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2 years ago

What type of scientists identify the sun and stars

Physics

2 answers:

Paraphin [41]2 years ago

8 0

Answer:

Astronomers

Explanation:

Astronomy

Field of study

DescriptionAstronomy is a natural science that studies celestial objects and phenomena. It uses mathematics, physics, and chemistry to try and explain their origin and evolution. Objects of interest include planets, moons, stars, nebulae, galaxies, and comets.

https://en.wikipedia.org/wiki/Astronomer

PIT_PIT [208]2 years ago

8 0

Astronomers identify sun and stars

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How the Law of Conservation of Matter is supported by the experimental demonstrations?

Delvig [45]

The law of conservation of matter states that matter cannot be created nor destroyed. It can only be transformed from one form to another. To state an example where this is shown, let's say a piece of paper is burning. Not having a scientific background, you would say that the matter is being destroyed. But in reality, the paper is simple being transformed to ash, carbon dioxide, and water vapor. Overall, the total mass would still remain the same.

6 0

2 years ago

An ore sample weighs 17.50 N in air. When the sample

zysi [14]

Answer with Explanation:

We are given that

Weight of an ore sample=17.5 N

Tension in the cord=11.2 N

We have to find the total volume and the density of the sample.

We know that

Tension, T= $W-F_b$

$F_b$ =buoyancy force

T=Tension force

W=Weight

By using the formula

$11.2=17.5-F_b$

$F_b=17.5-11.2=6.3$ N

$F_b=V_{object}\times \rho_{water}\cdot g$

Where

$V_{object}$ =Volume of object

$\rho_{water}=1000 kgm^{-3}$ =Density of water

$g=9.8 ms^{-2}$ =Acceleration due to gravity

Substitute the values then we get

$6.3=9.8\times 1000\times V_{object}$

$V_{object}=\frac{6.3}{9.8\times 1000}=6.43\times 10^{-4} m^3$

Volume of sample= $6.43\times 10^{-4} m^3$

Density of sample, $\rho_{object}=\frac{Mass}{volume_{object}}$

Where mass of ore sample=1.79 kg

Substitute the values then, we get

$\rho_{object}=\frac{1.79}{6.43\times 10^{-4}}=2.78\times 10^3 kg/m^3$

Density of the sample= $2.78\times 10^{3} kgm^{-3}$

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3 years ago

A circular grill of diameter 0.25 m has an emissivity of 0.8. If the surface temperature is maintained at 150°C, determine the r

Schach [20]

Answer:

the required electrical power when the room air and surroundings are at 30°C.= 52.51822 Watt

Explanation:

Power required to maintain the surface temperature at 150°C from 20°C

P= εσA(T^4-t^4)

P= power in watt

ε= emissivity

A= area of surface

T= 150°C= 423 K

t= 20°C= 303K

/sigma= 5.67×10^{-8} watt/m^2K^4

putting vales we get

$= 0.8\times5.67\times10^{-8} \pi\frac{0.25^4}{4}(423^4-303^4)$

P=52.51822 Watt

the required electrical power when the room air and surroundings are at 30°C.= 52.51822 Watt

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2 years ago

In Scientific Notation, numbers are written in the form a x 10b . In this form, b is called the

ivann1987 [24]

Answer:

b is called the power of base 10.

Explanation:

In the scientific notation form

$a\,\,10^b$

the exponent b is called the power of the base 10.

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2 years ago

If the speed of an object is tripled, its kinetic energy will be

Savatey [412]

Answer:

if the speed of the object triples the kinetic energy becomes nine times the initial kinetic energy. If the speed of an object changes by a factor k, the change in the kinetic energy is of a factor k^2.

Explanation:

7 0

3 years ago

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