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Oduvanchick [21]
3 years ago
7

A solid weighs 200N in air, 150N in water and 170N in a liquid. Find relative density of solid, relative density of liquid and d

ensity of liquid in Sl system.
Physics
1 answer:
fomenos3 years ago
3 0

Answer:

\rho_{s} = 4

\rho_{l} = 0.6

\rho{liq} = 600 kg/m^{3}

Given:

Weight of solid in air, w_{sa} = 200 N

Weight of solid in water, w_{sw} = 150 N

Weight of solid in liquid, w_{sl} = 170 N

Solution:

Calculation of:

1. Relative density of solid, \rho_{s}

\rho_{s} = \frac{w_{sa}}{w_{sa} - w_{sw}}

\rho_{s} = \frac{200}{200 - 150} = 4

2. Relative density of liquid, \rho_{l}

\rho_{l} = \frac{w_{sa} - w_{sl}}{w_{sa} - w_{sw}}

\rho_{l} = \frac{200 - 170}{200 - 150} = 0.6

3. Density of liquid in S.I units:

Also, we know:

\rho{l} = \frac{\rho_{liq}}{\rho_{w}}

where

= {\rho_{liq}} = density of liquid

= {\rho_{w}} = 1000 kg/m^{3} = density of water

Now, from the above formula:

0.6 = \frac{\rho_{liq}}{1000}

\rho{liq} = 600 kg/m^{3}

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A balloon is a sphere with a radius of 5.0 m. the force of air against the walls of the balloon is 45 n. what is the air pressur
Mazyrski [523]

The air pressure inside the balloon is: 0.1432 Pa

The formulas and procedures that we will use to solve this problem are:

  • a = 4 * π * r²
  • P = F/a

Where:

  • a = area of the sphere
  • r = radius
  • π = mathematical constant
  • P = Pressure
  • F = Force
  • a = surface area

Information about the problem:

  • r = 5.0 m
  • F = 45 N
  • 1 Pa = N/m²
  • 1 N = kg * m/s²
  • a=?
  • P=?

Using the formula of the sphere area we get:

a = 4 * π * r²

a = 4 * 3.1416 * (5.0 m)²

a = 314.16 m²

Applying the pressure formula we get:

P = F/a

P = 45 N/314.16 m²

P = 0.1432 Pa

<h3>What is pressure?</h3>

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Learn more about pressure at: brainly.com/question/26269477

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