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2 years ago

Which of the following astronomical bodies or collection of astronomical bodies is likely to be a part of the others listed?

2 answers:

5 0

The correct answer is: Sun

Explanation:
Sun is the only star in our solar system, and planets orbit around it. Our solar system is the part of Milky way (located in one of the outer bands of the Milky way), and the Milky way is the part of the Universe (Universe is made up of billions of galaxies, and Milky way is one of those galaxies). Hence, in the given options, the Sun is the part of all of the other options given.

storchak [24]2 years ago

3 0

Answer:

SUNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNN

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Light would most likely be transmitted through A. A mirror B. A stone C. A leaf D. A window

ohaa [14]

A window is the most transparent object from these, so that is the answer.

7 0

3 years ago

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State whether the following statement are true or false .

Klio2033 [76]

1. It’s true
4 ,7 ,8 is correct

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2 years ago

A merry-go-round of radius 2 m is rotating at one revolution every 5 s. A

galben [10]

Answer:

a) The angular speed of the child is approximately 1.257 rad/s

b) The angular speed of the teenager is approximately 1.257 rad/s

c) The tangential speed of the child is approximately 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager is approximately 2.513 m/s

Explanation:

The revolutions per minute, r.p.m. of the merry-go-round = 1 revolution/(5 s)

The radius of the merry-go-round = 2 m

The location of the child = 1 m from the axis

The location of the teenager = 2 m from the axis

1 revolution = 2·π radians

Therefore, we have;

The angular speed, ω = (Angle turned)/(Time elapsed) = (2·π radians)/(5 s)

∴ The angular speed of the merry-go-round, ω = 2·π/5 radians/second

a) The angular speed of the child = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

b) The angular speed of the teenager = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

c) The tangential speed, v = r × The angular speed, ω

Where;

r = The radius of rotation of the object

For the child, r = 1 m

The tangential speed of the child = 1 m × 2·π/5 radians/second = 2·π/5 m/s ≈ 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager = 2 m × 2·π/5 radians/second = 4·π/5 m/s ≈ 2.513 m/s

8 0

2 years ago

what volume of alcohol will have the same mass as 4.2m^3 of petrol? (density of alcohol 0.4kg/m^3 and petrol is 0.3kg/m^3)

Alexxx [7]

Answer:

3.15m³

Explanation:

To solve this problem, let us first find the mass of the petrol from the given dimension.

Mass = density x volume

Volume of petrol = 4.2m³

Density of petrol = 0.3kgm⁻³

Mass of petrol = 4.2 x 0.3 = 1.26kg

So;

We can now find the volume of the alcohol

Volume of alcohol = $\frac{mass}{density}$

Mass of alcohol = 1.26kg

Density of alcohol = 0.4kgm⁻³

Volume of alcohol = $\frac{1.26}{0.4}$ = 3.15m³

7 0

2 years ago

What is the magnitude and direction (right or left) of the

Sunny_sXe [5.5K]

Answer: 12 N to the right

Explanation:

If we calculate the net force acting on the box, we will have:

In y-component:

$Fy_{net}=F_{n}+F_{g}$ (1)

Where $F_{n}=12 N$ is the Normal force, directed upwards and $F_{g}=-12 N$ is the weight of the box (gravity force), directed downwards.

$Fy_{net}=12 N-12 N$ (2)

$Fy_{net}=0 N$ (3) Hence the net force in the vertical component is zero

In x-component:

$Fx_{net}=F_{left}+F_{right}$ (4)

Where $F_{left}=-3 N$ and $F_{right}= 15 N$

$Fx_{net}=-3 N + 15 N$ (5)

$Fx_{net}=12 N$ (6) This is the net force in the horizontal component

Therefore, the total net force acting on the box is 12 N directed to the right

5 0

2 years ago