Answer:
2
Explanation:
We know that in the Fraunhofer single-slit pattern,
maxima is given by

Given values
θ=2.12°
slit width a= 0.110 mm.
wavelength λ= 582 nm
Now plugging values to calculate N we get

Solving the above equation we get
we N= 2.313≅ 2
Answer:
1 m
Explanation:
L = 100 m
A = 1 mm^2 = 1 x 10^-6 m^2
Y = 1 x 10^11 N/m^2
F = 1000 N
Let the cable stretch be ΔL.
By the formula of Young's modulus



ΔL = 1 m
Thus, the cable stretches by 1 m.
Answer: 4nmeter
Explanation: The two observer a and b will measure the same wavelength since the speed of the space craft is very small compared with the speed of light c. That is
V which is the speed of space craft 15000km/s = 15000000m/s
Comparing this with the speed of light c 3*EXP(8)m/s we have
15000000/300000000
= 0.05=0.1
Therefore the speed of the space craft V in terms of the speed of light c is 0.1c special relativity does not apply to object moving at such speed. So the wavelength would not be contracted it will remain same for both observers.
The Renaissance, was the rebirth to science as well as many other advancements
1.A and 2.B there the answers