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7nadin3 [17]
3 years ago
5

How many electrons have been removed from a positively charged electroscope if it has a net charge of 6x10-11?

Physics
1 answer:
alina1380 [7]3 years ago
3 0

How much negative charge has been removed from a positively charged electroscope if it has a charge of 7.5x10-11C?

If C is Coulombs then, −1 C is equivalent to the charge of approximately 6.242×10^18 electrons.

You already gave the answer: -7.5x10–11C

It is like asking: What colour is my white horse?

Now if you want to know how many electrons have been removed, then multiplying the number of electrons in a coulomb by the coulombs you have, you get 4.6815 x 10^8 electrons. Which is almost the same answer as before.

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Suppose there are single-particle energy eigenvalues of 0, e, 2.::, and 3.:: which are non-degenerate. A total of 6.0 is to be s
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Answer:

so 40 ways for distinguishable

4 ways for bosons

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Explanation:

The explanation is attached below

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You are conducting an experiment inside a train car that may move along level rail tracks. A load is hung from the ceiling on a
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Answer:

b

c

e

h

Explanation:

Note that the swing direction was not giving in the question and direction could be sideways (in a turn) or in a track or both

The question show something in common ...acceleration

So let's look at the statements and pick the correct ones

a is false while b is correct as the train is accelerating

c is correct. The train is accelerating even thou the speed could not be ascertained

d is false and not feasible as the train is accelerating

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7 0
3 years ago
Thermopane window is constructed, using two layers of glass 4.0 mm thick, separated by an air space of 5.0 mm.
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To solve this problem it is necessary to apply the concepts related to rate of thermal conduction

\frac{Q}{t} = \frac{kA\Delta T}{d}

The letter Q represents the amount of heat transferred in a time t, k is the thermal conductivity constant for the material, A is the cross sectional area of the material transferring heat, \Delta T, T is the difference in temperature between one side of the material and the other, and d is the thickness of the material.

The change made between glass and air would be determined by:

(\frac{Q}{t})_{glass} = (\frac{Q}{t})_{air}

k_{glass}(\frac{A}{L})_{glass} \Delta T_{glass} = k_{air}(A/L)_{air} \Delta T_{air}

\Delta T_{air} = (\frac{k_{glass}}{k_{air}})(\frac{L_{air}}{L_{glass}}) \Delta T_{glass}

\Delta T_{air} = (\frac{0.84}{0.0234})(\frac{5}{4}) \Delta T_{glass}

\Delta T_{air} = 44.9 \Delta T_{glass}

There are two layers of Glass and one layer of Air so the total temperature would be given as,

\Delta T = \Delta T_{glass} +\Delta T_{air} +\Delta T_{glass}

\Delta T = 2\Delta T_{glass} +\Delta T_{air}

20\°C = 46.9\Delta T_{glass}

\Delta T_{glass} = 0.426\°C

Finally the rate of heat flow through this windows is given as,

\Delta {Q}{t} = k_{glass}\frac{A}{L_{glass}}\Delta T_{glass}

\Delta {Q}{t} = 0.84*24*10 -3*0.426

\Delta {Q}{t} = 179W

Therefore the correct answer is D. 180W.

3 0
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