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Yuki888 [10]
4 years ago
11

A stone of density 5500 kg/m3 is submerged in water. If the volume of the stone is 0.25 m3, the mass of water it displaces is___

_________.
Physics
1 answer:
vampirchik [111]4 years ago
4 0

Let us use the principle given in the relation of density as a function of mass and volume, but this time we will reorganize it in terms of mass so that said displaced mass is equivalent to the density of water by the volume of the stone , So,

\rho = \frac{m}{V} \rightarrow m = \rho V

Where,

\rho= Density

m = mass

V = Volume

Replacing,

m = 5500*0.25

m = 1375kg

Therefore the mass of water it displaces is 1375kg

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What is the pressure inside such a balloon (in atm) if it starts out at sea level with a temperature of 12.8°C and rises to an a
tatiyna

Answer:

The final pressure inside the balloon will be P_2=0.0359\ atm.

Explanation:

Given initial temperature of the balloon is T_1=12.8\°C

And the final temperature of the balloon is T_2=-47.7\°C

Initially, the balloon is at atmospheric pressure P_1=1\ atm and volume be V_1

And the final pressure in the balloon is P_2 and the volume be V_2

Also, it is given in the question that the final volume became twenty-two times the original volume.

We can write V_2=22V_1

Now, using ideal gas equation.

\frac{P_2V_2}{P_1V_1}=\frac{nRT_2}{nRT_2}

Where, R is the gas constant. And n is moles of substance inside the balloon.

In our problem the value of n will be the same for both cases. Also, R is the gas constant.

\frac{P_2V_2}{P_1V_1}=\frac{T_2}{T_1}

Also, we have V_2=22V_1 from the question.

\frac{P_2\times 22V_1}{P_1V_1}=\frac{T_2}{T_1}\\\\\frac{P_2\times 22}{P_1}=\frac{T_2}{T_1}

We need to convert the temperature from \°C to Kelvin.

T_1=12.8\°C\\T_1=12.8+273.15=285.95\\T_2=-47.7\°C\\T_2=-47.7+273.15=225.45

Plug these values we get,

\frac{P_2\times 22}{1}=\frac{225.45}{285.95}\\\\22\times P_2=0.789\\P_2=\frac{0.789}{22}\\P_2=0.0359\ atm.

So, the final pressure inside the balloon will be P_2=0.0359\ atm.

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Apply the formulas:

Vf1 = ?

vf1=(\frac{m1-m2}{m1+m2})Vi1+(\frac{2m2}{m1+m2})Vi2

Replacing:

Vf1=\frac{(1.1-2.4)}{(1.1+2.4)}2.7+\frac{(2\times2.4)}{(1.1+2.4)}-1.9Vf1=(\frac{-1.3}{3.5})2.7+(\frac{4.8}{3.5})-1.9Vf1=-1-2.6=-3.6\text{ m/s}

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Answer:

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