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slamgirl [31]
2 years ago
10

A uniformly charged spherical droplet of mercury has electric potential Vbig throughout the droplet. The droplet then breaks int

o n identical spherical droplets, each of which has electric potential Vsmall throughout its volume. The n small droplets are far enough apart from one another that they do not interact significantly.
Find (Vbig)/(Vsmall) , the ratio of Vbig, the electric potential of the initial drop, to Vsmallthe electric potential of one of the smaller drops.(The ratio should be dimensionless and should depend only on n)
Physics
1 answer:
mezya [45]2 years ago
7 0

Answer:

\frac{V_{big}}{V_{small}} = n^{2/3}

Explanation:

Let the total charge on the big drop is given as Q

now if the radius of the drop is R then electric potential of the big drop is given as

V_{big} = \frac{KQ}{R}

Now if it break into n identical drops

then let the charge on each drop is "q" and radius is "r"

by volume conservation

\frac{4}{3}\pi R^3 = n(\frac{4}{3}\pi r^3)

r = \frac{R}{n^{1/3}}

now we have potential of smaller drop given as

V_{small} = \frac{kq}{r}

V_{small} = \frac{K(Q/n)}{\frac{R}{n^{1/3}}}

V_{small} = \frac{1}{n^{2/3}}\frac{KQ}{R}

\frac{V_{big}}{V_{small}} = n^{2/3}

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Mariana [72]

Answer:

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Explanation:

Please see attachment below.

Given

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8 0
2 years ago
If one replaces the conducting cube with one that has positive charge carriers, in what direction does the induced electric fiel
Grace [21]

Answer:

There will be no change in the direction of the electric field .

Explanation:

The direction will remain the same because the sign of the charges has no effect on it.

When one replaces the conducting cube with one that has positive charge carriers there will be no change in the direction of the field as there is no defined relationship between the direction of the electric field and sign of the charge.

3 0
2 years ago
If you have to apply 40n of force on a crowbar to lift a rock that weights 400n, what is the actual mechanical advantage of the
Mrrafil [7]
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MA= \frac{F_{out}}{F_{in}}
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3 0
2 years ago
Calculate the number of moles in each of the following masses: 0.039 g of palladium 0.0073 kg of tantalum
marysya [2.9K]

Answer:

<em>The number of moles of palladium and tantalum are 0.00037 mole and 0.0000404 mole respectively</em>

Explanation:

Number of mole = reacting mass/molar mass

n = R.m/m.m......................... Equation 1

Where n = number of moles, R.m = reacting mass, m.m = molar mass.

For palladium,

R.m = 0.039 g and m.m = 106.42 g/mol

Substituting theses values into equation 1

n = 0.039/106.42

n = 0.00037 mole

For tantalum,

R.m = 0.0073 and m.m = 180.9 g/mol

Substituting these values into equation 1

n = 0.0073/180.9

n = 0.0000404 mole

<em>Therefore the number of moles of palladium and tantalum are 0.00037 mole and 0.0000404 mole respectively</em>

3 0
2 years ago
The velocity versus time graph of particle A is tangent to the velocity versus time graph for particle B at point O. What is the
lara [203]
As velocities are tangent, the value of both Particle A and Particle B would be same for that point O (Intersecting point)

a = v / t
Here, v = 7, t = 6
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So, a = -1.17 m/s²

In short, Your Answer would be Option C

Hope this helps!
7 0
2 years ago
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