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Lena [83]
2 years ago
5

Water, oil and silver are poured into the vessel

Physics
1 answer:
taurus [48]2 years ago
7 0

The height of the oil column above the water in the vessel is determined as 2 cm.

<h3>Pressure of the vessel</h3>

The pressure of the vessel due to water, oil and silver poured into the vessel is determined from mercury column.

let level of mercury = 20 cm + 0.5 cm = 20.5 cm

20.5 cmHg = 205 mmHg

1 mmHg = 133.32 Pa

205 mmHg = 27,330.6 Pa

<h3>Height of the liquids in the vessel</h3>

P = ρgh

where;

ρ is the density of water, oil and silver respectively

ρ = 1000 kg/m³ + 881 kg/m³ + 10,800 kg/m³ = 12,681 kg/m³

h = P/(ρg)

h = (27,330.6) / (12,681 x 9.8)

h = 0.22 m

h = 22 cm

<h3>Height of oil column</h3>

Oil is less dense than water and will float on water.

Height of oil column = 22 cm - 20 cm = 2 cm

Learn more about density here: brainly.com/question/6838128

#SPJ1

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Answer:

A. nuclear fusion reactions

C. it's still hot from the big bang

Explanation:

The inside of the earth is hot due to some reasons. This heat provides the internal energy the drives processes within the earth interior. Here are some of the ways in which the heat has accumulated:

  • Nuclear reactions within the earth interior by fusion and other radioactive processes releases a large amount of heat.
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These are some of the sources of the earth's internal heat.

3 0
3 years ago
Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a
34kurt

Answer:

E=\frac{KQ}{2\sqrt 2a^2}

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}

Substitute x=a and R=a

Then, we get

E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}

E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}

E=\frac{KQa}{2\sqrt 2a^3}

E=\frac{KQ}{2\sqrt 2a^2}

Where K=9\times 10^9 Nm^2/C^2

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=\frac{KQ}{2\sqrt 2a^2}

4 0
3 years ago
When you go camping, you burn wood. Are you contributing to air pollution?
Jet001 [13]
Yes because of the smoke you are creating in the air
3 0
3 years ago
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worty [1.4K]

Answer:

308 N-s

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a stone is thrown down off a bridge with a veloctiy of 5.6m/s. What is the velocity afer 3 seconds have passed?
Flura [38]
I think it's 16.8 because you multiply 5.6x3=16.8
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4 years ago
Read 2 more answers
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