All categories

2 years ago

What is the volume of 2400kg of gasoline (petrol) if the density of petrol is 0.7g/km3

Physics

1 answer:

Keith_Richards [23]2 years ago

4 0

Explanation:

Density = mass / volume

700 kg/m³ = 2400 kg / V

V ≈ 3.4 m³

You might be interested in

Two cylinders each contain 0.30 mol of a diatomic gas at 320 K and a pressure of 3.0 atm. Cylinder A expands isothermally and cy

Svetllana [295]

Answer :

(a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is $7.86\times10^{-3}\ m^3$

(d). The final volume of the gas in the cylinder B is $5.7\times10^{-3}\ m^3$

Explanation :

Given that,

Number of mole n = 0.30 mol

Initial temperature = 320 K

Pressure = 3.0 atm

Final pressure = 1.0 atm

We need to calculate the initial volume

Using formula of ideal gas

$P_{1}V_{1}=nRT$

$V_{1}=\dfrac{nRT}{P_{1}}$

Put the value into the formula

$V_{1}=\dfrac{0.30\times8.314\times320}{3.039\times10^{5}}$

$V_{1}=2.62\times10^{-3}\ m^3$

(a). We need to calculate the final temperature of the gas in the cylinder A

Using formula of ideal gas

In isothermally, the temperature is not change.

So, the final temperature of the gas in the cylinder A is 320 K.

(b). We need to calculate the final temperature of the gas in the cylinder B

Using formula of ideal gas

$T_{2}=T_{1}\times(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}-1}$

Put the value into the formula

$T_{2}=320\times(\dfrac{3}{1})^{\frac{1}{1.4}-1}$

$T_{2}=233.7\ K$

(c). We need to calculate the final volume of the gas in the cylinder A

Using formula of volume of the gas

$P_{1}V_{1}=P_{2}V_{2}$

$V_{2}=\dfrac{P_{1}V_{1}}{P_{2}}$

Put the value into the formula

$V_{2}=\dfrac{3\times2.62\times10^{-3}}{1}$

$V_{2}=0.00786\ m^3$

$V_{2}=7.86\times10^{-3}\ m^3$

(d). We need to calculate the final volume of the gas in the cylinder B

Using formula of volume of the gas

$V_{2}=V_{1}(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}}$

$V_{2}=2.62\times10^{-3}\times(\dfrac{3}{1})^{\frac{1}{1.4}}$

$V_{2}=0.0057\ m^3$

$V_{2}=5.7\times10^{-3}\ m^3$

Hence, (a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is $7.86\times10^{-3}\ m^3$

(d). The final volume of the gas in the cylinder B is $5.7\times10^{-3}\ m^3$

6 0

3 years ago

Two identical resistors connected in series have an equivalent resistance of 4 ohms. The same two resistors, when connected in p

irina1246 [14]

Answer:

1 ohm

Explanation:

since there are two identical resistors, one resistor will be

R = $\frac{4}{2}$ =2ohm [ proven as in series $R_{e} = 2 + 2 = 4ohm$ ]

to calculate the equivalent resistance when in parallel:

$\frac{1}{R_{e} } = \frac{1}{R_{1} } + \frac{1}{R_{2}}$

so,

$\frac{1}{R_{e} } = \frac{1}{2} + \frac{1}{2}$

$R_{e} = 1ohm$

4 0

2 years ago

Calculate the velocity of a wave that has a frequency of 60 Hz and wavelength of 2.0 m/s

mote1985 [20]

Answer:We have , a relation in frequency f and wavelength λ of a wave having the velocity v as ,

v=fλ ,

given f=60Hz , λ=20m ,

therefore velocity of wave , v=60×20=1200m/s

3 0

3 years ago

If a roller coaster train has a potential energy of 1,500 J and a kinetic energy of 500 J as it starts to travel downhill, its t

Lina20 [59]

$E_{total} = KE + PE \\ KE = 500 J \\ PE = 1500 J \\ E_{total} = 500J + 1500J \\ E_{total} = 2000J$

4 0

3 years ago

Read 2 more answers

What are each layer on an atom

Zarrin [17]

Protons, electrons, and neutrons. The nucleus (center) of the atom contains the protons (positively charged) and the neutrons (no charge).

5 0

3 years ago

Read 2 more answers