Let us consider the air with the index 1 and the lucite with index 2. Using the Snell's Secound Law, we have:
Entering the unknowns, remembering that the air refrective index is 1 and the lucite refrective index is 1.5, comes:
Using the arcsin properties, we get:
Obs: Approximate results, and the drawing is attached
If you notice any mistake in my english, let me know, because i am not native.
Precipitation has the greatest affect on wind speed.<span>
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As the spring returns to it's equilibrium position, it performs
1/2 (4975 N/m) (0.097 m)² ≈ 23 J
while the gravitational force (opposing the block's upward motion) performs
-(0.244 kg) g<em> </em>(0.097 m) ≈ -2.3 J
of work on the block. By the work energy theorem, the total work done on the block is equal to the change in its kinetic energy:
23 J - 2.3 J = 1/2 (0.244 kg) v² - 0
where v is the speed of the block at the moment it returns to the equilibrium position. Solve for v :
v² = (23 J - 2.3 J) / (1/2 (0.244 kg))
v = √((23 J - 2.3 J) / (1/2 (0.244 kg)))
v ≈ 44 m/s
After leaving the spring, block is in free fall, and at its maximum height h it has zero vertical velocity.
0² - (44 m/s)² = 2 (-g) h
Solve for h :
h = (44 m/s)² / (2g)
h ≈ 2.3 m
Explanation:
Given that,
The mass of rock, m = 2.35-kg
It was released from rest at a height of 21.4 m.
(a) The kinetic energy is given by :
As the rock was at rest initially, it means, its kinetic energy is equal to 0.
(b) The gravitational potential energy is given by :
It can be calculated as :
(c) The mechanical energy is equal to the sum of kinetic and potential energy such that,
M = 0 J + 492.84 J
M = 492.84 J
Hence, this is the required solution.