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3 years ago

What is the most fundamental property of a star in determining its evolution? luminosity temperature composition mass size?

1 answer:

5 0

The most fundamental property of a star that determines its evolution is the MASS of the star.

In fact, while low-mass stars usually end their life becoming red giants and then white or black dwarfs, the high-mass stars can evolve into red supergiants and eventually originate a supernova. After the explosion of the supernova, depending on their mass, they can becomes neutron stars or collapse into black holes.

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Which statements about braking a car are true? A)The greater the kinetic energy of a car, the longer it takes for the car to sto

NikAS [45]

Answer:

option c

Explanation:

Kinetic energy is due to the speed of a body.

$K.E = \frac{1}{2}mv^2$

When speed is doubled, the kinetic energy is quadruple.

From third equation of motion, braking distance is also proportional to square of speed. Thus, when speed is doubled, the braking distance is quadruple.

Thus, option c is correct.

3 0

3 years ago

Read 2 more answers

A car traveling in a straight line has a velocity of 5.0 m/s. After an acceleration of 0.75 m/s/s, the cars velocity is 8.0. In

Bogdan [553]

Vs - velocity on beginning
ve - velocity on ending. You've got:
$v_s = 5 \frac{m}{s} \\ v_e=8 \frac{m}{s} \\ \hbox{Then:} \\ \Delta v=v_e - v_s = 8 \frac{m}{s} - 5\frac{m}{s}=3 \frac{m}{s} \\ a=0,75 \frac{m}{s^2} \\ \hbox{And from formula:} \\ a=\frac{\Delta v}{\Delta t} \qquad \Rightarrow \qquad \Delta t= \frac{\Delta v}{a} \\ \hbox{Substitute:} \\ \Delta t=\frac{3\frac{m}{s}}{0,75 \frac{m}{s^2}}= \frac{3}{\frac{3}{4}} s= 3 \cdot \frac{4}{3} s= 4 s$
So he needed 4 second.

3 0

3 years ago

A 1.65 kg mass stretches a vertical spring 0.260 m If the spring is stretched an additional 0.130 m and released, how long does

Irina-Kira [14]

Answer:

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

Explanation:

We notice that block-spring system depicts a Simple Harmonic Motion, whose equation of motion is:

$y(t) = A\cdot \cos \left(\sqrt{\frac{k}{m} }\cdot t +\phi\right)$ (1)

Where:

$y(t)$ - Position of the mass as a function of time, measured in meters.

$A$ - Amplitude, measured in meters.

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass of the block, measured in kilograms.

$t$ - Time, measured in seconds.

$\phi$ - Phase, measured in radians.

The spring is now calculated by Hooke's Law, that is:

$k = \frac{m\cdot g}{\Delta y}$ (2)

Where:

$g$ - Gravitational acceleration, measured in meters per square second.

$\Delta y$ - Deformation of the spring due to gravity, measured in meters.

If we know that $m=1.65\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\Delta y = 0.260\,m$ , then the spring constant is:

$k = \frac{(1.65\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.260\,m}$

$k = 62.237\,\frac{N}{m}$

If we know that $A = 0.130\,m$ , $k = 62.237\,\frac{N}{m}$ , $m=1.65\,kg$ , $x(t) = 0\,m$ and $\phi = 0\,rad$ , then (1) is reduced into this form:

$0.130\cdot \cos (6.142\cdot t)=0$ (1)

And now we solve for $t$ . Given that cosine is a periodic function, we are only interested in the least value of $t$ such that mass reaches equilibrium position. Then:

$\cos (6.142\cdot t) = 0$

$6.142\cdot t = \cos^{-1} 0$

$t = \frac{1}{6.142}\cdot \left(\frac{\pi}{2} \right)\,s$

$t \approx 0.255\,s$

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

4 0

3 years ago

The star Vega has an apparent visual magnitude of 0.03, and the star HR 4374 has an apparent visual magnitude of 4.87. It has be

laila [671]

Answer:

(D) Vega must produce more energy than HR 4374.

Explanation:

The apparent visual magnitude is defined as the luminosity of a star seen from earth. The Greek astronomer Hipparcos was the first who invent a numerical scale to describe how bright a star appear to be in the sky to the naked eye, so he gave an apparent magnitude of 6 to the faintest star in the sky and an apparent magnitude of 1 to the brigthest one. Since the arrive of the telescope era this range was expanded to negative numbers for the most luminous one.

The human eye has a logaritmical response to the luminosity of stars, so as an example, if one star has an apparent magnitude of 6 and another one has an apparent magnitude of 1, the second one will be 100 times brigther than the first one. In the other hand, for a star that has apparent magnitude of 1 it will appear to be 2.512 times brigther than a star with apparent magnitude of 2.

The apparent magnitude of a star can variate as a consecuence of the distance from earth or for its energy production.

With all these facts clear enough described above, it is easier to conclude that as the two star are at the same distance the apparent visual magnitude will depend in the energy production. In the case of Vega it has an apparent visual magnitude of 0.03, while HR 4374 has an apparent visual magnitude of 4.87. According with the scale stablished for Hipparcos, Vega will be more luminous than HR 4374. This means that Vega produce more energy than HR 4374.

8 0

3 years ago

A 2 kg ball is released from rest at the top of a track and

xz_007 [3.2K]

A) The formula for kinetic energy is E = 1/2 mv^2, so the energy of the ball is 1/2 * 2 * 10^2 = 100J. b) Energy is always conserved, and so if no energy is lost to resistive forces then all 100J of kinetic energy came from its potential energy at the top of the track. c) The formula for potential energy is E = mgh, which we can rearrange for h = E/mg. We know the energy, the mass and the strength of gravity, so we can find h = 100 / (2*9.81) = 5.10m.

7 0

3 years ago