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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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At what distance from the wire is the magnitude of the electric field equal to 2.41 n/cn/c ?
Sladkaya [172]

The correct answer is 1.07m.

The area surrounding an electric charge where its impact may be felt is known as the electric field. When another charge enters the field, the presence of an electric field may be felt. The electric field will either attract or repel the charge depending on its makeup. Any electric charge has a property known as the electric field. The charge and electrical force working in the field determine the strength or intensity of the electric field.

Here, is the charge per unit length, r is the distance from the wire, and

is the free space permittivity  ε_0. Electric field due to the long straight wire is,

E= λ/2πε_0r

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r=λ/2πε_0E

Substitute 2.41 N/C for E,

E=1.44×10^-10C/m

λ=8.85×10^-12C^2/Nm^2

r=(1.44×10^-10C/m)/(2(3.14)(8.85×10^-12C^2/Nm^2)(2.41N/C))

r=1.07m

At a distance of 1.07 m the magnitude of electric field is 2.41 N/C.

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brainly.com/question/12821750

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A tennis ball travels the length of the court 24m in 0.5s. Find the average speed
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