Answer:
The distance between the places where the intensity is zero due to the double slit effect is 15 mm.
Explanation:
Given that,
Distance between the slits = 0.04 mm
Width = 0.01 mm
Distance between the slits and screen = 1 m
Wavelength = 600 nm
We need to calculate the distance between the places where the intensity is zero due to the double slit effect
For constructive fringe
First minima from center
Second minima from center
The distance between the places where the intensity is zero due to the double slit effect
Put the value into the formula
Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.
V = 310 m/s
f = 60 MHz = 60 × 10^6 Hz
v = xf
x = v/f
x = 310/(60 × 10^6) m
x = 5.166667 × 10^(−6) m
Q = mass water x specific heat water x delta T.
<span>714,000 = mass water x specific heat water x 30.
Substitute specific heat water and solve for mass water.</span>
Answer:
the distance that the object is raised above its initial position is 5.625 m.
Explanation:
Given;
applied effort, E = 15 N
load lifted by the ideal pulley system, L = 16 N
distance moved by the effort, d₁ = 6 m
let the distance moved by the object = d₂
For an ideal machine, the mechanical advantage is equal to the velocity ratio of the machine.
M.A = V.R
Therefore, the distance that the object is raised above its initial position is 5.625 m.
The middle or centre of the Earth is the core. However the middle of the layers from the surface to the centre of the Earth is known as mantle.
