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Anna11 [10]
4 years ago
6

Gravity and Electromagnetic force

Physics
2 answers:
nekit [7.7K]4 years ago
8 0

Answer:

1) Gravitational force and electromagnetic force both are field forces which means both forces exerted from long range of distance and we do not require any physical contact to apply these type of forces.

2) Now we also know that

F_g = \frac{Gm_1m_2}{r^2}

F_e = \frac{kq_1q_2}{r^2}

both the forces are inversely depends on the square of the distance between two charges or two masses

so both forces follow inverse square law.

Now two ways they are alike is

1) gravitational force never exist in form of repulsive force while electromagnetic force is either repulsive or attractive.

2) Gravitational force is independent of the medium between two masses while electromagnetic force is dependent on the medium between two charges

Sladkaya [172]4 years ago
6 0
Differences between gravitational and electromagnetic radiation

So far we have been emphasizing how, at a fundamental level, the generation and propagation of gravitational and electromagnetic radiation are basically quite similar. This is a major point in demystifying gravitational waves. But, on a more practical level, gravitational and electromagnetic waves are quite different: we see and use electromagnetic waves every day, while we have yet to make a confirmed direct detection of gravitational waves (which is why they seemed so mysterious in the first place).

There are two principal differences between gravity and electromagnetism, each with its own set of consequences for the nature and information content of its radiation, as described below.

<span><span><span>Gravity is a weak force, but has only one sign of charge.
Electromagnetism is much stronger, but comes in two opposing signs of charge.</span>
This is the most significant difference between gravity and electromagnetism, and is the main reason why we perceive these two phenomena so differently. It has several immediate consequences:<span>Significant gravitational fields are generated by accumulating bulk concentrations of matter. Electromagnetic fields are generated by slight imbalances caused by small (often microscopic) separations of charge.<span>Gravitational waves, similarly, are generated by the bulk motion of large masses, and will have wavelengths much longer than the objects themselves. Electromagnetic waves, meanwhile, are typically generated by small movements of charge pairs within objects, and have wavelengths much smaller than the objects themselves.</span><span>Gravitational waves are weakly interacting, making them extraordinarily difficult to detect; at the same time, they can travel unhindered through intervening matter of any density or composition. Electromagnetic waves are strongly interacting with normal matter, making them easy to detect; but they are readily absorbed or scattered by intervening matter. 

</span><span>Gravitational waves give holistic, sound-like information about the overall motions and vibrations of objects. Electromagnetic waves give images representing the aggregate properties of microscopic charges at the surfaces of objects.</span></span>
</span><span><span>Gravitational charge is equivalent to inertia.
Electromagnetic charge is unrelated to inertia. </span>
This is the more fundamental difference between electromagnetism and gravity, and influences many of the details of gravitational radiation, but in itself is not responsible for the dramatic differences in how we perceive these two types of radiation. Most of the consequences of the principle of equivalence in gravity have already be discussed, such as:<span><span>The fundamental field of gravity is a gravitational force gradient (or tidal) field, and requires an apparatus spread out over some distance in order to detect it. The fundamental field in electromagnetism is an electric force field, which can be felt by individual charges within an apparatus.</span><span>The dominant mode of gravitational radiation is quadrupolar: it has a quadratic dependence on the positions of the generating charges, and causes a relative "shearing" of the positions of receiving charges. The dominant mode of electromagnetic radiation is dipolar: it has a linear dependence on the positions of the generating charges, and creates a relative translation of the positions of receiving charges.</span></span></span></span>
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The voltage across the terminals of a 250nF capacitor is푣푣=�50푉푉, 푡푡≤0(푚푚1푒푒−4000푡푡+푚푚2푡푡푒푒−4000푡푡)푉푉, 푡푡 ≥0The initial current
olga2289 [7]

The first part of the question is not complete and it is;

The voltage across the terminals of a 250 nF capacitor is 50 V, A1e^(-4000t) + (A2)te^(-4000t) V, t0, What is the initial energy stored in the capacitor? Express your answer to three significant figures and include the appropriate units. t

Answer:

A) initial energy = 0.3125 mJ

B) A1 = 50 and A2 = 1,800,000

C) Capacitor Current is given by the expression;

I = e^(-4000t)[0.95 - 1800t]

Explanation:

A) In capacitors, Energy stored is given as;

U = (1/2)Cv²

Where C is capacitance and v is voltage.

So initial kinetic energy;

U(0) = (1/2)C(vo)²

From the question, C = 250 nF and v = 50V

So, U(0) = (1/2)(250 x 10^(-9))(50²) = 0.3125 x 10^(-3)J = 0.3125 mJ

B) from the question, we know that;

A1e^(-4000t) + (A2)te^(-4000t)

So, v(0) = A1e^(0) + A2(0)e^(0)

v(0) = 50

Thus;

50 = A1

Now for A2; let's differentiate the equation A1e^(-4000t) + (A2)te^(-4000t) ;

And so;

dv/dt = -4000A1e^(-4000t) + A2[e^(-4000t) - 4000e^(-4000t)

Simplifying this, we obtain;

dv/dt = e^(-4000t)[-4000A1 + A2 - 4000A2]

Current (I) = C(dv/dt)

I = (250 x 10^(-9))e^(-4000t)[-4000A1 + A2 - 4000tA2]

Thus, Initial current (Io) is;

Io = (250 x 10^(-9))[e^(0)[-4000A1 + A2]]

We know that Io = 400mA from the question or 0.4 A

Thus;

0.4 = (250 x 10^(-9))[-4000A1 + A2]

0.4 = 0.001A1 - (250 x 10^(-9)A2)

Substituting the value of A1 = 50V;

0.4 = 0.001(50) - (250 x 10^(-9)A2)

0.4 = 0.05 - (250 x 10^(-9)A2)

Thus, making A2 the subject, we obtain;

(0.4 + 0.05)/(250 x 10^(-9))= A2

A2 = 1,800,000

C) We have derived that ;

I = (250 x 10^(-9))e^(-4000t)[-4000A1 + A2 - 4000tA2]

So putting values of A1 = 50 and A2 = 1,800,000 we obtain;

I = (250 x 10^(-9))e^(-4000t)[(-4000 x 50) + 1,800,000 - 4000(1,800,000)t]

I = e^(-4000t)[0.05 + 0.45 - 1800t]

I = e^(-4000t)[0.95 - 1800t]

5 0
4 years ago
What is the streak of a mineral used for
ad-work [718]

The streak of a mineral is used for identifying what the mineral may be.

Hope this helps :)

5 0
3 years ago
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Jlenok [28]

Answer:

A) 0 miles

Explanation:

The displacement is the distance between the starting point and the end point of the route.

In this case, even if Matino takes a whole tour around the city, since<u> it ends in the same place where it started</u>, the difference between the starting and finishing point is zero, so its total displacement is zero.

Care must be taken to distinguish the terms of displacement and distance traveled, beacause they are not the same, since in this case the distance traveled would be 3.15miles, but the displacement is zero, because it ends at the point where it started.

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Part 1 :
tatuchka [14]

1) 9.57 N

We have two forces applied on the apple:

- The force of gravity, in the downward direction:

W = 9.42 N

- The force exerted by the wind, in the horizontal direction (to the right):

Fw = 1.68 N

The two forces are perpendicular to each other, so we can find the magnitude of the net force by using Pythagorean's theorem.

Therefore, we have:

F=\sqrt{W^2+F_w^2}=\sqrt{(9.42)^2+(1.68)^2}=9.57 N

2) 10^{\circ}

The direction of the net external force, measured from the downward vertical, can be measured using the following formula:

\theta = tan^{-1}(\frac{F_x}{F_y})

where

F_x is the force in the horizontal direction

F_y is the force in the vertical direction

In this problem,

F_x = F_w = 1.68 N

F_y = W = 9.42 N

and so we find:

\theta = tan^{-1}(\frac{1.68}{9.42})=10^{\circ}

4 0
3 years ago
The hard external covering of some invertebrates is called a what
IRISSAK [1]
That would be the exoskeleton!
6 0
3 years ago
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