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4 years ago

If a positively charged body is moved against an electric field it will gain?

Physics

1 answer:

Gnoma [55]4 years ago

5 0

Answer:

positive charge

Explanation:

Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field. There are two types of electric charge: positive and negative.

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If the distance between two masses is tripled, the gravitational force between changes by a factor of:_______

adell [148]

Answer:

option A

Explanation:

given,

distance between two masses is doubled

new distance, r' = 3 r

using gravitational force equation

$F = \dfrac{GMm}{r^2}$ ............(1)

new gravitational force

$F' = \dfrac{GMm}{r'^2}$

now from the given condition

$F' = \dfrac{GMm}{(3r)^2}$

$F' = \dfrac{GMm}{9r^2}$

$F' = \dfrac{1}{9}\dfrac{GMm}{r^2}$

now, from equation (1)

$F' = \dfrac{1}{9}F$

$\dfrac{F'}{F} = \dfrac{1}{9}$

now, the change in gravitational force factor is equal to $\dfrac{1}{9}$

Hence, the correct answer is option A

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3 years ago

A point charge of 6.0 nC is placed at the center of a hollow spherical conductor (inner radius = 1.0 cm, outer radius = 2.0 cm)

egoroff_w [7]

Explanation:

The given data is as follows.

q = 6.0 nC = $6 \times 10^{-9} C$

inner radius (r) = 1.0 cm = 0.01 m (as 1 cm = 100 m)

So, there will be same charge on the inner surface as the charge enclosed with an opposite sign.

Formula to calculate the charge density is as follows.

$\sigma = \frac{q_{in}}{A}$ .......... (1)

Since, area of the sphere is as follows.

A = $4 \pi r^{2}$ ........... (2)

Hence, substituting equation (2) in equation (1) as follows.

$\sigma = \frac{q_{in}}{4 \pi r^{2}}$

= $\frac{6 \times 10^{-9} C}{4 \times 3.14 \times (0.01)^{2}}$

= $0.477 \times 10^{-5}$

or, = 4.77 $\mu C/m^{2}$

Thus, we can conclude that the resulting charge density on the inner surface of the conducting sphere is 4.77 $\mu C/m^{2}$ .

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3 years ago

WHY IS SATURN ONE OF THE COLDEST PLANETS IN OUR SOLAR SYSTEM

Usimov [2.4K]

What are the answer choices?

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3 years ago

Interactive Solution 28.5 illustrates one way to model this problem. A 7.11-kg object oscillates back and forth at the end of a

____ [38]

Explanation:

Given that,

Mass of the object, m = 7.11 kg

Spring constant of the spring, k = 61.6 N/m

Speed of the observer, $v=2.79\times 10^8\ m/s$

We need to find the time period of oscillation observed by the observed. The time period of oscillation is given by :

$t_o=2\pi \sqrt{\dfrac{m}{k}} \\\\t_o=2\pi \sqrt{\dfrac{7.11}{61.6}} \\\\t_o=2.13\ s$

Time period of oscillation measured by the observer is :

$t=\dfrac{t_o}{\sqrt{1-\dfrac{v^2}{c^2}} }\\\\t=\dfrac{2.13}{\sqrt{1-\dfrac{(2.79\times 10^8)^2}{(3\times 10^8)^2}} }\\\\t=5.79\ s$

So, the time period of oscillation measured by the observer is 5.79 seconds.

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3 years ago

The lowest surface temperature in the solar system (-200°c occurs on

zimovet [89]

The lowest surface temperature in the solar system was recorded on Uranus (-224 degrees Celsius). The temperature of a planet does not only depend on the amount of solar radiation that it receives but also on the amount of heat that it gives off. Because of Uranus' orientation it absorbs little radiation which makes it colder than Neptune although Neptune is further away from the Sun.

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3 years ago