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victus00 [196]
3 years ago
10

Two observers are 300 ft apart on opposite sides of a flagpole. The angles of elevation from the observers to the top of the pol

e are 16degrees and 20degrees. Find the height of the flagpole.
Physics
1 answer:
valentina_108 [34]3 years ago
3 0

Answer:

h = 48.077 ft

Explanation:

given,

distance between two observer = 300 ft

angle of elevation to top pole = 16° and 20°

height of the flagpole = ?

now,

Let h be the height of the flagpole

Let x be the distance of the pole

tan 16^0 = \dfrac{h}{x}

x =\dfrac{h}{tan 16^0}

now,

again applying

tan 20^0 = \dfrac{h}{300-x}

300-x=\dfrac{h}{tan 20^0}

300-3.49 h=2.75 h

6.24h = 300

h = 48.077 ft

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4 0
2 years ago
Read 2 more answers
Can someone help?
laila [671]

Answer:

i) 21 cm

ii) At infinity behind the lens.

iii) A virtual, upright, enlarged image behind the object

Explanation:

First identify,

object distance (u) = 42 cm (distance between  object and lens, 50 cm - 8 cm)

image distance (v) = 42 cm (distance between  image and lens, 92 cm - 50 cm)

The lens formula,

\frac{1}{v} -\frac{1}{u} =\frac{1}{f}

Then applying the new Cartesian sign convention to it,

\frac{1}{v} +\frac{1}{u} =\frac{1}{f}

Where f is (-), u is (+) and  v is (-) in  all 3  cases. (If not values with signs have to considered, this method that need will not arise)

Substituting values you get,

i) \frac{1}{42} +\frac{1}{42} =\frac{1}{f}\\\frac{2}{42} =\frac{1}{f}

f = 21 cm

ii) u =21 cm, f = 21 cm v = ?

Substituting in same equation\frac{1}{v} =\frac{1}{21} =\frac{1}{21} \\\\\frac{1}{v} = 0\\

  v ⇒ ∞ and image will form behind the lens

iii) Now the object will be within the focal length of the lens. So like in the attachment, a virtual, upright, enlarged image behind the object.

7 0
3 years ago
Normal faults occur from _____ stress and reverse faults occur from _____ stress.
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Normal faults can be caused by physical stress. 
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