answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Contact [7]
1 year ago
5

Determine the correlation between coronal mass ejections from the Sun to the accumulation of the rare and valuable isotope He3 t

hat has a high concentration on the Moon.
its for a project i appreciate the help
its an open ended question so feel free to give lots of detail
because of the nature of this question and the need for a quality answer i will award brainliest where i see fit and this question is worth a lot of point
Physics
1 answer:
Reil [10]1 year ago
6 0

Most ejections originate from active regions on the Sun's surface, such as groupings of sunspots associated with frequent flares. These regions have closed magnetic field lines, in which the magnetic field strength is large enough to contain the plasma.

You might be interested in
A system contains a perfectly elastic spring, with an unstretched length of 20 cm and a spring constant of 4 N/cm.
mote1985 [20]

Answer:

a) When its length is 23 cm, the elastic potential energy of the spring is

0.18 J

b) When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J

Explanation:

Hi there!

a) The elastic potential energy (EPE) is calculated using the following equation:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = stretched lenght.

Let´s calculate the elastic potential energy of the spring when it is stretched 3 cm (0.03 m).

First, let´s convert the spring constant units into N/m:

4 N/cm · 100 cm/m = 400 N/m

EPE = 1/2 · 400 N/m · (0.03 m)²

EPE = 0.18 J

When its length is 23 cm, the elastic potential energy of the spring is 0.18 J

b) Now let´s calculate the elastic potential energy when the spring is stretched 0.06 m:

EPE = 1/2 · 400 N/m · (0.06 m)²

EPE = 0.72 J

When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J

7 0
2 years ago
In a 1.25-T magnetic field directed vertically upward, a particle having a charge of magnitude 8.50μC and initially moving north
Goshia [24]

Answer:

a) The sign of the charge is positive.

b) The magnetic force on the particle is 0.050 newtons.

Explanation:

The magnetic force F on a moving charge with velocity v passing through a magnetic field B is:

\overrightarrow{F}=q\overrightarrow{v}\times\overrightarrow{B}(1)

a)

Because it is a cross product, we can find the direction of the force using the right-hand rule, that is too the direction of the movement. We have two possibilities here because the velocity vector and magnetic field are perpendicular: the particle deflects towards east or toward west, which depends on the charge of the particle. Note that if you put your right hand fingers, except thumb, pointing towards north (direction of velocity) and later close them in the direction of the magnetic field, if you maintain your thumb perpendicular to this movement it will point towards east (See figure), so that will be de direction of the force if the charge is positive, but if the charge is negative, the direction will be opposite (towards west). So the charge has to be positive to deflects towards east.

b)

Now by 1:

F=qvB\sin\theta=(8.50\times10^{-6})(4750)(1.25)\sin90\simeq\mathbf{0.050\,N}

5 0
2 years ago
A lightning bolt transfers 6.0 coulombs of charge from a cloud to the ground in 2.0 x 10-3 second. what is the average current d
AlladinOne [14]
The current is defined as the amount of charge transferred through a certain point in a certain time interval:
I= \frac{Q}{\Delta t}
where
I is the current
Q is the charge
\Delta t is the time interval

For the lightning bolt in our problem, Q=6.0 C and \Delta t= 2.0 \cdot 10^{-3}s, so the average current during the event is
I= \frac{Q}{\Delta t} = \frac{6.0 C}{2.0 \cdot 10^{-3} s}=3000 A
4 0
2 years ago
A dog of mass 10 kg sits on a skateboard of mass 2 kg that is initially traveling south at 2 m/s. The dog jumps off with a veloc
Tasya [4]

Answer:

17 m/s south

Explanation:

m_1 = Mass of dog = 10 kg

m_2 = Mass of skateboard = 2 kg

v = Combined velocity = 2 m/s

u_1 = Velocity of dog = 1 m/s

u_2 = Velocity of skateboard

In this system the linear momentum is conserved

(m_1+m_2)v+m_1u_1+m_2u_2=0\\\Rightarrow u_2=-\dfrac{(m_1+m_2)v+m_1u_1}{m_2}\\\Rightarrow u_2=-\dfrac{(10+2)2+10\times 1}{2}\\\Rightarrow u_2=-17\ m/s

The velocity of the skateboard will be 17 m/s south as the north is taken as positive

3 0
2 years ago
A housefly walking across a surface may develop a significant electric charge through a process similar to frictional charging.
Umnica [9.8K]

Answer:

Number of electrons, n=3.87\times 10^8\ electrons

Explanation:

Given that,

Charge on the fly, q=62\ pC=62\times 10^{-12}\ C

Let n is the number of electrons lose to the surface it is walking across. It is case of quantization of electric charge. It is given by :

q=ne

n=\dfrac{q}{e}

n=\dfrac{62\times 10^{-12}}{1.6\times 10^{-19}}

n = 387500000 electrons

or

n=3.87\times 10^8\ electrons

So, there are 3.87\times 10^8 electrons. Hence, this is the required solution.

3 0
2 years ago
Other questions:
  • The charged particles in the beams that Thomson studied came from atoms. As these particles moved away from their original atoms
    8·2 answers
  • Object A with a mass of 500 kilograms hits stationary object B with a mass of 920 kilograms. If the collision is elastic, what h
    14·1 answer
  • 568 muons were counted by a detector on the top of Mount Washington in a one hour period of time. Assuming moving muons keep tim
    8·2 answers
  • Isabella drops a pen off her balcony by accident while celebrating the successful completion of a physics problem. assuming air
    6·1 answer
  • Match each force abbreviation to the correct description. Fg Fp Ff Fn force exerted by a push or pull. Support force at a right
    10·2 answers
  • A rocket sled accelerates at 21.5 m/s^2 for 8.75 s. (a) What's its velocity at the end of that time? (b) How far has it traveled
    8·1 answer
  • Assuming that you remain a finite distance from the origin, where in the X-Y plane could a point charge Q be placed, so that thi
    5·1 answer
  • In an experiment, students release a block from rest at the top of an inclined plane. The block slides down the plane through a
    11·1 answer
  • Write a hypothesis about the effect of the angle of the track on the acceleration of the cart. Use the "if . . . then . . . beca
    7·1 answer
  • A 3400 kg jet is flying at a constant speed of 170 m/s as it makes a vertical loop. At the top of the loop the pilot feels three
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!