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3 years ago

6

Could an average star, such as our sun, become a neutron star? Explain your answer.

1 answer:

larisa86 [58]3 years ago

5 0

<span>No. Neutron stars are the remnants of very large stars that have supernova'd. Anything below 1.44 solar masses becomes a dwarf, anything above 5 solar masses becomes a black hole. Everything in between becomes a neutron star (or quark star, but it's not proven).</span>

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A train starts at rest in a station and accelerates at a constant 0.987 m/s2 for 182 seconds. Then the train decelerates at a co

algol [13]

Answer:

$\displaystyle X_T=66.6\ km$

Explanation:

<u>Accelerated Motion </u>

When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

$\displaystyle V_f=V_o+a\ t$

where a is the acceleration, and vo is the initial speed .

The train has two different types of motion. It first starts from rest and has a constant acceleration of $0.987 m/s^2$ for 182 seconds. Then it brakes with a constant acceleration of $-0.321 m/s^2$ until it comes to a stop. We need to find the total distance traveled.

The equation for the distance is

$\displaystyle X=V_o\ t+\frac{a\ t^2}{2}$

Our data is

$\displaystyle V_o=0,a=0.987m/s^2,\ t=182\ sec$

Let's compute the first distance X1

$\displaystyle X_1=0+\frac{0.987\times 182^2}{2}$

$\displaystyle X_1=16,346.7\ m$

Now, we find the speed at the end of the first period of time

$\displaystyle V_{f1}=0+0.987\times 182$

$\displaystyle V_{f1}=179.6\ m/s$

That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0

$\displaystyle V_o=179.6,a=-0.321\ m/s^2\ ,V_f=0$

$\displaystyle t=\frac{v_f-v_o}{a}=\frac{0-179.6}{-0.321}$

$\displaystyle t=559.5\ sec$

Computing the second distance

$\displaystyle X_2=179.6\times559.5\ \frac{-0.321\times 559.5^2}{2}$

$\displaystyle X_2=50,243.2\ m$

The total distance is

$\displaystyle X_t=x_1+x_2=16,346.7+50,243.2$

$\displaystyle X_t=66,589.9\ m$

$\displaystyle \boxed{X_T=66.6\ km}$

8 0

4 years ago

Un cuerpo se mueve en línea recta segun la ecuación x=10+20t-4.9t2 (x está expresado en metros y t en segundos). ¿Cuál es la lon

lora16 [44]

Answer:

La longitud del camino recorrido es de 25.9 [m]

Explanation:

Se reemplaza el valor de tiempo en segundos en la ecuación dada de desplazamiento

x=10+20*(3) - 4.9*(3)^2

x= 25.9 [metros]

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3 years ago

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What is the difference between the pressures inside and outside a bicycle tire called?

Lena [83]

Its called gauge pressure

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3 years ago

"put a bowl of butter in the microwave and set a timer for twenty seconds. Observer the state and properties of the butter after

Kobotan [32]

Answer:

The physical change of the butter in the microwave is the butter melting

Explanation:

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2 years ago

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An object is placed 50cm in front of a concave mirror of radius 60cm. How far from the mirror is the image?

Shkiper50 [21]

Answer:To explian the exact calculation of this pruduct we must not try!

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3 years ago