The answer is solar flare. Solar flares are commonly accompanied by coronal mass ejection. Solar flares are periods when the sun is suddenly brighter at a spot at its surface. This is due to the ejection of electrons, ions, and atoms (plasma clouds) accompanied by electromagnetic waves.
What’s the problem what do u need help on
Transmission of information in ANY form can be done digitally
or analoguely.
Beginning about 30 years ago, everything slowly started changing
to digital. Today, all commercial satellite communication, all optical
fiber communication, all internet communication, all computer
communication, all commercial cable communication, all commercial
television, and much of the telephone system, are all digital.
On your computer ... .pdf, .jpg, .mp3 etc. are all digital methods of
moving and storing information.
AM and FM radio are an interesting subject. They're all still analog.
They could easily be changed to all digital, and it would be a big
improvement, both for the broadcasters and for the listeners.
BUT ... every AM and FM radio that anybody has now would be
obsolete. Every single radio would either need to be replaced,
OR you'd need to add a digital decoder to every radio, like we
had to do with our TV sets a few years ago when television
suddenly became all digital. With AM and FM radios, the decoders
would be bigger, and would cost more, than most of the radios.
And that's why commercial radio broadcasting is still analog.
Answer:
512.5 mJ
Explanation:
Let the two identical charges be q = +35 µC and distance between them be r₁ = 46 cm. A charge q' = +0.50 µC located mid-point between them is at r₂ = 46 cm/2 = 23 cm = 0.23 m.
The electric potential at this point due to the two charges q is thus
V = kq/r₂ + kq/r₂
= 2kq/r₂
= 2 × 9 × 10⁹ Nm²/C² × 35 × 10⁻⁶ C/0.23 m
= 630/0.23 × 10³ V
= 2739.13 × 10³ V
= 2.739 MV
When the charge q' is moved 12 cm closer to either of the two charges, its distance from each charge is now r₃ = r₂ + 12 cm = 23 cm + 12 = 35 cm = 0.35 m and r₄ = r₂ - 12 cm = 23 cm - 12 cm = 11 cm = 0.11 cm.
So, the new electric potential at this point is
V' = kq/r₃ + kq/r₄
= kq(1/r₃ + 1/r₄)
= 9 × 10⁹ Nm²/C² × 35 × 10⁻⁶ C(1/0.35 m + 1/0.11 m)
= 315 × 10³(2.857 + 9.091) V
= 315 × 10³ (11.948) V
= 3763.62 × 10³ V
= 3.764 MV
Now, the work done in moving the charge q' to the point 12 cm from either charge is
W = q'(V' - V)
= 0.5 × 10⁻⁶ C(3.764 MV - 2.739 MV)
= 0.5 × 10⁻⁶ C(1.025 × 10⁶) V
= 0.5125 J
= 512.5 mJ
Answer:
19.4 seconds
Explanation:
We have:
m: mass of the car = 1500 kg
v₀: is the initial speed = 19 m/s
: is the final speed = 0 (it stops)
: is the coefficient of kinetic friction = 0.100
First, we need to find the acceleration by using the second Newton's law:
Solving for a:
Now we can find the time until it stops:
Solving for t:
Therefore, the time until it stops is 19.4 seconds.
I hope it helps you!
