Answer:
Part a)
Width of the slit is

Part b)
Ratio of intensity is given as

Explanation:
Part a)
As we know by the formula of diffraction we will have

so we have


so we will have


Part b)
As we know that the intensity in diffraction pattern is given as


so for angle 45 degree



Answer:
ΔK.E = 14 nJ
Explanation:
Solution:
- The charge that moves under the influence of an Electric Field produced between a potential difference (V) stores electric potential energy U within that is converted to kinetic energy.
- We will use conservation of energy on the system that contains the charged particle with charge q loses its electric potential energy U as it moves towards positively charged object that converts into a gain in Kinetic energy of the charged particle ΔK.E:
ΔK.E = U
Where,
U = V*q
ΔK.E = V*q
ΔK.E = (7*10^-6)*(2*10^-3)
ΔK.E = 14 nJ
- The gain in kinetic energy is 14 nJ.
The answer is B.
This is because you add up all of the times (1.44s+1.70s+1.58s+1.76s) and you get 6.48 then you divide 6.48 by 4 to get the average of the times. Now you get the distance (200m) and because speed=distance/time you divide 200m/1.62s to get 123m/s. I hope this made sense :)
Answer:
the balls reached a height of 4.9985 m
Explanation:
Given the data in the question;
mass one m = 3.8 kg
mass two M = 2.1 kg
Initial velocities
u = 22 m/s
U = { moving downward} = 12 m/s
Now, using the law conservation of linear moment;
mu + MU = v( m + M )
we solve for "v" which is the velocity of the ball s after collision;
v = (mu + MU) / ( m + M )
so we substitute our given values into the equation
v = ( ( 3.8 × 22 ) + ( 2.1 × -12) ) / ( 3.8 + 2.1 )
v = ( 83.6 - 25.2 ) / 5.9
v = 58.4 / 5.9
v = 9.898 m/s
Now, we determine required height using the following relation;
v"² - v² = 2gh
where v" is the velocity at the top which is 0 m/s and g = -9.8 m/s²
0 - v² = 2gh
v² = -2gh
so we substitute
( 9.898 )² = -2 × -9.8 × h
97.97 = 19.6 × h
h = 97.97 / 19.6
h = 4.9985 m
Therefore, the balls reached a height of 4.9985 m
The true answer is the choice A
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