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4 years ago

12

All the stars in this photo are at about the same distance from Earth (some 25,000 light-years away). Which stars in this pictur

e are the largest in size (radius)?

1 answer:

givi [52]4 years ago

7 0

Answer:

You did not add the photo, but the bright red stars are the largest in size (radius).

Explanation:

Red stars are giant, luminous stars. They are the stars with the largest size (radius) among all the other stars and usually have very low temperatures (below 500k). The brighter these stars appear to human eyes, the greater their actual size.

Red stars have rays hundreds of times larger than the sun, literally being the biggest stars in space. Despite their gigantic size, these stars have low density.

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(b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of 1510 m/s? (c) With wh

makvit [3.9K]

Answers:

(a) $2509.98 m/s$

(b) 397042.215 m

(c) 1917.76 m/s

Explanation:

The question is incomplete, please remember to write the whole question :) However, part (a) is written below:

(a) What is the escape speed on a spherical asteroid whose radius is $700 km$ and whose gravitational acceleration at the surface is $a_{g}=4.5 m/s^{2}$

Knowing this, let's begin:

a) In this part we need to find the escape speed $V_{e}$ on the asteroid:

$V_{e}=\sqrt{\frac{2GM}{R}}$ (1)

Where:

$G$ is the universal gravitational constant

$M$ is the mass of the asteroid

$R=700 km=700(10)^{3} m$ is the radius of the asteroid

On the other hand we know the gravitational acceleration is $a_{g}=4.5 m/s^{2}$ , which is given by:

$a_{g}=\frac{GM}{R^{2}}$ (2)

Isolating $GM$ :

$GM=a_{g}R^{2}$ (3)

Substituting (3) in (1):

$V_{e}=\sqrt{\frac{2a_{g}R^{2}}{R}}=\sqrt{2a_{g}R}$ (4)

$V_{e}=\sqrt{2(4.5 m/s^{2})(700(10)^{3} m)}$ (5)

$V_{e}=2509.98 m/s$ (6) This is the escape velocity

b) In this part we will use the Conservation of mechanical energy principle:

$E_{o}=E_{f}$ (7)

Being:

$E_{o}=K_{o}+U_{o}=\frac{1}{2}m V^{2} - \frac{GMm}{R}$ (8)

$E_{f}=K_{f}+U_{f}=0 - \frac{GMm}{R+h}$ (9)

Where:

$E_{o}$ is the initial mechanical energy

$E_{f}$ is the final mechanical energy

$K_{o}$ is the initial kinetic energy

$K_{f}=0$ is the final kinetic energy

$U_{o}$ is the initial gravitational potential energy

$U_{f}$ is the final gravitational potential energy

$m$ is the mass of the object

$V=1510 m/s$ is the radial speed of the object

$h$ is the distance above the surface of the object

Then:

$\frac{1}{2}m V^{2} - \frac{GMm}{R}=- \frac{GMm}{R+h}$ (10)

Isolating $h$ :

$h=\frac{2 a_{g} R^{2}}{2a_{g}R-V^{2}}-R$ (11)

$h=\frac{2 (4.5 m/s^{2}) (700(10)^{3} m)^{2}}{2(4.5 m/s^{2})(700(10)^{3} m)-(1510 m/s)^{2}}-700(10)^{3} m$ (11)

$h=397042.215 m$ (12) This is the distance above the asteroid's surface

c) We will use the Conservation of mechanical energy principle again, but now the condition is that the object is dropped at a distance $h=981.8 km=981.8(10)^{3} m$ . This means that at the begining the object only has gravitational potential energy and then it has kinetic energy and gravitational potential energy:

$\frac{-GMm}{R+h}=\frac{-GMm}{R}+\frac{1}{2}mV^{2}$ (13)

Isolating $V$ :

$V=\sqrt{2a_{g} R(1-\frac{R}{R+h})}$ (14)

$V=\sqrt{2(4.5 m/s^{2}) (700(10)^{3} m)(1-\frac{700(10)^{3} m}{700(10)^{3} m+981.8(10)^{3} m})}$ (15)

Finally:

$V=1917.76 m/s$

7 0

3 years ago

What is the function of hammer and anvil in human ear

MArishka [77]

Answer:

Vibrations of the eardrum are in turn translated into oscillations of tiny bones (ossicles) found within the middle ear. The Hammer, Anvil, and Stirrup magnify the oscillations and relay this information to the chamber of the inner ear.

4 0

3 years ago

Read 2 more answers

How would the model change as the atom forms bonds? The third shell would have eight electrons after the atom gains seven electr

atroni [7]

Answer:

The third shell would be empty, so the eight electrons on the second level would be the outermost after the atom lost one electron

Explanation:

When an atom is bonded with other atoms, a more stable configuration must be reached, which is why the energy of the molecule is less than the energy of the individual atoms, for this to happen in general, electrons are shared or lost and gained in each atom, depending on the electronegative of the same.

If we analyze an atom within the molecule, its last shell is full, in the case of atoms with few electrons in this shell, they are lost and in the case of many electors in this shell, it gains electrons to have eight (8) in total.

When reviewing the different answers, the correct one is:

* The third shell would be empty, so the eight electrons on the second level would be the outermost after the atom lost one electron

4 0

3 years ago

Read 2 more answers

What is Physics? How is Physics used in your everyday life? Give at least two examples and explain your answer. You need at leas

ikadub [295]

Answers:

The transportation industry is no stranger to the manipulation of everyday physics. Cars and trains utilize the wheel, which provides a smooth, steady motion.

The ears hear sounds which occur through the movement of air molecules, and the chemistry that drives all of biology depends on the physics of energy and molecules. Every day, for example, plants absorb sunlight, water, and carbon dioxide, creating glucose and releasing oxygen as a byproduct.

Brainlist pls!

3 0

3 years ago

How can I solve this?

inysia [295]

There is no equation here

3 0

3 years ago