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bogdanovich [222]
1 year ago
15

A person exerts a 80 J of work by pulling a block along a horizontal surface with a

Mathematics
1 answer:
Gala2k [10]1 year ago
7 0

According to the solving the displacement  of the block is:

= 40m

<h3>What does force-displacement mean?</h3>

A shift inside of an object's position is referred to as "displacement." It has a magnitude and a direction, making it a vector quantity. An arrow pointing from of the starting point to the finishing point serves as its symbol. For instance, an object's position changes if it moves from position A to position B.

<h3>Why is there a shift in the workforce?</h3>

The scalar product of displacement and force is equivalent to work. The cosine of the angle between two vectors is added to their lengths to form the scalar product of the two vectors. Work is defined as the product of force, displacement, and the sine of the angle formed by the force's and displacement's directions.

<h3>According to the given data:</h3>

The formula for work is W=F × d.

Work done  = 80J

Force = 20N

Displacement = ?

So,

W=F × d.

80 = 20 x d

d = 80/20

d = 40 m

According to the solving the displacement  of the block is:

= 40m

To know more about displacement visit:

brainly.com/question/11934397

#SPJ9

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If cos() = − 2 3 and is in Quadrant III, find tan() cot() + csc(). Incorrect: Your answer is incorrect.
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Answer:

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}

Step-by-step explanation:

Given

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\theta \to Quadrant III

Required

Determine \tan(\theta) \cdot \cot(\theta) + \csc(\theta)

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We know that:

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This gives:

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\sin^2(\theta)  = \frac{9 -4}{9}

\sin^2(\theta)  = \frac{5}{9}

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Sin is negative in quadrant III. So:

\sin(\theta)  = -\frac{\sqrt 5}{3}

Calculate \csc(\theta)

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\csc(\theta) = \frac{1}{-\frac{\sqrt 5}{3}}

\csc(\theta) = \frac{-3}{\sqrt 5}

Rationalize

\csc(\theta) = \frac{-3}{\sqrt 5}*\frac{\sqrt 5}{\sqrt 5}

\csc(\theta) = \frac{-3\sqrt 5}{5}

So, we have:

\tan(\theta) \cdot \cot(\theta) + \csc(\theta)

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \tan(\theta) \cdot \frac{1}{\tan(\theta)} + \csc(\theta)

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 + \csc(\theta)

Substitute: \csc(\theta) = \frac{-3\sqrt 5}{5}

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 -\frac{3\sqrt 5}{5}

Take LCM

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}

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