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Pepsi [2]
3 years ago
6

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?
Physics
1 answer:
Lana71 [14]3 years ago
5 0

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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Answer:

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Explanation:

First, list all known and desired values/variables (initial vertical velocity is 0 as the plane is kept level and vertical acceleration is just gravity):

Vertical \ velocity \ (\frac{m}{s} ) =  u_{v} = 0 \\\\ Horizontal \ velocity \ (\frac{m}{s} ) =  u_{h} = 200\\\\ Vertical \ acceleration \ (\frac{m}{s^{2} } ) =  a_{v} =  9.8 \\\\ Horizontal \ acceleration \ (\frac{m}{s^{2} } ) =  a_{h} =  0 \\\\ Vertical \ displacement \ (m) = s_{v} = 800 \\\\ Horizontal \ displacement \ (m) = s_{h}

The horizontal displacement is going to be the distance travelled, horizontally of course, once the package is released;

First thing to understand is that the vertical and horizontal components are to be dealt with separately because they don't affect each other;

Since there is no horizontal acceleration (ignoring air resistance), we simply require a velocity and time to find the horizontal displacement, using the formula v = d/t (or speed = distance/time);

What we have is the horizontal velocity but we don't have the time taken;

One thing we know is that the time elapsed for the vertical fall of 800m and for the horizontal displacement must be the same;

What we do, therefore, is find the time taken for the vertical displacement using the formula, s = ut + ¹/₂·at², since we know the vertical velocity, height and acceleration:

800 = (0)t + ¹/₂·(9.8)t²

800 = 4.9t²

t² = 163.26...

t = 12.77...

We now have the time taken for the vertical fall and the horizontal displacement, we can use this with the horizontal velocity we know already and get the horizontal displacement:

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Hope this helps!
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4 years ago
Find the work w1 done on the block by the force of magnitude f1 = 95.0 n as the block moves from xi = -5.00 cm to xf = 4.00 cm .
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<h3><u>Answer;</u></h3>

= 8.55 Joules

<h3><u>Explanation;</u></h3>

Work done is the product of force and the distance moved by an object.

Work done = Force × distance

Force = 95 Newtons

Distance = X2 -X1

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Thus;

work done = 95 × 9/100

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5 0
3 years ago
7. A car moving at 10m/s (about 22.4 mph) crashes into a barrier and stops in 0.25 m.
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Answer:

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a)When car is stopped its final velocity become zero

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V- 0 m/s

S - 0.25 m

t -?

S = (v+u)*t/2

0.25 =(10+0)*t/2

t = 0.05s

b) If we happened to calculate the avarage force we have to consider about acceleration

V= 0

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F = m *a

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Answer:

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Q_{in} = 13.742\,kW

b) The amount of entropy generation is determined by the Second Law of Thermodynamics:

\Delta S = \frac{Q_{in}}{T_{in}}

\Delta S = \frac{13.742\,kJ}{370.15\,K}

\Delta S = 370.15\,\frac{kJ}{K}

3 0
4 years ago
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