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DENIUS [597]
3 years ago
5

A gardener pushes a wheelbarrow around the edges of a lawn. When she has finished, she is 5 m from her starting point. She has t

o pull upwards with a force of 300 N to lift the legs of the wheelbarrow off the ground. She pushes the wheelbarrow along the ground with a force of 100 N. Calculate how much work the gardener does in pushing the wheelbarrow around the lawn.
Physics
1 answer:
zvonat [6]3 years ago
3 0

Answer:

Work done by the gardner is 500 J

Explanation:

As we know that the gardner apply force perpendicular upward by magnitude 300 N and along the floor horizontal force is 100 N

so we have

F = 100 \hat i + 300 \hat j

now the displacement of the gardner along the floor is

d = 5\hat i

now work done is given as

W = F. d

so we have

W = (100 \hat i + 300 \hat j). (5\hat i)

W = 500 J

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Answer:

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As per the question:

Charge density of rod 1, \lambda = 1\ nC = 1\times 10^{- 9}\ C

Charge density of rod 2, \lambda = - 1\ nC = - 1\times 10^{- 9}\ C

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To calculate the electric field at origin:

We know that the electric field due to a long rod is given by:

\vec{E} = \frac{\lambda }{2\pi \epsilon_{o}{R}

Also,

\vec{E} = \frac{2K\lambda }{R}                  (1)

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At x = - 1 cm = - 0.01 m:

Using eqn (1):

\vec{E} = \frac{2\times 9\times 10^{9}\times 1\times 10^{- 9}}{0.01} = 1800\ N/C

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Using eqn (1):

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Now, the total field at the origin is the sum of both the fields:

\vec{E_{net}} = 1800 + 1800 = 3600\ N/C

7 0
3 years ago
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