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3 years ago

Where do magnetic fields occur?

Physics

1 answer:

Tresset [83]3 years ago

5 0

Answer:

<h2>The Magnetosphere and Magnets</h2>

Explanation:

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According to the National Geographic, "Earth’s magnetic field dominates a region called the magnetosphere, which wraps around the planet and its atmosphere. Solar wind, charged particles from the sun, presses the magnetosphere against the Earth on the side facing the sun and stretches it into a teardrop shape on the shadow side. The magnetosphere protects the Earth from most of the particles, but some leak through it and become trapped. When particles from the solar wind hit atoms of gas in the upper atmosphere around the geomagnetic poles, they produce light displays called auroras. These auroras appear over places like Alaska, Canada and Scandinavia, where they are sometimes called “Northern Lights.” The “Southern Lights” can be seen in Antarctica and New Zealand. The magnetic field is the area around a magnet that has magnetic force. All magnets have north and south poles."

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Hope this helps! <3

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Refer to the drawing that accompanies Check Your Understanding Question 14. Suppose that the voltage of the battery in the circu

Gnoma [55]

Answer:

Maximum speed of the rod, v = 10.34 m/s

Explanation:

It is given that,

Voltage of the battery, V = 2.7 V

The magnetic field perpendicularly into the plane of the paper is, B = 0.9 T

Length of the rod between the rails, l = 0.29 m

Due to the motion of the rails inside the magnetic field, an emf will induced in it which is given by :

$\epsilon=Blv$

v is the speed attained by the rod

$v=\dfrac{\epsilon}{Bl}$

$v=\dfrac{V}{Bl}$

$v=\dfrac{2.7}{0.9\times 0.29}$

v = 10.34 m/s

So, the maximum speed attained by the rod after the switch is closed is 10.34 m/s. Hence, this is the required solution.

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3 years ago

Why is it more helpful to know a tornadoes velocity rather than its speed

victus00 [196]

So you not only know how fast it is going, but where it is going.

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3 years ago

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Oscilloscopes have parallel metal plates inside them to deflect the electron beam. These plates are called the deflecting plates

alexdok [17]

Answer:

1.6 pF

Explanation:

The capacitance of a parallel-plate capacitor in air is given by:

$C=\frac{\epsilon_0 A}{d}$

where

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A is the area of each plate

d is the separation between the plates

In this problem we have:

- Separation between the plates: d = 5.00 mm = 0.005 m

- Area of the plates: $A = 3.00 cm \cdot 3.00 cm = 9.00 cm^2 = 9\cdot 10^{-4}m^2$

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4 years ago

To properly operate the laboratory thermometer Group of answer choices it needs to be shaken but make sure you have enough room

ruslelena [56]

Answer:

it needs to be shaken but make sure you have enough room to shake it safely

Explanation:

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3 years ago

You are traveling in a car toward a hill at a speed of 36.4 mph. The car's horn emits sound waves of frequency 231 Hz, which mov

Marina CMI [18]

Answer:

a. The frequency with which the waves strike the hill is 242.61 Hz

b. The frequency of the reflected sound wave is 254.23 Hz

c. The beat frequency produced by the direct and reflected sound is

11.62 Hz

Explanation:

Part A

The car is the source of our sound, and the frequency of the sound wave it emits is given as 231 Hz. The speed of sound given can be used to determine the other frequencies, as expressed below;

$f_{1} = f[\frac{v_{s} }{v_{s} -v} ]$ ..............................1

where $f_{1}$ is the frequency of the wave as it strikes the hill;

f is the frequency of the produced by the horn of the car = 231 Hz;

$v_{s}$ is the speed of sound = 340 m/s;

v is the speed of the car = 36.4 mph

Converting the speed of the car from mph to m/s we have ;

hint (1 mile = 1609 m, 1 hr = 3600 secs)

v = $36.4 mph *\frac{1609 m}{1 mile} *\frac{1 hr}{3600 secs}$

v = 16.27 m/s

Substituting into equation 1 we have

$f_{1}$ = $231 Hz (\frac{340 m/s}{340 m/s - 16.27 m/s})$

$f_{1}$ = 242.61 Hz.

Therefore, the frequency which the wave strikes the hill is 242.61 Hz.

Part B

At this point, the hill is the stationary point while the driver is the observer moving towards the hill that is stationary. The frequency of the sound waves reflecting the driver can be obtained using equation 2;

$f_{2} = f_{1} [\frac{v_{s}+v }{v_{s} } ]$

where $f_{2}$ is the frequency of the reflected sound;

$f_{1}$ is the frequency which the wave strikes the hill = 242.61 Hz;

$v_{s}$ is the speed of sound = 340 m/s;

v is the speed of the car = 16.27 m/s.

Substituting our values into equation 1 we have;

$f_{2} = 242.61 Hz [\frac{340 m/s+16.27 m/s }{340 m/s } ]$

$f_{2}$ = 254.23 Hz.

Therefore, the frequency of the reflected sound is 254.23 Hz.

Part C

The beat frequency is the change in frequency between the frequency of the direct sound and the reflected sound. This can be obtained as follows;

Δf = $f_{2}$ - $f_{1}$

The parameters as specified in Part A and B;

Δf = 254.23 Hz - 242.61 Hz

Δf = 11.62 Hz

Therefore the beat frequency produced by the direct and reflected sound is 11.62 Hz

3 0

3 years ago