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Firdavs [7]
3 years ago
8

Calculate the electric field strength where a test charge of 2 coulombs is repelled by a force of 2 newtons. ________ N/C

Physics
1 answer:
Nitella [24]3 years ago
7 0

Answer:

1 N/C

Explanation:

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Hope This Help??

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If we keep cutting down tree's and if there are enough forest fires then it will turn into a desert....Hope it helps! 
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A car that increase its speed from 20 km/h to 100 km/h undergoes -------acceleration,
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negative acceleration- deceleration

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A person is lying on a diving board 2.00 m above the surface of the water in a swimming pool. She looks at a penny that is on th
wlad13 [49]

Answer:

7.98 m

Explanation:

In the given question,

distance above surface= 2 m

Distance penny from person = 8 m

Since the swimming pool is filled with water and atmosphere has air therefore the refractive index phenomenon will occur.

The refractive index of water: air is 4/3 (1.33).

Using the formula, 4/3 = real depth, apparent depth

real depth= 4/3 x apparent depth

Now, calculating apparent depth = 8 - 2

= 6 m

therefore, real depth =  4/3 x apparent depth

= 1.33 x 6

= 7.98

thus, 7.98 m is the real depth of water.

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3 years ago
The physical phenomenon that MRI is based on does not depend on ionizing radiation, but on other properties of atoms instead. Wh
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4 0
3 years ago
Read 2 more answers
Positive Charge is distributed along the entire x axis with a uniform density 12 nC/m. A proton is placed at a position of 1.00
lions [1.4K]

Answer:

b.  \Delta KE = 390 eV

Explanation:

As we know that the electric field due to infinite line charge is given as

E =\frac{\lambda}{2\pi \epsilon_0 r}

here we can find potential difference between two points using the relation

\Delta V = \int E.dr

now we have

\Delta V = \int(\frac{\lambda}{2\pi \epsilon_0 r}).dr

now we have

\Delta V = \frac{\lambda}{2\pi \epsilon_0}ln(\frac{r_2}{r_1})

now plug in all values in it

\Delta V = \frac{12\times 10^{-9}}{2\pi \epsilon_0}ln(\frac{1+5}{1})

\Delta V = 216ln6 = 387 V

now we know by energy conservation

\Delta KE = q\Delta V

\Delta KE = (e)(387V) = 387 eV

3 0
3 years ago
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