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MAXImum [283]
3 years ago
11

A lawn mower is pushed a distance of 100 ft. Along a horizontal path by a constant force of 60 lbs. Yhe handle of the lawn mower

is held at a downward angle of 45° toward the horizontal. Explain what each value in this situation represents in relation to the formula w=F•PQ=||F||•||PQ||costheta. Then find the work done pushing the lawn mower
Mathematics
1 answer:
olga2289 [7]3 years ago
6 0
For this case we have the following equation:
 w = || F || • || PQ || costheta
 Where,
 || F ||: force vector module
 || PQ ||: distance module
 costheta: cosine of the angle between the force vector and the distance vector.
 Substituting values:
 w = (60) * (100) * (cos (45))
 w = 4242.640687 lb.ft
 Answer:
 
The work done pushing the lawn mower is: 
 w = 4242.640687 lb.ft
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A)

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A ship sails 250km due North qnd then 150km on a bearing of 075°.1)How far North is the ship now? 2)How far East is the ship now
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Answer:

1)  288.8 km due North

2)  144.9 km due East

3)  323.1 km

4)  207°

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<u>Bearing</u>: The angle (in degrees) measured clockwise from north.

<u>Trigonometric ratios</u>

\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}

where:

  • \theta is the angle
  • O is the side opposite the angle
  • A is the side adjacent the angle
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<u>Cosine rule</u>

c^2=a^2+b^2-2ab \cos C

where a, b and c are the sides and C is the angle opposite side c

-----------------------------------------------------------------------------------------------

Draw a diagram using the given information (see attached).

Create a right triangle (blue on attached diagram).

This right triangle can be used to calculate the additional vertical and horizontal distance the ship sailed after sailing north for 250 km.

<u>Question 1</u>

To find how far North the ship is now, find the measure of the short leg of the right triangle (labelled y on the attached diagram):

\implies \sf \cos(75^{\circ})=\dfrac{y}{150}

\implies \sf y=150\cos(75^{\circ})

\implies \sf y=38.92285677

Then add it to the first portion of the journey:

⇒ 250 + 38.92285677... = 288.8 km

Therefore, the ship is now 288.8 km due North.

<u>Question 2</u>

To find how far East the ship is now, find the measure of the long leg of the right triangle (labelled x on the attached diagram):

\implies \sf \sin(75^{\circ})=\dfrac{x}{150}

\implies \sf x=150\sin(75^{\circ})

\implies \sf x=144.8888739

Therefore, the ship is now 144.9 km due East.

<u>Question 3</u>

To find how far the ship is from its starting point (labelled in red as d on the attached diagram), use the cosine rule:

\sf \implies d^2=250^2+150^2-2(250)(150) \cos (180-75)

\implies \sf d=\sqrt{250^2+150^2-2(250)(150) \cos (180-75)}

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To find the bearing that the ship is now from its original position, find the angle labelled green on the attached diagram.

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\implies \sf Bearing=180^{\circ}+26.64077...^{\circ}

\implies \sf Bearing=207^{\circ}

Therefore, as bearings are usually given as a three-figure bearings, the bearing of the ship from its original position is 207°

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