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2 years ago

Three forces of 300 N in the direction of N30E, 400N in the direction of N60E and

Mathematics

1 answer:

andreyandreev [35.5K]2 years ago

4 0

Split up each force into horizontal and vertical components.

• 300 N at N30°E :

(300 N) (cos(30°) i + sin(30°) j)

• 400 N at N60°E :

(400 N) (cos(60°) i + sin(60°) j)

• 500 N at N80°E :

(500 N) (cos(80°) i + sin(80°) j)

The resultant force is the sum of these forces,

∑ F = (300 cos(30°) + 400 cos(60°) + 500 cos(80°)) i

… … … + (300 sin(30°) + 400 sin(60°) + 500 sin(80°)) j N

∑ F ≈ (546.632 i + 988.814 j) N

so ∑ F has a magnitude of approximately 1129.85 N and points in the direction of approximately N61.0655°E.

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Before conducting some experiments, a scientist mixes 1/2 gram of Substance A with 3/4 grams of Substance B. If the scientist us

Tju [1.3M]

Answer: 10

<u>Step-by-step explanation:</u>

$(\frac{1}{2}+\frac{3}{4})$ ÷ $\frac{1}{8}$

= $(\frac{1}{2}(\frac{2}{2})+\frac{3}{4})$ ÷ $\frac{1}{8}$

= $(\frac{2}{4}+\frac{3}{4})$ ÷ $\frac{1}{8}$

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3 years ago

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Gre4nikov [31]

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2 years ago

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klemol [59]

Answer:

It should be 1,628 $m^{2}$ .

Step-by-step explanation:

We see two shapes in the figure and those shapes are a rectangle and a circle. Let's find the area of the circle first. 25m won't help us find the area of the circle so let's pretend that 25m isn't there for now. 40m seems to be the diameter of the circle and to find the area of the circle we need to multiply the radius squared by Pi or 3.14. Half of 40 is 20 so we can multiply 20 squared by Pi to give us 1,256.63706. Do not round this number yet. As you can see this circle isn't a full circle. It's a semicircle. We can divide 1,256.63706 by 2 to find the area of a semicircle. You should get 628.31853 and round that to the nearest tenth and we get 628.32. Now let's keep that number in mind and find the area of the rectangle. Using a calculator we can easily multiply 40 and 25 to find the area of the rectangle, which is 1,000. Add 1,000 and 628.32 and our final results should be 1,628.32. I don't see this number in your options but option B is closely related to this answer. I hope this helps and if this answer is wrong then please give me some feedback on what I did wrong! Thank you!

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3 years ago

At a high school, students can choose between three art electives, four history electives, and five computer electives. Each stu

V125BC [204]

Answer:

$\frac{^3C_1\times ^4C_1}{^{12}C_2}$

Step-by-step explanation:

Given,

Art electives = 3,

History electives = 4,

Computer electives = 5,

Total number of electives = 3 + 4 + 5 = 12,

Since, if a student chooses an art elective and a history elective,

So, the total combination of choosing an art elective and a history elective = $^3C_1\times ^4C_1$

Also, the total combination of choosing any 2 subjects out of 12 subjects = $^{12}C_2$

Hence, the probability that a student chooses an art elective and a history elective = $\frac{\text{Total combination of choosing an art elective and a history}}{\text{ Total combination of choosing any 2 subjects}}$