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Kryger [21]
3 years ago
14

A Texas rancher wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in

the figure below, where A = 4.90 km and θC = 15°.

Physics
1 answer:
xz_007 [3.2K]3 years ago
4 0
The answer:
the full question is as follow:
 <span>A Texas rancher wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in Figure below , where A = 4.90 km and θC = 15°. He then correctly calculates the length and orientation of the fourth side D. What is the magnitude and direction of vector D? 

As shown in the figure, 
A + B + C + D = 0, so to find the </span>magnitude and direction of vector D, we should follow the following method:
 D = 0 - (A + B + C) , 
let  W = - (A + B + C), so the magnitude and direction of vector D is the same of the vector W characteristics

Magnitude
 A + B + C = <span> (4.90cos7.5 - 2.48sin16 - 3.02cos15)I</span>
<span>+ (-4.9sin7.5 + 2.48cos16 + 3.02sin15)J
</span>= 1.25I +2.53J
the magnitude of W= abs value of (A + B + C) = sqrt (1.25² + 2.53²)
= 2.82
 
the direction of D can be found by using Dx and Dy value
we know that     tan<span>θo = Dx / Dy = 1.25 / 2.53 =0.49
</span>tanθo =0.49 it implies θo = arctan 0.49 = 26.02°

direction is 26.02°

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Answer:

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Explanation:

The work done by a force on an object is given by the following formula:

W = F.d

W = F d Cosθ

where,

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Since, force P is parallel to the motion of the box. Therefore, θ = 0°

Hence,

W = P d Cos 0°

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<u>Therefore, work done by force (P) is Positive.</u>

<u></u>

<u>FOR NORMAL FORCE (Fn) AND WEIGHT (W)</u>:

Since, normal force and weight are perpendicular to the motion of the box. Therefore, θ = 90°

Hence,

W = Fn d Cos 90°= mg d Cos 90°

W = Fn d(0) = mg d (0)

W = 0

<u>Therefore, work done by Normal Force (Fn) and Weight (mg) is Zero.</u>

<u></u>

<u>FOR KINETIC FRICTIONAL FORCE (fk)</u>:

Since, kinetic frictional force acts in the opposite direction of motion of the box. Therefore, θ = 180°

Hence,

W = fk d Cos 180°

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<u>Therefore, work done by Kinetic Frictional Force (fk) is Negative.</u>

<u></u>

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