A force of 50 pounds is exerted at an angle of 80 with the horizontal. What is its horizontal component?

Mathematics

1 answer:

Shalnov [3]3 years ago

6 0

Answer:

Option A is the correct answer.

Explanation:

Any vector can be resolved in to two components. Horizontal component and vertical component.

Consider a vector F which is at angle θ⁰ to the horizontal, we can resolve this vector in to two.

Horizontal component = F cos θ

Vertical component = F sin θ

Here we have Force , F = 50 pounds

Angle with horizontal = 80°

Horizontal component = F cos θ = 50* cos 80 = 8.68 pounds

≅ 9 lb.

Option A is the correct answer.

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alexira [117]

Answer:

=7/6

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5 0

3 years ago

Read 2 more answers

Multiply.

Crank

The main rules that we use here are :

i) $\sqrt{a \cdot b}= \sqrt{a} \cdot \sqrt{b}$ for nonnegative values a and b.

ii) $\sqrt{a} \cdot \sqrt{a} =a$ .

Thus, first 'decompose' the numbers in the radicals into prime factors:

$4 \sqrt{3} \cdot 10 \cdot \sqrt{2\cdot2\cdot3}\cdot \sqrt{2\cdot 3}\cdot \sqrt{2}.$ .

By rule (i) we write:

$4 \sqrt{3} \cdot 10 \cdot \sqrt{2}\cdot \sqrt{2}\cdot \sqrt{\cdot3}\cdot \sqrt{2} \cdot \sqrt{3} \cdot \sqrt{2}$ .

We can collect these terms as follows:

$40 \sqrt{3}\cdot (\sqrt{3}\cdot \sqrt{3}) \cdot (\sqrt{2}\cdot \sqrt{2}) \cdot( \sqrt{2} \cdot \sqrt{2})$ , and by rule (ii) we have:

$40 \sqrt{3}\cdot 3 \cdot 2 \cdot2=40\cdot12\cdot \sqrt{3}=480 \sqrt{3}.$

Answer: $480 \sqrt{3}$ .