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3 years ago

Deduce an expression for the gravitational potential

Physics

1 answer:

bonufazy [111]3 years ago

6 0

M*g*h

G is the gravity which is usually a fixed number (10)

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(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th

julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of r from a point charge, Q, is given by

$V = -\dfrac{kQ}{r}$

where $k = 9\times 10^9 \text{ F/m}$

Difference in potential between the points is

$kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}$

$PD = 135\times 10^3\text{ V} = 135\text{ kV}$

(b)

If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be x.

$270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]$

$10 - \dfrac{1}{x} = \dfrac{270000}{9\times10^9\times3\times10^{-6}} = 10$

$\dfrac{1}{x} = 0$

$x = \infty$

The charge chould be moved to infinity

7 0

3 years ago

What are the main activities involved in studying physics?

Firlakuza [10]

The main activity that is involved in studying of physics is the study of natural laws. The study of physics has to do with many aspects of the universe. Physics majorly looks into the natural laws that operate in the universe and describe how they affect matter in relation to time.

7 0

3 years ago

A small object of mass 3.82 g and charge -16.5 µC is suspended motionless above the ground when immersed in a uniform electric f

horrorfan [7]

Answer:

2271.16N/C upward

Explanation:

The diagram well illustrate all the forces acting on the mass. The weight is acting downward and the force is acting upward in other to balance the weight.since the question says it is motionless, then indeed the forces are balanced.

First we determine the downward weight using

$W=mg\\g=9.81m/s^{2}$

Hence for a mass of 3.82g 0r 0.00382kg we have the weight to be

$W=0.00382kg*9.81m/s^{2}$

$W=0.0375N$

To calculate the electric field,

$E=f/q\\E=0.0375/16.5*10^{-6} \\E=2271.16N/C$

Since the charge on the mass is negative, in order to generate upward force, there must be a like charge below it that is repelling it, Hebce we can conclude that the electric field lines are upward.

Hence the magnitude of the electric force is 2271.16N/C and the direction is upward