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givi [52]
3 years ago
15

The Density of a liquid is 1.0g/cm³. What is its density in kg/m? Show your workings clearly.​

Physics
1 answer:
Ipatiy [6.2K]3 years ago
6 0

Answer:

\displaystyle 1 \frac{gr}{cm^3}=1,000\frac{kg}{m^3}

Explanation:

Note: We are assuming the unit 'm' is 'm^3' to be compatible with the given unit.

<u>Unit Conversions </u>

Is the procedure followed to change the units of a certain magnitude to other units of the same magnitude. Example: centimeters to meters, hours to minutes, Kg to gr.

When dealing with composite units, each unit must be converted separately.

The density of some liquid is

\displaystyle 1 \frac{gr}{cm^3}

We are required to convert to Kg/m^3, so we convert separately

We know that

\displaystyle 1 gr=\frac{1}{1000}Kg

\displaystyle 1 cm^3=\frac{1}{1,000,000}m^3

The conversion is

\displaystyle 1 \frac{gr}{cm^3}=\frac{\frac{1}{1000}Kg}{\frac{1}{1,000,000}m^3}

\displaystyle 1 \frac{gr}{cm^3}=\frac{1,000,000\ kg}{1000\ m^3}=1,000\frac{kg}{m^3}

Thus

\boxed{\displaystyle 1 \frac{gr}{cm^3}=1,000\frac{kg}{m^3}}

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If the depth of water in a well is 10m, what is the pressure exerted by it the bottom of the well ? ( Use g = 10 m/s2)
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Answer:

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Since the density of the water is 1,000 kg/m³, the mass of water directly above area A is (1,000 kg/m³) × (10×a m³) = (1000×10×a kg) = 10,000×a kg.

Since g = 9.8 m/s², the force of gravity acting on the water directly above area A is (9.8 m/s²) × (10,000×a kg) = 9.8×10,000×a N (newtons) = 98,000×a N.

So the pressure of water acting on area A is (98,000×a N)/(a m²) = (98,000×a)/a N/m² = 98,000 pascals (pa). And since A could be any given area at the bottom of the well, this is the pressure at any point at the bottom of the well.

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What is the relationship between the applied force of a hanging mass on a spring and the spring force of the spring?
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Answer:

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