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3 years ago

What population of stars does our star belong to?

Physics

1 answer:

Dvinal [7]3 years ago

4 0

Answer:

Our star ,that is, Sun belong to Population I.

Explanation:

In 1940s, stars were divided into two categories by Wlter Baade:

a. Population I (Metal-rich)

b. Population II (Metal-poor)

Population I stars tend to be luminous, concentrated in the disks of spiral galaxies. They are still hot and young containing rich amount of heavy metals.

Sun falls in this category.

Population II stars tend to be less luminous found in the nucleus of the galaxy.

They are less hot. They are poor in heavy metals because they are old.

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What is Circular Motion?

nevsk [136]

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5 0

3 years ago

Read 2 more answers

Frank has a sample of steel that has a mass of 80 grams if the density is 8g/cm3 what is the volume

nikklg [1K]

Density is mass divided by volume. rho=m/v. So, v=m/rho. In frank's case this is 80/8 = 10 cm^3.

7 0

3 years ago

Two boats start together and race across a 48-km-wide lake and back. Boat A goes across at 48 km/h and returns at 48 km/h. Boat

nika2105 [10]

Answer:

Part 1)

Boat A will win the race

Part 2)

Boat A will win the race by 48 km as the 2nd boat will reach the other end while boat A will just touches the finish line

Part 3)

average velocity must be zero

Explanation:

As we know that the distance moved by the boat is given as

$d = 48 km$

now the time taken by the boat to move to and fro is given as

$t = \frac{d}{v}$

$t = \frac{48 + 48}{48}$

$t = 2 hrs$

Time taken by Boat B to cover the distance

$t = \frac{48}{24} + \frac{48}{72}$

$t = 2.66 h$

Part 1)

Boat A will win the race

Part 2)

Boat A will win the race by 48 km as the 2nd boat will reach the other end while boat A will just touches the finish line

Part 3)

Since the displacement of Boat A is zero

so average velocity must be zero

3 0

3 years ago

Urgent!

Masja [62]

mass of iron block given as

$m_1 = 1.90 kg$

density of iron block is

$\rho = 7860 kg/m^3$

now the volume of the iron piece is given as

$V = \frac{m}{\rho}$

$V = \frac{1.90}{7860} = 2.42* 10^{-4} m^3$

Now when this iron block is complete submerged in oil inside the beaker the buoyancy force on the iron block will be given as

$F_b = \rho_L V g$

here we know that

$\rho_L$ = density of liquid = 916 kg/m^3

$F_b = 916* 2.42 * 10^{-4} * 9.8$

$F_b = 2.17 N$

Now for the reading of spring balance we can say the spring force and buoyancy force on the block will counter balance the weight of the block at equilibrium

$F_s + F_b = mg$

$F_s + 2.17 = 1.90* 9.8$

$F_s = 16.45 N$

So reading of spring balance will be 16.45 N

Now for other scale which will read the normal force of the surface we can write that normal force on the container will balance weight of liquid + container and buoyancy force on block

$F_n = F_g + F_b$