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.
Since the product of P·Vis constant along an isotherm, an expansion to twice the volume implies a pressure reduction to half the original pressure. I hope my answer has come to your help. God bless and have a nice day ahead!<span>
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Answer:
Mass, m = 2.2 kg
Explanation:
It is given that,
Frequency of the piano, f = 440 Hz
Length of the piano, L = 38.9 cm = 0.389 m
Tension in the spring, T = 667 N
The frequency in the spring is given by :
is the linear mass density
On rearranging, we get the value of m as follows :
m = 0.0022 kg
or
m = 2.2 grams
So, the mass of the object is 2.2 grams. Hence, this is the required solution.
Answer:
The torque in the coil is 4.9 × 10⁻⁵ N.m
Explanation:
T = NIABsinθ
Where;
T is the torque on the coil
N is the number of loops = 9
I is the current = 7.8 A
A is the area of the circular coil = ?
B is the Earth's magnetic field = 5.5 × 10⁻⁵ T
θ is the angle of inclination = 90 - 56 = 34°
Area of the circular coil is calculated as follows;
T = 9 × 7.8 × 0.0227 × 5.5×10⁻⁵ × sin34°
T = 4.9 × 10⁻⁵ N.m
Therefore, the torque in the coil is 4.9 × 10⁻⁵ N.m
We get the clearest image if there is no magnification. When we have no magnification the image and real object have the same size.
If we look at the diagram that I attached we can see that:
Two triangles that I marked are similar and from this we get:
The image and the object must have the same height so we get:
This tells how far the screen should be from the lens.
The position of the screen on the optical bench is:
