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Luden [163]
3 years ago
7

A physics student stands on the rim of the canyon and drops a rock. The student measures the time for it to reach the bottom to

be 3.2 seconds how deep is the canyon
Physics
1 answer:
zysi [14]3 years ago
5 0

Answer:

50.2 m

Explanation:

We can solve the problem by using the following SUVAT equation for the vertical position of the rock:

y(t)=h+ut+\frac{1}{2}gt^2

where

h is the initial height (the depth of the canyon), taking the bottom of the canyon as reference position

u = 0 is the initial velocity of the rock

t is the time

g=-9.8 m/s^2 is the acceleration of gravity

When the rock reaches the bottom, t = 3.2 s and y = 0. Substituting these numbers and solving for h, we find the depth of the canyon:

h=\frac{1}{2}gt^2 = -\frac{1}{2}(-9.8)(3.2)^2=50.2 m

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Light of wavelength 600nm illuminates a diffraction grating. The second-order maximum is at angle 39°. How many lines per millim
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Explanation:

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Recall the expression for the relation between the wavelength, angle and central maxima distance.

Recall the expression for the relation between the wavelength, angle and central maxima distance.

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