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stiks02
1 month ago
7

Two spherical drops of mercury each have a charge of 0.10 nC and a potential of 300 V at the surface. The two drops merge to for

m a single drop. What is the potential at the surface of the new drop?Two spherical drops of mercury each have a charge of 0.10 nC and a potential of 300 V at the surface. The two drops merge to form a single drop. What is the potential at the surface of the new drop?
Physics
1 answer:
Yuliya22 [3.3K]1 month ago
6 0

Answer:

V = 228\ V

Explanation:

the provided information includes,

the charge of each of the two spherical drops = 0.1 nC

the potential on the surface = 300 V

when the drops combine into a larger drop

what is the surface potential of the new combined drop =?

V = \dfrac{kq}{r}

r = \dfrac{9\times 10^9\times 0.1 \times 10^{-9}}{300}

radius = 0.003 m

volume = 2 \times \dfac{4}{3}\pi r^3

= 2 \times \dfac{4}{3}\pi \times 0.003^3

= 2.612 × 10⁻⁷ m³

\dfac{4}{3}\pi R^3 = 2.612\times 10^{-7}

R =\sqrt[3]{\dfrac{2.612 \times 10^{-7}\times 3}{4\times \pi}}

R = 0.00396 m

V = \dfrac{kq}{r}

V = \dfrac{9\times 10^9 \times 0.1 \times 10^{-9}}{0.00396}

V = 227.27

V = 228\ V

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Softa [3030]

Answer:

a) ∆x∆v = 5.78*10^-5

∆v = 1157.08 m/s

b) 4.32*10^{-11}

Explanation:

This problem can be addressed using Heisenberg's uncertainty principle, which is expressed as:

\Delta x\Delta p \geq \frac{\hbar}{2}

Where h represents Planck’s constant (6.62*10^-34 J s).

Assuming that the electron's mass remains the same, we proceed as follows:

\Delta x \Delta (m_ev)=m_e\Delta x\Delta v \geq \frac{\hbar}{2}

Utilizing the electron's mass (9.61*10^-31 kg) and the uncertainty in position (50 nm), we can compute ∆x∆v and ∆v:

\Delta x \Delta v\geq\frac{\hbar}{2m_e}=\frac{(1.055*10^{-34}Js)}{2(9.1*10^{-31}kg)}=5.78*10^{-5}\ m^2/s

\Delta v\geq\frac{5.78*10^{-5}}{50*10^{-9}m}=1157.08\frac{m}{s}

If we treat the electron like a classic particle, the time required to cross the channel is determined using the upper limit of the uncertainty in velocity:

t=\frac{x}{v}=\frac{50*10^{-9}m}{1157.08m/s}=4.32*10^{-11}s

6 0
1 month ago
Observe: Up until now, all the problems you have solved have involved converting only one unit. However, some conversion problem
Softa [3030]

Response:

It is important to note that for units like time and angles, the transformations aren't based on ten, as shown below:

        60 s = 1 min

        60 min = 1 h

        24 h = 1 day

Hence, for this conversion, special attention is needed

the length conversion is based on ten

Clarification:

In numerous problems, the units utilized undergo transformations via equations into alternate units referred to as derived units, and the transformation of these derived units generally results from the multiplication of the transformations of the basic units.

For instance, in the context of velocity, its derived unit is expressed as m / s; therefore, the transformation operates similarly to that of length and time, meaning if they are multiplied in the equation, it continues to be multiplied, and if divided, it remains divided.

It is important to note that for units like time and angles, the transformations aren't based on ten, as shown below:

        60 s = 1 min

        60 min = 1 h

        24 h = 1 day

Hence, for this conversion, special attention is needed

the length conversion is based on ten

      1000 m = 1 km

7 0
19 days ago
A sphere of radius 0.03m has a point charge of q= 7.6 micro C located at it’s centre. Find the electric flux through it?
Sav [3153]

Answer:

The electric flux going through the sphere is 8.58 *10^{5} \frac{Nm^2}{C}

Explanation:

Given

Radius,\ r = 0.03m\\Charge,\ q =7.6\µC

Required

Calculate the electric flux

Electric flux can be computed using the formula;

Ф = q/ε

Where ε stands for the electric constant permittivity

ε = 8.8542 * 10^{-12}

Substituting ε = 8.8542 * 10^{-12} and q =7.6\µC; the formula simplifies to

Ф = \frac{7.6\µC}{8.8542 * 10^{-12}}

Ф = \frac{7.6 * 10^{-6}}{8.8542 * 10^{-12}}

Ф = \frac{7.6}{8.8542} *\frac{10^{-6}}{10^{-12}}

Ф = \frac{7.6}{8.8542} *10^{12-6}}

Ф = 0.85834970974 *10^{12-6}}

Ф = 0.85834970974 *10^{6}}

Ф = 8.5834970974 *10^{5}}

Ф = 8.58 *10^{5} \frac{Nm^2}{C}

Thus, the electric flux through the sphere is 8.58 *10^{5} \frac{Nm^2}{C}

7 0
2 months ago
A proton has mass 1.7 ✕ 10−27 kg. what is the magnitude of the impulse required to increase its speed from 0.991c to 0.994c?
Maru [3345]

Impulse can be equated to the force associated with momentum change, expressed as F*t = mv - mu
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initial velocity, u = 0.991 x (3 x 10^8) = 2.973 x 10^8
final velocity, v = 0.994 x (3 x 10^8) = 2.982 x 10^8
Consequently, the calculation is:
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= (5.0694 x 10^35) - (5.0694 x 10^35)
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6 0
1 month ago
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Yuliya22 [3333]
This should cover all your needs:)
7 0
13 days ago
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