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2 years ago

11

Please help me asap i will mark a branlist number 7

2 answers:

Paha777 [63]2 years ago

8 0

Answer:

B

Step-by-step explanation:

12^2=144

13^2=169

and 150 is more than 12^2 and less than 13^2

SVETLANKA909090 [29]2 years ago

6 0

Answer:

When doing square roots the opposite is squaring so, to find one that is more than 12 and less that 13 would be finding one that is in between √144 and √169

That leaves your answer as √150, because it is more than √144 (which is 12) and less than √169 (which is 13)

B √150

Hope this helps ;)

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Answer with Step-by-step explanation:

We are given that

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The component of force is divided into two direction

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1.The vector parallel to the plane will be= $r=cos\frac{\pi}{3}i-sin\frac{\pi}{3}j=\frac{1}{2}i-\frac{\sqrt 3}{2}j$

By using $cos\frac{\pi}{3}=\frac{1}{2},sin\frac{\pi}{3}=\frac{\sqrt 3}{2}$

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By using $i\cdot i=j\cdoty j=k\cdot k=1,i\cdot j=j\cdot k=k\cdot i=j\cdot i=k\cdot j=i\cdot k=0$

Therefore, force along the plane= $\mid F_x\mid(\frac{1}{2}i-\frac{\sqrt 3}{2}j)=4\sqrt 3(\frac{1}{2}i-\frac{\sqrt 3}{2}j)$

2.The vector perpendicular to the plane= $r=-sin\frac{\pi}{3}-cos\frac{\pi}{3}=-\frac{\sqrt 3}{2}i-\frac{1}{2}j$

The force perpendicular to the plane= $\mid F_y\mid=F\cdot r=-8j(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)$

The force perpendicular to the plane= $4$ N

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Sum of two component of force= $F_x+F_y=4\sqrt 3(\frac{1}{2}i-\frac{\sqrt 3}{2}j)+4(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)$

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Step-by-step explanation:

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Step-by-step explanation:

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Step-by-step explanation:

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