Two very narrow slits are spaced 1.80 mm apart and are placed 35.0 cm from a screen. What is the distance between the first and
1 answer:
The distance between the first and second dark lines of the interference pattern is mathematically given as
d= 0.107 m
<h3>What is the distance between the first and second dark lines of the interference pattern?</h3>Question Parameters:
Two very narrow slits are spaced 1.80 mm apart and are placed 35.0 cm from a screen
coherent light with 550 nm
Generally, the equation for the distance between the first and second dark lines is mathematically given as
Therefore
d= 550*10^-9*0.35/(1.8*10^-6)
d= 0.107 m
In conclusion, the distance between the first and second dark lines
d= 0.107 m
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Answer:
v = 14.41 m/s
Explanation:
It is given that,
mass of the ball, m = 200 g = 0.2 kg
Height of the roof, h = 12 m
The ball is tossed 1.4 m above the ground, h' = 1.4 m
Let v is the minimum speed with which the ball is tossed. Using the conservation of energy to find it as :
v = 14.41 m/s
So, the minimum speed with which the ball is thrown straight up is 14.41 m/s. Hence, this is the required solution.
Answer:
v = 8.1 m/s
θ = -36.4º (36.4º South of East).
Explanation:
- Assuming no external forces acting during the collision (due to the infinitesimal collision time) total momentum must be conserved.
- Since momentum is a vector, if we project it along two axes perpendicular each other, like the N-S axis (y-axis, positive aiming to the north) and W-E axis (x-axis, positive aiming to the east), momentum must be conserved for these components also.
- Since the collision is inelastic, we can write these two equations for the momentum conservation, for the x- and the y-axes:
- We can go with the x-axis first:
⇒
- Replacing by the givens, we can find vfx as follows:
- We can repeat the process for the y-axis:
⇒
- Replacing by the givens, we can find vfy as follows:
- The magnitude of the velocity vector of the wreckage immediately after the impact, can be found applying the Pythagorean Theorem to vfx and vfy, as follows:
- In order to get the compass heading, we can apply the definition of tangent, as follows:
⇒ tg θ = vfy/vfx = (-4.8m/s) / (6.5m/s) = -0.738 (9)
⇒ θ = tg⁻¹ (-0.738) = -36.4º
- Since it's negative, it's counted clockwise from the positive x-axis, so this means that it's 36.4º South of East.