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lidiya [134]
3 years ago
8

From an h = 53 feet observation tower on the coast, a Coast Guard officer sights a boat in difficulty. The angle of depression o

f the boat is θ = 4 ◦ . How far (in feet) is the boat from the shoreline? Answer in units of feet.

Physics
1 answer:
maksim [4K]3 years ago
4 0

Answer:

757,93 feets

Explanation:

We can make a right triangle between the boat (A), the Coast Guard officer (B) and the base of the observation tower (C), like in the graph attached. Now, you could also made a rectangle, adding the horizontal at the height of the Coast Guard, starting in B and ending in D, the vertex opossing C.

The angle of depression, its O in the graph.

Now, as we got an rectangle, of course, the segment AD its the same length as CB, and CA, the distance from the boat to shoreline, its the same length as DB.

ADB its an right triangle, with AB, the hypothenuse, and BD and DA, the catheti (or <em>legs</em>).

Now, we know the lenght BC, the height of the tower, 53 feets, so we also know the lenght of DA. DA its the opposite cathetus to the angle O. We wish to know the length AC, equal to the lenght DB, the adjacent cathetus of the angle O.

Know, the trigonometric function that connects the adjacent cathetus with the opossite cathetus its the tangent.

tangent( O ) = \frac{opposite}{adjacent}

We can take that the angle O = 4 °, and knowing that the opossite cathetus its 53 feets, we got:

tangent( 4) = \frac{53 feets}{DB}

DB=  \frac{53 feets}{tangent( 4)}

DB=  757,93 feets

This its equal to the distance from the boat to the shoreline.

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Some car manufacturers claim that their vehicles could climb a slope of 42 ∘. For this to be possible, what must be the minimum
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Answer:

D. 0.9

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Calculating minimum coefficient of static friction, we first resolve the forces (normal and frictional) acting on the vehicle at an angle to the horizontal into their x and y components. After this, we can now substitute the values of x and y components into equation of static friction. Diagrammatic illustration is attached.

Resolving into x component:

                        ∑F_{x} = F_{s} - mgsin\alpha =0

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Resolving into y component:

                        ∑F_{y} = F_{n} - mgcos\alpha =0

                          F_{n} = mgcos\alpha      ------(2)

Static frictional force, F_{s} \leq μ F_{n}       ------(3)

substituting F_{s} from equation (1) and F_{n} from equation (2) into equation (3)

                         mgsin\alpha \leq μ mgcos\alpha

                         sin\alpha \leq μ cos\alpha

                         μ \geq \frac {sin\alpha}{cos\alpha}

                         μ \geq tan\alpha

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A researcher measures the thickness of a layer of benzene (nn = 1.50) floating on water by shining monochromatic light onto the
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Answer:

The minimum thickness is t= 8.75*10^{-8} m

Explanation:

generally the equation for thin film interference is mathematically represented as

            2nt = (m + \frac{1}{2} )  \lambda

Where t the  thickness  

           m is any  integer

            n is the refractive index of the film

            \lambda is the wavelength of light

Since we are looking for the thickness we make t the subject of the  formula

          t = \frac{(m+ \frac{1}{2} ) \lambda}{2n}

m= 0 cause the thickness is minimum at m=0

   Substituting values

                    t = \frac{(0 +\frac{1}{2}) 8525*10^{-9} }{2 *1.5}

                       t= 8.75*10^{-8} m

8 0
3 years ago
Read 2 more answers
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