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MA_775_DIABLO [31]
4 years ago
14

Where do the electrons that form the auroras enter the magnetosphere? a. Through holes c. At the equator b. Between the magnetic

field lines d. In the magnetotail
Physics
2 answers:
marta [7]4 years ago
7 0
I think the answer is d. In the magnetotail. I hope this helps! :)
sergiy2304 [10]4 years ago
3 0

Answer:

Between the magnetic field lines.

Explanation:

The electron which forms the auroras can enter the magnetosphere through the invisible magnetic field lines. These charge particles comes from the sun as in the form of wind.

Therefore, the electrons which forms the auroras can enter the magnetosphere in between the magnetic field lines which are invisible.

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A fisherman is fishing from a bridge and is using a "44.0-N test line." In other words, the line will sustain a maximum force of
avanturin [10]

Answer:

(a) W= 44N

(b)W= 31.65 N

Explanation:

Data

T=44 N : Maximum force that the rope can withstand without breaking

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

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W: heaviest fish that can be pulled up vertically

∑F = 0

T-W =0  

W = T

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(b)  We apply the formula (1) , a= 1.26 m/s²

W: heaviest fish that can be pulled up vertically

W= m*g

m= W/g

g= 9.8 m/s² : acceleration due to gravity

∑F = 0

T-W = m*a

T= W+(W/g)*a

44=W*(1+1/9.8)* (1.26 )

44= W* 1.39

W= 44/1.39

W= 31.65 N

7 0
4 years ago
A force of 5.25 newtons acts on an object of unknown mass at a distance of 6.9 × 108 meters from the center of Earth. To increas
NemiM [27]
The force between two objects is calculated through the equation, 
                        F = Gm₁m₂/d²
where m₁ and m₂ are the masses of the objects. In this case, an unknown mass and Earth. d is the distance between them and G is universal gravitation constant. 

In the second case, if the force is to become 2.5 times the original and all the variables are constant except d then,
                       2.5F = Gm₁m₂ / (D²)
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Subsituting the known value of d,
                                D = 0.623(6.9 x 10^8) = <em>4.298 x 10^8 m</em>
                           
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By the help of newtons law of gravitation we can derive keplers third law of planetary motion.
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Write a function to accept a vector of masses (m) from the user and gives the corresponding energy to them. Energy vector is the
jarptica [38.1K]

Answer:

Written in Python

def energyvector(mass):

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Explanation:

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def energyvector(mass):

This line initializes the speed of light

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This line calculates the corresponding energy

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