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tatuchka [14]
3 years ago
8

A device has power -2 D . The device is

Physics
1 answer:
slavikrds [6]3 years ago
7 0

Answer:

b.

Explanation:

-vesign shows the lens is <em><u>C</u></em><em><u>O</u></em><em><u>N</u></em><em><u>C</u></em><em><u>A</u></em><em><u>V</u></em><em><u>E</u></em>

<em><u>f</u></em><em><u>=</u></em>1/power

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Paolo says that his bicycle is hard to pedal.Mia looks at the bicycle and tells him he needs to oil the chain.explain why oiling
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Which substance cannot be seperated physically or chemically?
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3 years 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

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The net Electric field is given by:

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

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