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3 years ago

A box is dragged 16 m across a level floor by a 25-N force at an angle of 45° to the floor. It is then dragged by

Mathematics

1 answer:

WARRIOR [948]3 years ago

6 0

<u>solution</u>

W=f×d×cos theta

=25N×16m×cos 45°

=400Nm×cos 45°

=282.84J

W=f×d×cos theta

=25N×7m×cos 10°

=175Nm×cos10°

=172.34J

Total workdone=282.84J+172.34J=455.18J

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Answer:

Its true ⇒ answer (a)

Step-by-step explanation:

* The solution of two linear equation means that this solution

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* To check if the point is a solution of an equation

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* Lets study our problem

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One of the same-side exterior angles formed by two lines and a transversal is equal to 1/6 of the right angle and is 11 times sm

sergey [27]

One of the same-side exterior angles formed by two lines and a transversal is equal to 1/6 of the right angle and is 11 times smaller than the other angle. Then the lines are parallel

<h3><u>Solution:</u></h3>

Given that, One of the same-side exterior angles formed by two lines and a transversal is equal to 1/6 of the right angle and is 11 times smaller than the other angle.

We have to prove that the lines are parallel.

If they are parallel, sum of the described angles should be equal to 180 as they are same side exterior angles.

Now, the 1st angle will be 1/6 of right angle is given as:

$\begin{array}{l}{\rightarrow 1^{\text {st }} \text { angle }=\frac{1}{6} \times 90} \\\\ {\rightarrow 1^{\text {st }} \text { angle }=15 \text { degrees }}\end{array}$

And now, 15 degrees is 11 times smaller than the other

Then other angle = 11 times of 15 degrees

$\text {Other angle }=11 \times 15=165 \text { degrees }$

Now, sum of angles = 15 + 165 = 180 degrees.

As we expected their sum is 180 degrees. So the lines are parallel.

Hence, the given lines are parallel

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