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lawyer [7]
3 years ago
11

Johnathan wants to lift a 200N box onto a 5m ledge. How much work is done?

Physics
2 answers:
Y_Kistochka [10]3 years ago
8 0

Answer:

1000 J which agrees with answer a in your list of options

Explanation:

Recall the formula for work (W) as the force (F) applied times the distance (d) covered in the same direction of the force:

W = F * d

W = 200 N * 5 m = 1000 J

lara31 [8.8K]3 years ago
3 0
The answer is - a - 100 j
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A laptop can convert 400J of electrical energy into a 240J of light and sound. What is the efficiency? Where does the rest of th
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Efficiency = (Wanted) energy out ÷ energy in × 100

Energy in = 400J
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Efficiency = 240 ÷ 400 × 100
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The index of refraction is the ratio of the speed of light in a vacuum to the speed of light in a medium. Which true statement i
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Answer:

A. Materials with a low index of refraction cause light to refract very little.

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At a distance r from a charge e on a particle of mass m the electric field value is
mel-nik [20]

At a distance r from a charge e on a particle of mass m the electric field value is  8.9876 × 10⁹ N·m²/C². Divide the magnitude of the charge by the square of the distance of the charge from the point. Multiply the value from step 1 with Coulomb's constant.

<h3>what is magnitude ?</h3>

Magnitude can be defined as the maximum extent of size and the direction of an object.

It is used as a common factor in vector and scalar quantities, as we know scalar quantities are those quantities that have magnitude only and vector quantities are those quantities have both magnitude and direction.

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7 0
1 year ago
A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these c
viva [34]

Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC = 71\times 10^{- 9} C

Q' = + 42 nC = 42\times 10^{- 9} C

Separation distance, d = 1.9 m

Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E} = 708.03 N/C

Now,

Electric field at the mid-point due to charge Q' is given by:

\vec{E'} = \frac{Q'}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E'} = \frac{42\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E'} = 418.84 N/C

Now,

The net Electric field is given by:

\vec{E_{net}} = \vec{E} - \vec{E'}

\vec{E_{net}} = 708.03 - 418.84 = 289.19 N/C

5 0
3 years ago
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