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3 years ago

11

The density of water is 1.00 g/cm3. What is its density in kg/m3?

1 answer:

r-ruslan [8.4K]3 years ago

3 0

1 g = 1 ÷ 1000 kg
= 0.001 kg

1 cm³ = 1 ÷ 100 ÷ 100 ÷ 100 m³
= 0.000001 m³

1 g/cm³ = 1 g / 1 cm³
= 0.001 kg / 0.000001 m³
= 1000 kg/m³

The density is 1000 kg/m³.

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A quantity of gas has a volume of 0.20 cubic meter and an absolute temperature of 333 degrees kelvin. When the temperature of th

Elena L [17]

P1v1/t1 = p2v2/t2

p1=p2, v1=.2, t1=333, t2=533
we can find v2 from this

be aware, temperature must be in Kelvin.

7 0

3 years ago

An ideal monatomic gas initially has a temperature of 300 K and a pressure of 5.79 atm. It is to expand from volume 420 cm3 to v

maxonik [38]

Answer:

a) The final pressure is 1.68 atm.

b) The work done by the gas is 305.3 J.

Explanation:

a) The final pressure of an isothermal expansion is given by:

$T = \frac{PV}{nR}$

$T_{i} = T_{f}$

$\frac{P_{i}V_{i}}{nR} = \frac{P_{f}V_{f}}{nR}$

Where:

$P_{i}$ : is the initial pressure = 5.79 atm

$P_{f}$ : is the final pressure =?

$V_{i}$ : is the initial volume = 420 cm³

$V_{f}$ : is the final volume = 1450 cm³

n: is the number of moles of the gas

R: is the gas constant

$P_{f} = \frac{P_{i}V_{i}}{V_{f}} = \frac{5.79 atm*420 cm^{3}}{1450 cm^{3}} = 1.68 atm$

Hence, the final pressure is 1.68 atm.

b) The work done by the isothermal expansion is:

$W = P_{i}V_{i}ln(\frac{V_{f}}{V_{i}}) = 5.79 atm*\frac{101325 Pa}{1 atm}*420 cm^{3}*\frac{1 m^{3}}{(100 cm)^{3}}ln(\frac{1450 cm^{3}}{420 cm^{3}}) = 305.3 J$

Therefore, the work done by the gas is 305.3 J.

I hope it helps you!

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3 years ago

A gasoline engine receives 200 J of energy from combustion and loses 150 J as heat to the exhaust. What is its efficiency? 75% 2

choli [55]

50/200×100%=25% is answer the formula is usefull energy output divided by total energy provided into 100%

8 0

3 years ago

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A sphere of radius R contains charge Q spread uniformly throughout its volume. Find an expression for the electrostatic energy c

tensa zangetsu [6.8K]

Answer:

$E = \frac{3kQ^2}{5R}$

Explanation:

Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as

$V = \frac{kq}{r}$

now we can say that

$q = \frac{Q}{\frac{4}{3}\pi R^3} (\frac{4}{3}\pi r^3)$

$q = \frac{Qr^3}{R^3}$

now electric potential is given as

$V = \frac{k\frac{Qr^3}{R^3}}{r}$

$V = \frac{kQr^2}{R^3}$

now work done to bring a small charge from infinite to the surface of this sphere is given as

$dW = V dq$

$dW = \frac{kQr^2}{R^3} dq$

here we know that

$dq = \frac{3Qr^2dr}{R^3}$

now the total energy of the sphere is given as

$E = \int dW$

$E = \int_0^R \frac{kQr^2}{R^3} (\frac{3Qr^2dr}{R^3})$

$E = \frac{3kQ^2}{R^6} (\frac{R^5}{5} - 0)$

$E = \frac{3kQ^2}{5R}$

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4 years ago

An object has a mass of 60 kg. It decelerates from 50 m/s to 20 m/s when a resultant force of 300 N acts on it. For how long doe

Pavlova-9 [17]

Force=300N
Mass=60kg

Find Acceleration

$\\ \rm\longmapsto F=ma$

$\\ \rm\longmapsto a=\dfrac{F}{m}$

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$\\ \rm\longmapsto a=-5m/s^2$

Now

u=50m/s
v=20m/s

Using 1st equation of kinematics

$\\ \rm\longmapsto v=u+at$

$\\ \rm\longmapsto t=\dfrac{v-u}{a}$

$\\ \rm\longmapsto t=\dfrac{20-50}{-5}$

$\\ \rm\longmapsto t=\dfrac{-30}{-5}$

$\\ \rm\longmapsto t=6$

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3 years ago