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2 years ago

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A person is watching a boat from the top of a lighthouse. The boat is approaching the lighthouse directly. When first noticed th

e angle of depression to the boat is 18°33'. When the boat stops, the angle of depression is 51°33'. The lighthouse is 200 feet tall. How far did the boat travel from when it was first noticed until it stopped? Round your answer to the hundredths place.

1 answer:

dexar [7]2 years ago

8 0

$\tan(51^o 33')= \frac{200}{base1}$
<span>
base1= 158.802

</span> $\tan(18^o 33')= \frac{200}{base2}$
<span>
base2=596.008

the distance travelled = 596.008 - 158.802 = 438 m</span>

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Answer:

The option which is:

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Step-by-step explanation:

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Given:

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Let us take number of $5 bills = x and

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Complete Question

The complete question is shown on the first uploaded image

Answer:

A

is dimensionally consistent

B

is not dimensionally consistent

C

is dimensionally consistent

D

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E

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F

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G

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H

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Step-by-step explanation:

From the question we are told that

The equation are

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$B) \ \ x = t$

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$H \ \ \ a = \frac{v}{t} + \frac{xv^2}{2}$

Generally in dimension

x - length is represented as L

t - time is represented as T

m = mass is represented as M

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$a^3 = (\frac{L}{T^2} )^3 = L^3\cdot T^{-6}$

and $\frac{x^2v}{t^5 } = \frac{L^2 L T^{-1}}{T^5} = L^3 \cdot T^{-6}$

Hence

$a^3 = \frac{x^2 v}{t^5}$ is dimensionally consistent

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$x = L$

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$t = T$

Hence

$x = t$ is not dimensionally consistent

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$v = LT^{-1}$

and

$\frac{x^2 }{at^3} = \frac{L^2}{LT^{-2} T^{3}} = LT^{-1}$

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$x = L$

;

$vt = LT^{-1} T = L$

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$x = L$

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$v = LT^{-1}$

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$at = LT^{-2}T = LT^{-1}$

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$a = LT^{-2}$

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$\frac{v}{t} = \frac{LT^{-1}}{T} = LT^{-2}$

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Answer:

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