All categories

3 years ago

The particles ejected from the sun during a coronal mass ejection is called

Physics

1 answer:

ipn [44]3 years ago

7 0

Answer:

Plasma

Explanation:

A Coronal Mass Ejection (CME) is an outburst of energy that occurs near the outer part of the sun's atmosphere which causes a production of plasma along with a magnetic field.

The outermost part of the sun's atmosphere is called the Solar Corona Although difficult to see, the corona can be seen during a total solar eclipse.

Plasma from CME are clouds of magnetized electrically charged particles which the solar wind causes to travel at a speed of 1.6 million km/hr.

You might be interested in

A boy starts from point A and moves 5 units toward the east, then turns back and moves 3 units toward the west. What is the disp

Flura [38]

The displacement of the boy is 2 units

4 0

3 years ago

Calculate the kinetic energy of a moving 4 kg object traveling at a velocity of 3 m/a

Alina [70]

Answer:

Explanation:KE = 18 J

7 0

3 years ago

What are organelles?<br>Which part of a cell holds organelles?

DanielleElmas [232]

Answer:

An organelle is a structure in a cell that has specific jobs to do. Organelles are held in the cytoplasm

3 0

3 years ago

Read 2 more answers

Current is produced when charges are accelerated by an electric field to move to a position of lower

Vaselesa [24]

Current is created when charges are quickened by an electric field to move where the position of lower temperature. An electric current is a stream of electric charge. In electric circuits, this charge is regularly conveyed by moving electrons in a wire.

8 0

3 years ago

Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.

shusha [124]

Answer:

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

$r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2 }$

- Then compute the angle that Force makes with the y axis:

cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

F_net = F_1,2 + kq / y^2

- Hence,

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

- Now for the limit y >>a:

$F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2} )$

- Insert limit i.e a/y = 0

$F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2}) \\\\F_n = 3*k*q/y^2$

Hence the Electric Field is off a point charge of magnitude 3q.

8 0

3 years ago