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3 years ago

State two condition necessary for a solid to float in a liquid

Physics

1 answer:

hoa [83]3 years ago

4 0

Answer:

it's density must be less than water

law of floatation

wt of the immerged body = wt of the water displaced by it

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Because white dwarfs are small, as their name implies, they are hard to see. What is a way astronomers have to find white dwarfs

Bumek [7]

Astronomers find white dwarfs that distinguish them from main sequence stars because white dwarfs get really hot, we can search for their ultraviolet radiation.

<h3>What is a white dwarf?</h3>

A white dwarf is a very hot star that radiated much energy in the form of ultraviolet radiation.

This UV radiation is initially very bright and then these stars become red with time.

In conclusion, Astronomers find white dwarfs they can search for their ultraviolet radiation.

Learn more about white dwarfs here:

brainly.com/question/19602278

#SPJ1

3 0

2 years ago

In a circuit, energy is transferred to a charge.<br> Where is this energy transferred from?

vladimir1956 [14]

Answer:

Electric current.

Explanation:

The energy result from electric current resulting from potential differences between terminals which form an Electric circuit. This energy could come from different sources like chemical, wind, light

An electric circuit is one where there is movement of electrons;this electrons acquire charge which is energy. The electrons flow due to a potential difference; you have heard water flows from a higher position to a lower one freely. The highest height is said to be at higher potential and the lower point low potential.

So it's the same with electrons.

The formular for energy on charge is Q= I × t where I is electric current and t is time.

7 0

3 years ago

How large is the Milky Way Galaxy? A. It is the largest galaxy ever observed. B. It takes up over half of the known universe. C.

lisabon 2012 [21]

Answer:

D. It is very small when compared to the universe

Explanation:

The Milky Way can be regarded galaxy which has Solar System in it. Milky way gives the description of appearance of galaxy from Earth, it is a hazy band of light that's been formed from the stars which can be visualized in the sky during the night, though it cannot be sorted by mere human eyes. Milky Way has existed for about 13.51 billion years with the radius of 52,850 light years. the Number of stars in milky way is about 100-400 billion. It should be noted that themilky way galaxy is a very large galaxy but It is very small when compared to the universe

4 0

3 years ago

Un the way to the moon, the Apollo astro-

kherson [118]

Answer:

Distance = 345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$