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2 years ago

What do the arrows around the positive point charge illustrate?

Physics

1 answer:

yawa3891 [41]2 years ago

3 0

The arrows around the positive point charge illustrates the direction of the electric force/field.

<h3>What is Electric field?</h3>

This is described as an area in which force is exerted om other charged particles or objects.

Arrows could be around the positive or negative point charge and helps to depict the direction of the electric force.

Read more about Electric field here brainly.com/question/14372859

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Answer:

Energy Level K

Explanation:

In electronic configuration there are four shells named <em>KLMN</em><em> </em> which hold 2 , 8 , 18 , 32 electrons respectively.

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3 years ago

A cartoon plane with four engines can accelerate at 8.9 meters per second squared

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The answer is 2728283 69

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3 years ago

What volume (in liters) of gasoline has a total heat of combustion equal to the energy obtained in part (a)? (see section 17.6;

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<h3><u>Answer;</u></h3>

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3 years ago

QUESTION 2 DOK 3 A Thompson's gazelle has a maximum acceleration of 4.5 m/s2 At this acceleration, how much time is required for

Dominik [7]

Answer:

2.47 s

Explanation:

Convert the final velocity to m/s.

40 km/h → 11.1111 m/s

We have the acceleration of the gazelle, 4.5 m/s².

We can assume the gazelle starts at an initial velocity of 0 m/s in order to determine how much time it requires to reach a final velocity of 11.1111 m/s.

We want to find the time t.

Find the constant acceleration equation that contains all four of these variables.

v = v₀ + at

Substitute the known values into the equation.

11.1111 = 0 + (4.5)t
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The Thompson's gazelle requires a time of 2.47 s to reach a speed of 40 km/h (11.1111 m/s).

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3 years ago

The masses of the earth and moon are 5.98 x 1024 and 7.35 x 1022 kg, respectively. Identical amounts of charge are placed on eac

dybincka [34]

Answer:

The magnitude of charge on each is $5.707\times 10^{13} C$

Solution:

As per the question:

Mass of Earth, $M_{E} = 5.98\times 10^{24} kg$

Mass of Moon, $M_{M} = 7.35\times 10^{22} kg$

Now,

The gravitational force of attraction between the earth and the moon, if 'd' be the separation distance between them is:

$F_{G} = \frac{GM_{E}M_{M}}{d^{2}}$ (1)

Now,

If an identical charge 'Q' be placed on each, then the Electro static repulsive force is given by:

$F_{E} = \frac{1}{4\pi\epsilon_{o}}\frac{Q^{2}}{d^{2}}$ (2)

Now, when the net gravitational force is zero, the both the gravitational force and electro static force mut be equal:

Equating eqn (1) and (2):

$\frac{GM_{E}M_{M}}{d^{2}} = \frac{1}{4\pi\epsilon_{o}}\frac{Q^{2}}{d^{2}}$

$(6.67\times 10^{- 11})\times (5.98\times 10^{24})\times (7.35\times 10^{22}) = (9\times 10^{9}){Q^{2}}$

$\sqrt{\farc{(6.67\times 10^{- 11})\times (5.98\times 10^{24})\times (7.35\times 10^{22})}{9\times 10^{9}}} = Q$

Q = $\pm 5.707\times 10^{13} C$

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3 years ago