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daser333 [38]
4 years ago
6

What factor determines how long a star lives?

Physics
2 answers:
Makovka662 [10]4 years ago
8 0
Their lifespan is determined by how large they are and their weight.
german4 years ago
5 0
The factor that determines how long a star can live is the mass of the star, or how big the star is. 
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Can Magnetic poles effects each other​
pochemuha

Answer:

Explanation:

But the reality is that: Multiple magnetic fields would fight each other. This could weaken Earth's protective magnetic field by up to 90% during a polar flip. Earth's magnetic field is what shields us from harmful space radiation which can damage cells, cause cancer, and fry electronic circuits and electrical grids.

How do magnetic poles interact? Magnetic poles that are alike repel each other, and magnetic poles that are unlike attract each other. The area of magnetic force around a magnet. The magnetic field lines spread out from the north pole, curve around, and return to the south pole.

When two magnets are brought together, the opposite poles will attract one another, but the like poles will repel one another. This is similar to electric charges. The earth is like a giant magnet, but unlike two free hanging magnets, the north pole of a magnet is attracted to the north pole of the earth.

Magnetic forces are non contact forces; they pull or push on objects without touching them. Magnets are only attracted to a few 'magnetic' metals and not all matter. Magnets are attracted to and repel other magnets.

(hope this helps can i plz have brainlist :D hehe)

3 0
3 years ago
A sinusoidal, transverse wave that propagates in the positive x-direction can be described with the wave function y of xt is equ
myrzilka [38]

Answer:

\omega =\frac{2 \pi }{T} rad/s

Explanation:

The wave function is:

y(xt) = Acos(kx- \omega t)

where :

k = wave number

x = position of a point on the wave

\omega = angular frequency

t = time

What is another way to express the angular frequency (omega)

Angular frequency (omega) can be express as :

\omega =\frac{2 \pi }{T} rad/s ( i.e one repetition that it takes to repeats itself)

4 0
3 years ago
Particle A of charge 2.70 10-4 C is at the origin, particle B of charge -6.36 10-4 C is at (4.00 m, 0), and particle C of charge
Serhud [2]

Answer:

FC vector representation

F_{C} =(19.03i+13.78j)N

Magnitude of FC

F_{C}=23.495N

Vector direction FC

\beta=35.91 degrees: angle that forms FC with the horizontal

Explanation:

Conceptual analysis

Because the particle C is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

The directions of the individual forces exerted by qA and qB on qC are shown in the attached figure; The force (FAC) of qA over qC is repulsive because they have equal signs and the force (FBC) of qB over qC is attractive because they have opposite signs.

The FAC force is up in the positive direction and the FBC force forms an α angle with respect to the x axis.

\alpha = tan^{-1}(\frac{3}{4}) = 36.86 degrees

To calculate the magnitudes of the forces we apply Coulomb's law:

F_{AC} = \frac{k*q_{A}*q_{C}}{r_{AC}^2} Equation (1): Magnitude of the electric force of the charge qA over the charge qC

F_{BC} = \frac{k*q_{B}*q_{C}}{r_{BC}^2} Equation (2) : Magnitude of the electric force of the charge qB over the charge qC

Known data

k=8.99*10^9 \frac{N*m^2}{C^2}

q_{A}=2.70*10^{-4} C

q_{B}=-6.36*10^{-4} C

q_{C}=1.04*10^{-4} C

r_{AC} =3

r_{BC}=\sqrt{4^2+3^2} = 5

Problem development

In the equations (1) and (2) to calculate FAC Y FBC:

F_{AC} =8.99*10^9*(2.70*10^{-4}* 1.04*10^{-4})/(3)^2=28.05N

F_{BC} =8.99*10^9*(6.36*10^{-4}* 1.04*10^{-4})/(5)^2=23.785N

Components of the FBC force at x and y:

F_{BCx}=23.785 *Cos(36.86)=19.03N

F_{BCy}=23.785 *Sin(36.86)=14.27N

Components of the resulting force acting on qC:

F_{Cx} = F_{ACx}+ F_{BCx}=0+19.03=19.03N

F_{Cy} = F_{ACy}+ F_{BCy}=28.05-14.27=13.78N

FC vector representation

F_{C} =(19.03i+13.78j)N

Magnitude of FC

F_{C}= \sqrt{19.03^2+13.78^2} =23.495N

Vector direction FC

\beta = tan^{-1} (\frac{13.78}{19.03})=35.91 degrees: angle that forms FC with the horizontal

7 0
3 years ago
Please have a look at the image!
EastWind [94]
Nice problem!

Look. At point A the road changes from a straight line to a circular line, then the motion changes from  rectilinear to circular motion. So at that point, there is a sudden change in the acceleration, which for sure the passengers feel.

At point B, the circular motion changes from anticlockwise to clockwise, then there is another sudden change in the acceleration.

At point C, the road changes from circular to straihgt line, then the motion changes from rectilinear to circular, which produces a new sudeen change in acceleration.

Then the answeer is the last option, at points A, B and C.
3 0
3 years ago
You are playing baseball and hit a single, running to first base. Your distance and displacement are
Harrizon [31]

Answer:

C. Distance 90 feet Displacement 0 feet

8 0
3 years ago
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