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gregori [183]
2 years ago
13

You are pulling a child in a wagon. The rope handle is inclined upward at a 60∘ angle. The tension in the handle is 20 N.

Physics
2 answers:
dem82 [27]2 years ago
6 0
  • Angle (θ) = 60°
  • Force (F) = 20 N
  • Distance (s) = 200 m
  • Therefore, work done
  • = Fs Cos θ
  • = (20 × 200 × Cos 60°) J
  • = (20 × 200 × 1/2) J
  • = (20 × 100) J
  • = 2000 J

<u>Answer</u><u>:</u>

<u>2</u><u>0</u><u>0</u><u>0</u><u> </u><u>J</u>

Hope you could get an idea from here.

Doubt clarification - use comment section.

max2010maxim [7]2 years ago
5 0

S O L U T I O N:

Here we've been provided with the information i.e

  • Force (F) = 20N

  • Distance (s) = 200m

  • Angle (θ) = 60°

Ans we have given to find out what is work done (W) = ?

The standard formula to calculate work done is given by,

:\implies\sf{w = f \times s \times cos \theta}

:\implies\sf{w = 20 \times 200 \times cos60}

:\implies\sf{w = 20 \times 200 \times  \frac{1}{2} }

:\implies\sf{w = 20 \times 100 \times 1}

:\implies\sf{w = 2000J}

  • Total work done done by rope is 2000J.
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a vertical solid steel post 29cm in diameter and 2.0m long is required to support a load of 8200kg, ignore the weight of the pos
erik [133]

Answer:

The stress is  \sigma  =  1.218*10^{6} \  N/m^2

Explanation:

From the question we are told that

   The diameter of the post is  d =  29 \ cm  =  0.29 \  m

   The length is L  =  2.0 \  m

    The weight of the loading mass

Generally  the  radius of the post is mathematically represented as

     r =  \frac{0.29}{2}

=>   r = 0.145 \  m

Generally the area of the post is  

       A =  \pi r^2

=>     A =  3.14 *  0.145 ^2

=>     A =  0.066 \ m^2

Generally the weight exerted by the load is mathematically represented as

        F =  m  *  g

=>      F =  8200  *  9.8

=>      F =  80360 \  N

Generally the stress is mathematically represented as

         \sigma  =  \frac{F}{A}

=>      \sigma  =  \frac{80360 }{0.066}

=>      \sigma  =  1.218*10^{6} \  N/m^2

7 0
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satela [25.4K]

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3 years ago
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If a diffraction grating has 3700 lines per cm, what is the spacing d between lines
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So first we find the gap between the slits by the formula d=1/N 

<span>N is number of lines per metre so 3700 line/cm = 370000 lines/m </span>
<span>So d=2.7*10^-6 </span>

<span>Now we use the formula dsin(angle)=n(wavelength) </span>

<span>d is the same </span>
<span>n is the order of the diffraction pattern </span>

<span>so wavelenth=dsin(angle)/n </span>
<span>=[(2.7*10^-6)*sin30]/3 </span>
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7 0
2 years ago
a stone is dropped from top of a tower of 50m high ,simltaneously another stone is thrown upward with a speed of 20m/s find the
aleksklad [387]
At the time that I'll call ' Q ', the height of the stone that was
dropped from the tower is

             H = 50 - (1/2 G Q²) ,

and the height of the stone that was tossed straight up
from the ground is

             H  =  20Q - (1/2 G Q²) .

The stones meet when them's heights are equal,
so that's the time when

                         <span>50 - (1/2 G Q²)  =  20Q - (1/2 G Q²) .

This is looking like it's going to be easy.

Add  </span><span>(1/2 G Q²)  to each side.
Then it says
                                             50  =  20Q

Divide each side by 20:          2.5  =  Q .

And there we are.  The stones pass each other

                                       2.5 seconds

after they are simultaneously launched.
</span>
5 0
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