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Alchen [17]
4 years ago
12

She sights two sailboats going due east from the tower. The angles of depression to the two boats are 42o and 29o. If the observ

ation deck is 1,353 feet high, how far apart are the boats?

Physics
1 answer:
Reika [66]4 years ago
3 0

Answer:

The boats are  934.65 feet apart

Explanation:

Given:

The angles of depression to the two boats are 42 degrees and 29 degrees

Height of the observation deck i =  1,353 feet

To Find:

How far apart are the boats (y )= ?

Solution:

<em><u>Step 1 : Finding the value of x(Refer the figure attached)</u></em>

We can use the tangent ratio to find the x value

tan(42^{\circ}) = \frac{1353}{x}

x = \frac{1353}{tan(42^{\circ}) }

x = 590.47 feet

<em><u>Step 2 : Finding the value of  z (Refer the figure attached)</u></em>

tan(29^{\circ}) = \frac{1353}{z }

z  = \frac{1353}{tan(29^{\circ})}

z = 1525.12  feet

<em><u>Step 3 : Finding the value of  y (Refer the figure attached</u></em>)

y =  z -x

y = 1525.12 - 590.47

y = 934.65 feet

Thus the two boats are 934.65 feet apart

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Answer:

Efficiency of the machine = 75%

Explanation:

Given:

Input work = 8,000 J

Output work = 6,000 J

Find:

Efficiency of the machine

Computation:

Efficiency of the machine = [Output work / Input work]100

Efficiency of the machine = [6,000 / 8,000]100

Efficiency of the machine = 75%

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A box with mass m = 8 kg is pushed x = 10 m across a level floor by a constant applied force F P = 16.27 N. The coefficient of k
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Answer:

Final speed of the box after it has moved to x = 10 is given as

v = 3.35 m/s

Explanation:

As we know by work energy theorem that work done by all the forces is equal to the change in its kinetic energy

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so we have

W_{ex} + W_{fric} = \frac{1}{2}mv^2

F x - \mu mg x = \frac{1]{2}mv^2

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4. Calculate the kinetic energy of a 4.7 kg object moving at a speed of 7 m/s. SHOW YOUR WORK
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Answer:

\boxed {\boxed {\sf 115.15 \ J}}

Explanation:

Kinetic energy is the energy an object possesses due to motion. It is calculated with the following formula.

E_K= \frac{1}{2} mv^2

The mass of the object is 4.7 kilograms. The velocity of the object is 7 meters per second.

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  • v= 7 m/s

Substitute the values into the formula.

E_K= \frac{1}{2} (4.7 \ kg)(7 \ m/s)^2

Solve the exponent.

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E_K= \frac{1}{2} (4.7 \ kg)(49 \ m^2/s^2)

Multiply the numbers together.

E_K = 2.35 \ kg * 49 \ m^2/s^2

E_K= 115.15 \ kg*m^2/s^2

Convert the units. 1 kilogram square meter per square second is equal to 1 Joule.

E_K= 115.15 \ J

The object has <u>115.15 Joules</u> of kinetic energy.

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