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slavikrds [6]
4 years ago
7

Where is most plasma located?

Physics
1 answer:
adell [148]4 years ago
4 0

Answer:

Plasma is by far the most common form of matter. Plasma in the stars and in the tenuous space between them makes up over 99% of the visible universe and perhaps most of that which is not visible. On earth we live upon an island of "ordinary" matter.

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YOU READING THIS PLEASE GIVE ME A EASY 7TH GRADE TOPIC NO VOLCANIOS
Kipish [7]
Photothysenis. or fossils they are very easy
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4 years ago
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A CD has to rotate under the readout-laser with a constant linear velocity of 1.25 m/s. If the laser is at a position 3.7 cm fro
Savatey [412]

Answer:N=322.53 rpm

Explanation:

Given

Linear velocity (v)=1.25 m/s

Position from center is 3.7 cm

we know

v=\omega \times r

1.25\times 100=\omega \times 3.7

\omega =\frac{125}{3.7}=33.78

and \frac{2\pi N}{60}=\omega

N=\frac{\omega \times 60}{2\pi }

N=\frac{33.78\times 60}{2\pi }

N=322.53 rpm

8 0
4 years ago
А______<br> is a unit of measurement for energy. (7 Letters)
a_sh-v [17]

Answer:

Joule or kilowatt/hour

7 0
3 years ago
The aurora is caused when electrons and protons, moving in the earth’s magnetic field of ≈5.0×10−5T, collide with molecules of t
ollegr [7]

Answer:

8.79*10^6 rad/s

Explanation:

To find the frequency of the circular orbit for an electron you use the following expression, for the radius of the trajectory of an electron, that travels trough a constant magnetic field:

r=\frac{mv}{qB}         (1)

r: radius of the trajectory

m: mass of the electron = 9.1*10^-31 kg

v: speed of the electron = 1.0*10^6 m/s

q: charge of the electron = 1.6*10^-19 C

B: magnitude of the magnetic field = 5.0*10^-5 T

You use the fact that the angular frequency in a circular motion is given by:

\omega=\frac{v}{r}

Then, you solve the equation (1) in order to obtain v/r:

\frac{v}{r}=\omega=\frac{qB}{m}

Finally, you replace the values of the parameters:

\omega=\frac{(1.6*10^{-19}C)(5.0*10^{-5}T)}{9.1*10^{-31}kg}\\\\\omega=8.79*10^6\frac{rad}{s}

hence, the angular frequency is 8.79*10^6 rad/s

The frequency is:

f=2\pi \omega=5.5*10^7Hz

5 0
3 years ago
A proton is propelled at 4×106 m/s perpendicular to a uniform magnetic field. 1) If it experiences a magnetic force of 4.8×10−13
tia_tia [17]

Answer:

B = 0.75 T

Explanation:

As we know that the force on a moving charge in magnetic field is given by the formula

F = qvB

here we have

B = \frac{F}{qv}

here we know that

F = 4.8 \times 10^{-13} N

q = 1.6 \times 10^{-19} C

v = 4 \times 10^6 m/s

now from above equation we have

B = \frac{4.8 \times 10^{-13}}{(1.6 \times 10^{-19})(4 \times 10^6)}

B = 0.75 T

8 0
4 years ago
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