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3 years ago

The mass of a star can be determined by studying ___.

Physics

2 answers:

oksian1 [2.3K]3 years ago

7 0

Answer:

C. Binary star systems

Explanation:

Citrus2011 [14]3 years ago

4 0

Answer: C. Binary star systems

Let’s see your options:

a) The color of the star: the color is not used in calculating the mass of a star, because it has no relation to it. Think about a red supergiant and a red dwarf: they have the same color, but they are completely different stars, with respectively a big and a small mass.

b) Kepler’s laws: these laws can be applied in what is called the “approximation of 1 body”, which means that is assumed that one body has a much bigger mass than the other and can be considered at rest. This is the case of a star-planet system and the mass that can be calculated is that of the planet.

c) Binary star systems: these are the only cases in which is possible the direct measure of the mass of the stars. Binary systems are classified as follows:
- Visual binaries: each star can be resolved and the motion around the center of mass can be measured.
- Astrometric binaries: only one star is visible, but the presence of the companion can be inferred by the movement of the first star around the system’s center of mass.
- Eclipse binaries: the two stars are not resolved (separated), but the luminosity varies periodically when one star eclipses the other.
- Spectroscopic binaries: the two stars are not resolved, but their spectrum reveals that they are a binary system.
In all these cases we have a “two-body problem” that can be solved by changing system of reference: the motion of bodies 1 and 2 is equivalent to the motion of a body of mass equal to the system’s reduced mass $\mu = \frac{M_{1} \cdot M_{2}}{M_{1} + M_{2}}$ moving in the potential generated by the total mass (M1 + M2) considered at rest. Hence, we can determine the masses of the two stars.

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A 2.7-kg block is released from rest and allowed to slide down a frictionless surface and into a spring. The far end of the spri

exis [7]

a) The speed of the block at a height of 0.25 m is 2.38 m/s

b) The compression of the spring is 0.25 m

c) The final height of the block is 0.54 m

Explanation:

We can solve the problem by using the law of conservation of energy. In fact, the total mechanical energy (sum of kinetic+gravitational potential energy) must be conserved in absence of friction. So we can write:

$U_i +K_i = U_f + K_f$

where

$U_i$ is the initial potential energy, at the top

$K_i$ is the initial kinetic energy, at the top

$U_f$ is the final potential energy, at halfway

$K_f$ is the final kinetic energy, at halfway

The equation can be rewritten as

$mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2$

where:

m = 2.7 kg is the mass of the block

$g=9.8 m/s^2$ is the acceleration of gravity

$h_i = 0.54$ is the initial height

u = 0 is the initial speed

$h_f = 0.25 m$ is the final height of the block

v is the final speed when the block is at a height of 0.25 m

Solving for v,

$v=\sqrt{u^2+2g(h_i-h_f)}=\sqrt{0+2(9.8)(0.54-0.25)}=2.38 m/s$

The total mechanical energy of the block can be calculated from the initial conditions, and it is

$E=K_i + U_i = 0 + mgh_i = (2.7)(9.8)(0.54)=14.3 J$

At the bottom of the ramp, the gravitational potential energy has become zero (because the final heigth is zero), and all the energy has been converted into kinetic energy. However, then the block compresses the spring, and the maximum compression of the spring occurs when the block stops: at that moment, all the energy of the block has been converted into elastic potential energy of the spring. So we can write

$E=E_e = \frac{1}{2}kx^2$

where

k = 453 N/m is the spring constant

x is the compression of the spring

And solving for x, we find

$x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(14.3)}{453}}=0.25 m$

If there is no friction acting on the block, we can apply again the law of conservation of energy. This time, the initial energy is the elastic potential energy stored in the spring:

$E=E_e = 14.3 J$