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3 years ago

What are some of the physical properties of stars?

Physics

1 answer:

Arisa [49]3 years ago

8 0

Distance, luminosity, brightness, radius, chemical composition and temperature

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A proton is moved so that its electric potential energy increases from 4.0 × 10-14 J to 9.0 × 10-14 J. The magnitude of the char

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B. 3.1 × 10^5 V

Explanation:

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2 years ago

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Which of the following is the most dense?

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The answer is solid because gas is a chemical and water is a compound

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3 years ago

A light beam has speed c in vacuum and speed v in a certain plastic. The index of refraction n of this plastic is

Mnenie [13.5K]

Answer:

$n = \frac{c}{v}$

Explanation:

Refractive Index: It is a measure to find how fast the light travels through a medium. It is ration of the speed of light in vacuum to speed of light in the medium. Speed of light is not constant and varies depending on the density of the medium.

In vacuum the speed of light is 300000 km/s and is denoted by c. When the light beam enters any medium the speed will decrease. Here it is given that the speed in plastic is v. Thus the refractive index(n) is given as:

$n = \frac{c}{v}$

It is a dimensionless no.

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Given the hypothesis that the element zinc prevents cancer, which of the following procedures BEST exemplifies the nature of sci

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3 years ago

Block A has a mass of 0.5kg, and block B has a mass of 2kg. Block is is released at a height of 0.75 meters above B. The coeffic

VikaD [51]

Answer:

0.075 m

Explanation:

The picture of the problem is missing: find it in attachment.

At first, block A is released at a distance of

h = 0.75 m

above block B. According to the law of conservation of energy, its initial potential energy is converted into kinetic energy, so we can write:

$m_Agh=\frac{1}{2}m_Av_A^2$

where

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

$m_A=0.5 kg$ is the mass of the block

$v_A$ is the speed of the block A just before touching block B

Solving for the speed,

$v_A=\sqrt{2gh}=\sqrt{2(9.8)(0.75)}=3.83 m/s$

Then, block A collides with block B. The coefficient of restitution in the collision is given by:

$e=\frac{v'_B-v'_A}{v_A-v_B}$

where:

e = 0.7 is the coefficient of restitution in this case

$v_B'$ is the final velocity of block B

$v_A'$ is the final velocity of block A

$v_A=3.83 m/s$

$v_B=0$ is the initial velocity of block B

Solving,

$v_B'-v_A'=e(v_A-v_B)=0.7(3.83)=2.68 m/s$

Re-arranging it,

$v_A'=v_B'-2.68$ (1)

Also, the total momentum must be conserved, so we can write:

$m_A v_A + m_B v_B = m_A v'_A + m_B v'_B$

where

$m_B=2 kg$

And substituting (1) and all the other values,

$m_A v_A = m_A (v_B'-2.68) + m_B v_B'\\v_B' = \frac{m_A v_A +2.68 m_A}{m_A + m_B}=1.30 m/s$

This is the velocity of block B after the collision. Then, its kinetic energy is converted into elastic potential energy of the spring when it comes to rest, according to

$\frac{1}{2}m_B v_B'^2 = \frac{1}{2}kx^2$

where

k = 600 N/m is the spring constant

x is the compression of the spring

And solving for x,

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(2)(1.30)^2}{600}}=0.075 m$

5 0

3 years ago