B4 the tackle:
<span>The linebacker's momentum = 115 x 8.5 = 977.5 kg m/s north </span>
<span>and the halfback's momentum = 89 x 6.7 = 596.3 kg m/s east </span>
<span>After the tackle they move together with a momentum equal to the vector sum of their separate momentums b4 the tackle </span>
<span>The vector triangle is right angled: </span>
<span>magnitude of final momentum = √(977.5² + 596.3²) = 1145.034 kg m/s </span>
<span>so (115 + 89)v(f) = 1145.034 ←←[b/c p = mv] </span>
<span>v(f) = 5.6 m/s (to 2 sig figs) </span>
<span>direction of v(f) is the same as the direction of the final momentum </span>
<span>so direction of v(f) = arctan (596.3 / 977.5) = N 31° E (to 2 sig figs) </span>
<span>so the velocity of the two players after the tackle is 5.6 m/s in the direction N 31° E </span>
<span>btw ... The direction can be given heaps of different ways ... N 31° E is probably the easiest way to express it when using the vector triangle to find it</span>
Answer:
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Explanation:
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Explanation:
We need to calculate the speed of light in each materials
(I). Gallium phosphide,
The index of refraction of Gallium phosphide is 3.50
Using formula of speed of light
....(I)
Where,
= index of refraction
c = speed of light
Put the value into the formula


(II) Carbon disulfide,
The index of refraction of Gallium phosphide is 1.63
Put the value in the equation (I)


(III). Benzene,
The index of refraction of Gallium phosphide is 1.50
Put the value in the equation (I)


Hence, This is the required solution.