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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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FromTheMoon [43]

Answer:

T=7.4 N hence T<30 N

Explanation:

The figure is likely to be similar to the one attached. Writing the equation for forces we have

F-T=Fa/g where F is the force, T is tension, a is acceleration and g is acceleration due to gravity. Substituting the figures we have the first equation as

30 N - T = (30/9.81)a

Also, we know that T=F*a/g and substituting  10N for F we obtain the second equation as

T = (10/9.81)a

Adding the first and second equations we obtain

30 = 4.077471967

a Hence

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and T=a hence

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2 years ago
Read 2 more answers
A golfer hits a golf ball at an angle of 25.0° to the ground. if the golf ball covers a horizontal distance of 301.5 m, what is
kvasek [131]

<u>Answer:</u>

 Maximum height reached = 35.15 meter.

<u>Explanation:</u>

Projectile motion has two types of motion Horizontal and Vertical motion.

Vertical motion:

         We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

         Considering upward vertical motion of projectile.

         In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g m/s^2 and final velocity = 0 m/s.

        0 = u sin θ - gt

         t = u sin θ/g

    Total time for vertical motion is two times time taken for upward vertical motion of projectile.

    So total travel time of projectile = 2u sin θ/g

Horizontal motion:

  We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

  In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 m/s^2 and time taken = 2u sin θ /g

 So range of projectile,  R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}

 Vertical motion (Maximum height reached, H) :

     We have equation of motion, v^2=u^2+2as, where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.

   Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H

   0^2=(usin\theta) ^2-2gH\\ \\ H=\frac{u^2sin^2\theta}{2g}

In the give problem we have R = 301.5 m,  θ = 25° we need to find H.

So  \frac{u^2sin2\theta}{g}=301.5\\ \\ \frac{u^2sin(2*25)}{g}=301.5\\ \\ u^2=393.58g

Now we have H=\frac{u^2sin^2\theta}{2g}=\frac{393.58*g*sin^2 25}{2g}=35.15m

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Explanation:

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Generally, most light do not cause hardness of a rubber. But an ultraviolet light can cause rubber to become solid over a period of time. This is possible if there is a prolong exposure of the rubber, and because of the evaporation of volatiles in the polymer material. Ultraviolet light are known to cause a rubber to become solid.

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Answer:

From the relation above we can conclude that the  as the distance between the two plate increases the electric field strength decreases

Explanation:

I cannot  find any attached photo, but we can proceed anyways theoretically.

The electric field strength (E) at any point in an electric field is the force experienced by a unit positive charge (Q) at that point

i.e

E=\frac{F}{Q}

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F= \frac{kQ1Q2}{r^2}

But the electric field intensity due to a point charge Q at a distance r meters away is given by

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