Answer:
distance between the two second-order minima is 2.8 cm
Explanation:
Given data
distance = 1.60 m
central maximum = 1.40 cm
first-order diffraction minima = 1.40 cm
to find out
distance between the two second-order minima
solution
we know that fringe width = first-order diffraction minima /2
fringe width = 1.40 /2 = 0.7 cm
and
we know fringe width of first order we calculate slit d
β1 = m1λD/d
d = m1λD/β1
and
fringe width of second order
β2 = m2λD/d
β2 = m2λD / ( m1λD/β1 )
β2 = ( m2 / m1 ) β1
we know the two first-order diffraction minima are separated by 1.40 cm
so
y = 2β2 = 2 ( m2 / m1 ) β1
put here value
y = 2 ( 2 / 1 ) 0.7
y = 2.8 cm
so distance between the two second-order minima is 2.8 cm
Answer:
(A)
Explanation:
P1 = MVcos 30
P2 = MVcos 30
️P = -2 mvcos 30
️P = - square root of 3 mv
f1 = 2mv/ ️t
force on the wall = 2
f2 = square root of 3mv/ ️t, so f1 > f2
It's not so much a "contradiction" as an approximation. Newton's law of gravitation is an inverse square law whose range is large. It keeps people on the ground, and it keeps satellites in orbit and that's some thousands of km. The force on someone on the ground - their weight - is probably a lot larger than the centripetal force keeping a satellite in orbit (though I've not actually done a calculation to totally verify this). The distance a falling body - a coin, say - travels is very small, and over such a small distance gravity is assumed/approximated to be constant.
Density = (mass) / (volume)
Mass = 0.150 kg = 150 grams
Volume = (length) · (width) · (height)
Volume = (2.5 cm) · (4.3 cm) · (1.98 cm)
Volume = 21.285 cm³
Density = (150 g) / (21.285 cm³)
Density = 7.05 g/cm³
Answer:
For a plane mirror, the image distance equals the object distance, so the image distance will increase as the object distance increases
The height of the image stays the same and the image distance increases.)
Explanation:
For plane mirrors, the object distance (is equal to the image distance. That is the image is the same distance behind the mirror as the object is in front of the mirror. If you stand a distance of 2 meters from a plane mirror, you must look at a location 2 meters behind the mirror in order to view your image
