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3 years ago

What temperature can a solar flare reach

Physics

1 answer:

balandron [24]3 years ago

5 0

It reaches 10 or 20 million degrees kelvin but it can get as high as 10 million degrees kelvin

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What is the time constant of a series circuit where the capacitor is 0.330μF and the resistor is 10Ω ?

PtichkaEL [24]

Answer:

$\tau=3.3*10^{-6}s$

Explanation:

Take at look to the picture I attached you, using Kirchhoff's current law we get:

$C*\frac{dV}{dt}+\frac{V}{R}=0$

This is a separable first order differential equation, let's solve it step by step:

Express the equation this way:

$\frac{dV}{V}=-\frac{1}{RC}dt$

integrate both sides, the left side will be integrated from an initial voltage v to a final voltage V, and the right side from an initial time 0 to a final time t:

$\int\limits^V_v {\frac{dV}{V} } =-\int\limits^t_0 {\frac{1}{RC} } \, dt$

Evaluating the integrals:

$ln(\frac{V}{v})=e^{\frac{-t}{RC} }$

natural logarithm to both sides in order to isolate V:

$V(t)=ve^{-\frac{t}{RC} }$

Where the term RC is called time constant and is given by:

$\tau=R*C=10*(0.330*10^{-6})=3.3*10^{-6}s$

3 0

2 years ago

An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi

Veronika [31]

Answer:

$s=5.79\ km$

$\theta=47^{\circ}$ east of south

Explanation:

Given:

distance of the person form the initial position, $d'=8.4\ km$

direction of the person from the initial position, $47^{\circ}$ north of east

distance supposed to travel form the initial position, $d=5.3\ km$

direction supposed to travel from the initial position, due North

Now refer the schematic for visualization of situation:

$y=d'.\sin47^{\circ}-d$

$y=8.4\times \sin47-5.3$ ...............(1)

$x=d'.\cos47^{\circ}$

$x=8.4\times \cos47^{\circ}$ .................(2)

Now the direction of the desired position with respect to south:

$\tan\theta=\frac{y}{x}$

$\tan\theta=\frac{8.4\times \sin47}{8.4\times \cos47}$

$\theta=47^{\circ}$ east of south

Now the distance from the current position to the desired position:

$s=\sqrt{x^2+y^2}$

$s=\sqrt{(8.4\times \cos47)^2+(8.4\times \sin47-5.3)^2}$

$s=5.79\ km$

4 0

3 years ago

HELP PLEASE!! GIVING BRAINLIEST!! If you answer this correctly ill answer some of your questions you have posted! (20pts)

anyanavicka [17]

Explanation:

potential energy =360800J

mass(m)=?

height (h)=25m

g=9.8m/s²

we have

potential energy =360800J

mgh=360800J

m×9.8×25=360800

m=360800/(9.8×25)=1472.653061kg

8 0

3 years ago

Read 2 more answers

A very long straight current-carrying wire produces a magnetic field of 25 µT at a distance d from the wire. How far will the ma

daser333 [38]

The magnetic field strength of a very long current-carrying wire is proportional to the inverse of the distance from the wire. The farther you go from the wire, the weaker the magnetic field becomes.

B ∝ 1/d

B = magnetic field strength, d = distance from wire

Calculate the scaling factor for d required to change B from 25μT to 2.8μT:

2.8μT/25μT = 1/k

k = 8.9

You must go to a distance of 8.9d to observe a magnetic field strength of 2.8μT

6 0

3 years ago

The table shows data for the planet Uranus. A 2 column table with 4 rows. The first column is labeled Quantity with entries, Esc

prohojiy [21]

Answer:

The answer is 218

Explanation:

Weight = mass * gravitational acceleration

weight is represented by F

F = 25kg (8.7)

(I'm pretty sure that you don't have to include the meters per second/per second thing)

4 0

3 years ago