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MatroZZZ [7]
3 years ago
7

I WILL FAIL IF I DONT GET THIS DONE, CAN SOMEBODY PLEASE HELP ME?

Physics
1 answer:
ki77a [65]3 years ago
3 0
Plants that died a thousands of years ago. 
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3 years ago
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Three forces act on a flange as shown below. Determine the magnitude of the unknown force F (in lb) such that the net force acti
Tems11 [23]

Answer:

The unknown force will be 18.116 lb.

Explanation:

Given that,

Three forces act on a flange as shown in figure.

The net force acting on the flange is a minimum.

\dfrac{dF_{net}}{df}=0

We need to calculate the unknown force

Using formula of net force

\vec{F_{net}}=\vec{F_{x}}+\vec{F_{y}}

Put the value into the formula

\vec{F_{net}}=(F\cos45+70\cos30-40)\hat{i}+(70\sin30-F\sin45)\hat{j}

\vec{F_{net}}=(F\cos45+70\times\dfrac{\sqrt{3}}{2})\hat{i}+(70\times\dfrac{1}{2}-F\sin45)\hat{j}

The magnitude of net force,

F_{net}=\sqrt{F_{x}^2+F_{y}^2}

F_{net}=\sqrt{(F\times\dfrac{1}{\sqrt{2}}+60.62)^2+(35-F\times\dfrac{1}{\sqrt{2}})^2}

F_{net}=\sqrt{F^2+(60.62)^2+121.24\times\dfrac{F}{\sqrt{2}}+(35)^2-70\times\dfrac{F}{\sqrt{2}}}

F_{net}=\sqrt{F^2+4899.78+36.232F}

On differentiating w.r.to F

(\dfrac{dF_{net}}{dF})^2=2F+36.232

0=2F+36.232

F=-\dfrac{36.232}{2}

F=-18.116\ lb

Negative sign shows the direction of force which is downward.

Hence, The unknown force will be 18.116 lb.

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3 years ago
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Answer: 13inch

Explanation:

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4 years ago
Which song form has alternating repetitions of a main theme which two or more contrasting sections?
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The answer to this item is rondo, letter A. As already mentioned in the above's definition, this has a recurring or repetitive lead them two contrasting sections. The contrasting sections are more commonly referred to as "episodes" and occasionally as "digressions" or "couplets". 
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A block of 200 g is attached to a light spring with a force constant of 5 N / m and freely in a horizontal plane vibrates. The m
Finger [1]

Answer:

m  200 g , T  0.250 s,E 2.00 J

;

2 2 25.1 rad s

T 0.250

 

   

(a)

 

2 2

k m    0.200 kg 25.1 rad s 126 N m

(b)

 

2

2 2 2.00 0.178 mm  200 g , T  0.250 s,E 2.00 J

;

2 2 25.1 rad s

T 0.250

 

   

(a)

 

2 2

k m    0.200 kg 25.1 rad s 126 N m

(b)

 

2

2 2 2.00 0.178 m

Explanation:

That is a reason

8 0
3 years ago
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