = 0m is the Minimum distance from the axis at which the block can remain in place wothout skidding.
Explanation:
From a sum of forces:
where Ff = μ * N and 
N - m*g = 0 So, N = m*g. Replacing everything on the original equation:
(eq2)
Solving for r:

If we analyze eq2 you can conclude that as r grows, the friction has to grow (assuming that ω is constant), so the smallest distance would be 0 and the greatest 1.41m. Beyond that distance, μ has to be greater than 0.83.
1. 0.2 g/mL
The relationship between mass, density and volume of an object is

where
d is the density
m is the mass
V is the volume
For the object in this problem, we have
m = 10 g
V = 50 mL
Substituting into the equation,

2. 10 mL
In this exercise we know:
- The density of the object: d = 2 g/mL
- The mass of the object: m = 20 g
Therefore, we can re-arrange the previous equation to find the volume:

And substituting values into the equation, we find
