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Andru [333]
3 years ago
10

To measure the height of the cloud cover at an airport, a worker shines a spotlight upward at an angle 75° from the horizontal.

An observer D = 510 m away measures the angle of elevation to the spot of light to be 45°. Find the height h of the cloud cover. (Round your answer to the nearest meter.)
Physics
1 answer:
Yuri [45]3 years ago
4 0

Answer: The height of the cloud = 394.55 m

Explanation:

The observer is 500m away from the spotlight.

Let x be the distance from the observer to the interception of the segment of the height, h with the floor. The equations are thus:

Tan 45° = h/x ... eq1

Tan 75° = h/(500- x ) ... eq2

From eq 1, Tan 45° = 1, therefore eq1 becomes:

h = x ... eq3

Put eq3 into eq2

Tan 75° = h/(500- h)

h = ( 500 - h ) Tan 75°

h = 500Tan 75° - hTan75°

h + h Tan 75° = 500 Tan 75°

h ( 1 + Tan 75° ) = 500 Tan75°

h = 500Tan75°/ (1 + Tan 75°)

h= 1866.02 / 4.73

h = 394.55m

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The component of the force due to gravity perpendicular and parallel to the slope is  113.4 N and 277.8 N respectively.

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<u><em>The component of the force due to gravity perpendicular and parallel to the slope is  113.4 N and 277.8 N respectively.</em></u>

6 0
3 years ago
A charged box ( m = 495 g, q = + 2.50 μ C ) is placed on a frictionless incline plane. Another charged box ( Q = + 75.0 μ C ) is
Rudik [331]

Answer:

a=16.2m/s^{2}

Explanation:

From the attached file diagram, the total force acting on the charged box is the downward weight and the repulsive force acting in opposite to the weight force . Hence we can write the total force as

F=masin\alpha -\frac{kQq}{r^{2}} \\\alpha =35^{0}, m=0.495kg, r=0.61m, Q=2.5*10^{-6}, q=75.0*10^{-6}\\

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The value of the acceleration is 16.2m/s^2

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