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3 years ago

5750 milliliters converted to Liters

Physics

2 answers:

EleoNora [17]3 years ago

8 0

Answer:

5.75 liters

Explanation:

5750 milliliters converted to liters is 5.75

Triss [41]3 years ago

5 0

Answer:

5.75 L

Explanation:

Step 1: Find conversion

1000 mL = 1 L

Step 2: Use Dimensional Analysis

$5750 \hspace{3} mL(\frac{1 \hspace{3} L}{1000 \hspace{3} mL} )$ = 5.75 L

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3 years ago

A man standing on a bus remains still when the bus is at rest. When the bus moves forward and then

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The density of gasoline is 730 kg/m3 at 0°C. Its average coefficient of volume expansion is 9.60 10-4(°C)−1. Assume 1.00 gal of

kipiarov [429]

Answer: 0.4911 kg

Explanation:

We have the following data:

$\rho_{0\°C}= 730 kg/m^{3}$ is the density of gasoline at $0\°C$

$\beta=9.60(10)^{-4} \°C^{-1}$ is the average coefficient of volume expansion

We need to find the extra kilograms of gasoline.

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Knowing density is given by: $\rho=\frac{m}{V}$ , we can find the mass $m_{1}$ of 8.50 gallons:

$m_{1}=\rho_{0\°C}V$

$m_{1}=(730 kg/m^{3})(0.0323 m^{3})=23.579 kg$ (2)

Now, we have to calculate the factor $f$ by which the volume of gasoline is increased with the temperature, which is given by:

$f=(1+\beta(T_{f}-T_{o}))$ (3)

Where $T_{o}=0\°C$ is the initial temperature and $T_{f}=21.7\°C$ is the final temperature.

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$f=1.020832$ (5)

With this, we can calculate the density of gasoline at $21.7\°C$ :

$\rho_{21.7\°C}=730 kg/m^{3} f=(730 kg/m^{3})(1.020832)$

$\rho_{21.7\°C}=745.207 kg/m^{3}$ (6)

Now we can calculate the mass of gasoline at this temperature:

$m_{2}=\rho_{21.7\°C}V$ (7)

$m_{2}=(745.207 kg/m^{3})(0.0323 m^{3})$ (8)

$m_{2}=24.070 kg$ (9)

And finally calculate the mass difference $\Delta m$ :

$\Delta m=m_{2}-m_{1}=24.070 kg-23.579 kg$ (10)

$\Delta m=0.4911 kg$ (11) This is the extra mass of gasoline

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lora16 [44]

There are 3

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