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Elan Coil [88]
3 years ago
10

A parallel-plate capacitor with circular plates of radius 0.19 m is being discharged. A circular loop of radius 0.28 m is concen

tric with the capacitor and halfway between the plates. The displacement current through the loop is 2.6 A. At what rate is the electric field between the plates changing?
Physics
1 answer:
aliina [53]3 years ago
7 0

Answer:

\dfrac{dE}{dt}=2.59\times 10^{12}\ V/m.s

Explanation:

Given that

R= 0.19 m

r= 0.28 m

I= 2.6 A

We know that

I_D=\epsilon _oA\dfrac{dE}{dt}

A= Area of loop

dE/dt= rate of change of electric filed

I=Displacement current

Here r>R

So A=π R²

Now by putting the values

I=\epsilon _oA\dfrac{dE}{dt}

2.6=8.85\times 10^{-12}\times \pi\times 0.19^2\dfrac{dE}{dt}

\dfrac{dE}{dt}=2.59\times 10^{12}\ V/m.s

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Answer:

The answer is C.) 300,000,000 m/s

Explanation:

Light travels at a constant rate of 300,000,000 m/s. this can be determined by dividing the distance (144,000,000,000 meters) by the time (480 seconds). that's pretty fast. in fact, nothing can travel any faster than this. ever.

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3 years ago
A helium nucleus is composed of two protons (positively charged) and two neutrons (charge neutral). The distance between the two
lora16 [44]

Answer:

Using the given values

F = K q^2 / r^2 = 9 * 10E9 * (1.6 * E-19)^2 / (5.18 * E-15)^2 N

E = 9 * 1.6^2 / 5.18^2 * 10 = 8.5 N

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2 years ago
A short-wave radio antenna is supported by two guy wires, 155 ft and 175 ft long. Each wire is attached to the top of the antenn
Vladimir79 [104]

Answer:

163.8 ft

Explanation:

In triangle ABD

AB = 155 ft

Cos63 = \frac{BD}{AB} = \frac{BD}{155}\\BD = 155 Cos63 \\BD = 70.4 ft

Sin63 = \frac{AD}{AB} = \frac{AD}{155} \\AD = 166 Sin63\\AD = 148 ft

Using Pythagorean theorem in triangle ADC

AC^{2} = AD^{2} + DC^{2} \\175^{2} = 148^{2} + DC^{2} \\DC = 93.4 ft

d = distance between the anchor points

distance between the anchor points is given as

d = BD + CD = 70.4 + 93.4\\d = 163.8 ft

5 0
3 years ago
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Every few hundred years most of the planets line up on the same side of the Sun.(Figure 1)Calculate the total force on the Earth
mylen [45]

Answer: 3.7 \times 10^{-4} N

Explanation:

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F = G\frac{Mm}{r^2}

Where M and m are the masses of the object, r is the distance between the masses and G = 6.67× 10⁻¹¹ m³kg⁻¹ s⁻² is the gravitational constant.

We have to calculate the net force on Earth due to Venus, Jupiter and Saturn when they are in one line. It means when they are the closest distance.

F_{net] = G\frac{M_eM_v}{r_v^2}+G\frac{M_eM_j}{r_j^2}+G\frac{M_eM_s}{r_s^2}

Mass of Earth, Me = 5.98 × 10²⁴ kg

Mass of Venus, Mv = 0.815 Me

Mass of Jupiter, Mj = 318 Me

Mass of Saturn, Ms = 95.1 Me

closest distance between Earth and Venus, rv = 38 × 10⁶ km = 0.25 AU

closest distance between Jupiter and Earth, rj = 588 × 10⁶ km = 3.93 AU

closest distance between Earth and Saturn, rs = 1.2 × 10⁹ km = 8.0 AU

where 1 AU = 1.5 × 10¹¹ m

Inserting the values:

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4 0
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? Which statement is true of an object in equilibrium?
Degger [83]
The answer is C,<span> The sum of all forces acting on the object is zero. hope that helps!!</span>
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