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adelina 88 [10]
2 years ago
6

Randall is able to paddle his canoe at 2.0 m/s in still water. He drops the canoe into a river which is flowing north at 1.5 m/s

. Draw sketches showing how to determine the direction of the magnitude and direction of the velocity of the canoe relative to the Earth using vector addition
a. If he paddles north
b. If he paddles south
c. If he paddles east
Physics
1 answer:
Alecsey [184]2 years ago
4 0
South I’msure he travels south
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An observer at the top of a 462-ft cliff measures the angle of depression from the top of the cliff to a point on the ground to
Lana71 [14]

complete question:

An observer at the top of a 462-ft cliff measures the angle of depression from the top of the cliff to a point on the ground to be 5°. What is the distance from the base of the cliff to the point on the ground? Round to the nearest foot

Answer:

a ≈  5281 ft

Explanation:

The observer at the top of a 462 ft cliff  measures the angle of depression from the top of the cliff to a point on the ground to be 5°.

The angle of depression form the top of the cliff = 5°

The 5° is outside the triangle formed . To find the angle in the triangle we have to subtract 5° from 90°.  90° - 5° = 85° Note sum of an angle on a right angle is 90°.  

using SOHCAHTOA  principle we can solve for the distance from the base of the cliff to the point on the ground(a)

tan  85° = opposite / adjacent

tan 85°  = a / 462

cross multiply

462 × tan 85° = a

a = 11.4300523 × 462

a =  5280.66  ft

a ≈  5281 ft

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3 years ago
Monochromatic light of wavelength 385 nm is incident on a narrow slit. On a screen 3.00 m away, the distance between the second
LiRa [457]

To solve this problem it is necessary to apply the concepts related to the concept of overlap and constructive interference.

For this purpose we have that the constructive interference in waves can be expressed under the function

a sin\theta = m\lambda

Where

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d = Distance of slit to screen

m = Number of order which represent the number of repetition of the spectrum

\theta = Angle between incident rays and scatter planes

At the same time the distance on the screen from the central point, would be

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Where y = Represents the distance on the screen from the central point

PART A ) From the previous equation if we arrange to find the angle we have that

\theta = sin^{-1}(\frac{y}{d})

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PART B) Equation both equations we have

a sin\theta = m\lambda

a \frac{y}{d} = m\lambda

Re-arrange to find a,

a = \frac{(2)(385*10^{-9})(3)}{(1.4*10^{-2})}

a = 1.65*10^{-4}m

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