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.
A pure substance that is made up of only one kind of atom is called an element
Answer:
(i) Relative velocity of B w.r.t A= Sum of speeds of trains
=54+90
=144kmph
(ii)Relative velocity of B w.r.t Ground(G)=v
B/G
=−90kmph
v
G
=0
Relative velocity of ground(G) w.r.t B =v
G/B
=v
G
−v
B/G
v
G/B
=0−(−90)
v
G/B
=90kmph
Answer:
F = 1.6*10⁴ N
Explanation:
Given distance x = 0.15m, mass m = 1200kg, velocity v = 2m/s.
Unknown: force F
Force is given by Newton's law:
(1)
The average force to stop an object from a velocity will be the same force necessary to accelerate an object from rest to the same velocity.
The distance for an object starting from rest for a constant acceleration is given by:
(2)
The velocity for an object starting from rest for a constant acceleration:
(3)
Using equation 2 and 3 to eliminate time t:
(4)
Solving equation 4 for the acceleration a:
(5)
Using equation1 to solve for the force F:
(6)
The sound wave is series of rarefaction and compressions traveling through substance . and the waves travel through medium like solid , liquid , and gases