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Reptile [31]
2 years ago
10

A force F = (2xî + 4yĵ), where F is in newtons and x and y are in meters, acts on an object as the object moves in the x-directi

on from the origin to x = 4.99 m. Find the work W = ∫ F · dr done by the force on the object (in J).
Physics
1 answer:
blondinia [14]2 years ago
4 0

Answer:

Work done, W = 24.9001 J

Explanation:

Given that,

Force, F = (2xî + 4yĵ) Where x and y are in meters

This force acts on an object as the object moves in the x-direction from the origin to x = 4.99 m.

We need to find the work done by the force on the object. It is given by :

W=\int\limits^a_b {F dr} \\\\W=\int\limits^{4.99}_0 {(2xi+4yj)dx} \\\\W=\int\limits^{4.99}_0 {2xi\ dx} \\\\W=x^2|_0^{4.99}\\\\\text{Applying limits}\\\\W=(4.99)^2-0^2\\\\W=24.9001\ J

So, the work done is 24.9001 J.

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Calculate the electric field at the center of a square
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Answer:

E_y=1175510.2\ N.C^{-1}

The Magnitude of electric field is in the upward direction as shown directly towards the charge q_1.

Explanation:

Given:

  • side of a square, a=52.5\ cm
  • charge on one corner of the square, q_1=+45\times 10^{-6}\ C
  • charge on the remaining 3 corners of the square,q_2=q_3=q_4=-27\times 10^{-6}\ C

<u>Distance of the center from each corners</u>=\frac{1}{2} \times diagonals

diagonal=\sqrt{52.5^2+52.5^2}

diagonal=74.25\cm=0.7425\ m

∴Distance of center from corners, b=0.3712\ m

Now, electric field due to charges is given as:

E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}

<u>For charge q_1 we have the field lines emerging out of the charge since it is positively charged:</u>

E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}

  • E_1=2938775.5\ N.C^{-1}

<u>Force by each of the charges at the remaining corners:</u>

E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}

  • E_2=E_3=E_4=1763265.3\ N.C^{-1}

<u> Now, net electric field in the vertical direction:</u>

E_y=E_1-E_4

E_y=1175510.2\ N.C^{-1}

<u>Now, net electric field in the horizontal direction:</u>

E_y=E_2-E_3

E_y=0\ N.C^{-1}

So the Magnitude of electric field is in the upward direction as shown directly towards the charge q_1.

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1 answer

8 0
3 years ago
an empty propane tank dropped from a hot air balloon hits the ground with a speed of 143.8 m/s. from what height was the tank re
vodomira [7]
What you know:
Vi=0m/s
Vf=143.8m/s
A=-9.8m/s
d=???
Use the equation Vf^2=Vi^2+2A(d)
Rearrange to isolate d: d=Vf^2/2A
d=(143.8)^2/2(-9.8)
d=20678.4/-19.6
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