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Komok [63]
3 years ago
12

What does the n stand for? Is it north?

Physics
1 answer:
Aleksandr [31]3 years ago
4 0
That would be newton's. You'll be studying or have studied about newton's laws of physics. Both will help you. Hope this helps!
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In the equation for Ohm’s law what does I stand for?
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A) current 

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3 years ago
A 220 kg crate hangs from the end of a rope of length L = 14.0 m. You push horizontally on the crate with a varying force F to m
kifflom [539]

<u>Answer</u>:

(a) magnitude of F = 797 N

(b)the total work done  W = 0

(c)work done by the gravitational force =  -1.55 kJ

(d)the work done by the pull  = 0

(e) work your force F does on the crate = 1.55 kJ

<u>Explanation</u>:

<u>Given</u>:

Mass of the crate, m =  220 kg

Length of the rope, L = 14.0m

Distance, d =  4.00m

<u>(a) What is the magnitude of F when the crate is in this final position</u>

Let us first determine vertical angle as follows

=>Sin \theta = \frac{d }{L}

=> \theta = Sin^{-1} \frac{d}{L} =

Now substituting thje values

=> \theta = Sin^{-1} \frac{4}{12} =

=> \theta = Sin^{-1} \frac{1}{3}

=> \theta = Sin^{-1}(0.333)

=> \theta = 19.5^{\circ}

Now the tension in the string resolve into components

The vertical component supports the weight

=>Tcos\theta = mg

=>T = \frac{mg}{cos\theta}

=>T = \frac{230 \times 9.8 }{cos(19.5)}

=>T = \frac{2254 }{cos(19.5)}

=>T = \frac{2254 }{0.9426}

=>T =2391N

Therefore the horizontal force

F = TSin(19.5)

F = 797 N

b) The total work done on it

As there is no change in Kinetic energy

The total work done W = 0

<u>c) The work done by the gravitational force on the crate</u>

The work done by gravity

Wg = Fs.d = - mgh

Wg = - mgL ( 1 - Cosθ )

Substituting the values                                                            

= -230 \times 9.8\times 12 ( 1 - cos(19.5) )

= -230 \times 9.8\times 12 ( 1 - 0.9426) )

= -230 \times 9.8\times 12 (0.0574)

= -230 \times 9.8\times 0.6888

=  -230 \times 6.750

= -1552.55 J

The work done by gravity = -1.55 kJ

<u>d) the work done by the pull on the crate from the rope</u>

Since the pull  is perpendicular to the direction of motion,

The work done = 0

e)Find the work your force F does on the crate.

Work done by the Force on the crate

WF = - Wg  

WF = -(-1.55)

WF = 1.55 kJ

<u>(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)</u>

Here the work done by force is not equal to F*d  

and it is equal to product of the cos angle and F*d

So, it is not equal to the product of the horizontal displacement and the answer to (a)      

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A car changes velocity at a constant acceleration of 2.5m/s to reach 43.7m/s in 2.7 s how fast was the car moving when it began
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The formula we can use in this case is:

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