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Leni [432]
3 years ago
8

In this section we considered a circular parallel-plate capacitor with a changing electric field. Describe the induced magnetic

field lines?
Physics
1 answer:
Soloha48 [4]3 years ago
7 0

Answer:

The study of Maxwell's equations and Ampere's law are used for analysis

Explanation:

Though There's a magnetic field associated with a changing electric field in TEM propagation of an EM wave through space (which is how it propogates, the changing E field begets the M field, the changeing M field begets the E field, leapfrogging each other).

But between capacitor plates, according to Maxwell and Ampere law, the displacement current in Ampere law was exactly to solve cases like that of a capacitor. A magnetic field cannot have discontinuities, unlike the electric field because there are electric charges, but there are no magnetic monopoles, at least as far as we know in the Universe in its current state. We can therefore conclude that there cannot be a magnetic field outside the capacitor and nothing inside

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A 55 kg track and field athlete has an average power output of 5.4 kW during the 200 meter dash. How quickly did she finish the
olga_2 [115]

The time taken for the athlete to finish the race is 20 s (Option A)

<h3>What is power? </h3>

Power is simply defined as the rate at which work is done. It can be expressed mathematically as

Power (P) = work (W) / time (t)

But

Work = weight × distance

Therefore,

Power = (weight × distance ) / time

<h3>How to determine the time </h3>
  • Mass (m) = 55 Kg
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Weight = mg = 55 × 9.8 = 539 N
  • Power (P) = 5.4 KW = 5.4 × 1000 = 5400 W
  • Distance (d) = 200 m
  • Time (t) =?

Power = (weight × distance ) / time

5400 = (539 × 200) / t

5400 = 107800 / t

Cross multiply

5400 × t = 107800

Divide both side by 5400

t = 107800 / 5400

t = 20 s

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brainly.com/question/5684937

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5 0
2 years ago
How many times can mechanical energy become chemical energy
maria [59]

Answer:

Chemical energy, Energy stored in the bonds of chemical compounds. Chemical energy may be released during a chemical reaction, often in the form of heat; such reactions are called exothermic.

3 0
3 years ago
A man stands at the midpoint between two speakers that are broadcasting an amplified static hiss uniformly in all directions. Th
Illusion [34]

Answer:

Explanation:

The power of each of the speakers is 0.535 W. At a distance d intensity of sound can be found by the following formula

Intensity of sound =  Power / 4π d²

= .535 / 4 x 3.14 x (27.3/2)²

= 2.286 x 10⁻⁴ J m⁻² s⁻¹

Intensity of sound due to other source = 5.715 x 10⁻⁵J m⁻² s⁻¹

Total intensity = 2 x 2.286 x 10⁻⁴J m⁻² s⁻¹

= 4.57 x 10⁻⁴J m⁻² s⁻¹

b ) In this case, man is standing at distances 18.15 m and 9.15 m from the sources .

The total intensity of sound reaching him is as follows

0.535 / (4 π x18.15² ) + 0.535 /  (4 π x9.15² )

= 1.293 x 10⁻⁴ + 5.087 x 10⁻⁴

= 6.38 x 10⁻⁴J m⁻² s⁻¹

5 0
3 years ago
What is one benefit of lifelong physical activity?
kvasek [131]
The answer would be C
5 0
3 years ago
A radar for tracking aircraft broadcasts a 12 GHz microwave beam from a 2.0-m-diameter circular radar antenna. From a wave persp
NISA [10]

A) 750 m

First of all, let's find the wavelength of the microwave. We have

f=12GHz=12\cdot 10^9 Hz is the frequency

c=3.0\cdot 10^8 m/s is the speed of light

So the wavelength of the beam is

\lambda=\frac{c}{f}=\frac{3\cdot 10^8 m/s}{12\cdot 10^9 Hz}=0.025 m

Now we can use the formula of the single-slit diffraction to find the radius of aperture of the beam:

y=\frac{m\lambda D}{a}

where

m = 1 since we are interested only in the central fringe

D = 30 km = 30,000 m

a = 2.0 m is the aperture of the antenna (which corresponds to the width of the slit)

Substituting, we find

y=\frac{(1)(0.025 m)(30000 m)}{2.0 m}=375 m

and so, the diameter is

d=2y = 750 m

B) 0.23 W/m^2

First we calculate the area of the surface of the microwave at a distance of 30 km. Since the diameter of the circle is 750 m, the radius is

r=\frac{750 m}{2}=375 m

So the area is

A=\pi r^2 = \pi (375 m)^2=4.42\cdot 10^5 m^2

And since the power is

P=100 kW = 1\cdot 10^5 W

The average intensity is

I=\frac{P}{A}=\frac{1\cdot 10^5 W}{4.42\cdot 10^5 m^2}=0.23 W/m^2

4 0
3 years ago
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