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cupoosta [38]
1 year ago
14

A street light is at the top of a 19 foot tall pole. A woman 5.25 feet tall walks towards the pole with a speed of 7ft/sec along

a straight path. How fast is the tip of her shadow moving when she is 35ft from the base of the pole?
Physics
1 answer:
bezimeni [28]1 year ago
7 0
Answer:

The tip of her shadow is moving at the speed of 9.66 ft/sec

Explanation:

The height of the street light = 19 feet

The height of the woman = 5.25 feet

Distance between the woman and the base of the pole, x = 35 ft

The speed of the woman towards the pole, dx/dt = 7ft/sec

The distance from the base of the streetlight to the tip of the woman's shadow = y

The distance from the woman to the tip of her shadow = y - x

The diagram illustrating this description is shown below

Using similar triangle:

\begin{gathered} \frac{19}{5.25}=\text{ }\frac{y}{y-x} \\ 19(y-x)\text{ = 5.25y} \\ 19y-19x\text{ = 5.25y} \\ 19y-5.25y\text{ = 19x} \\ 13.75y\text{ = 19x} \\ y\text{ = }\frac{19}{13.75}x \\ y\text{ = }1.38x \end{gathered}

Find the derivative of both sides with respect to time, t

\begin{gathered} \frac{dy}{dt}=\text{ 1.38}\frac{dx}{dt} \\ \frac{dy}{dt}=\text{ 1.38(7)} \\ \frac{dy}{dt}\text{ = }9.66\text{ ft/sec} \end{gathered}

The tip of her shadow is moving at the speed of 9.66 ft/sec

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Answer:

Here is the answer.

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3 years ago
A college student is working on her physics homework in her dorm room. her room contains a total of 6.0×1026 gas molecules. as s
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3 years ago
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.

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If one body is positive, and the other is negative, they tend to attract! 
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