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Annette [7]
3 years ago
5

A 2.2kg block of ice slides across a rough floor. Its initial velocity is 2.5m/s and its final velocity is 0.50m/s. How much of

the ice block melted as a result of the work done by friction? (Latent Heat of water is 3.3*10^5J/kg)
Physics
1 answer:
kirill [66]3 years ago
7 0

Answer:A 2.2kg block of ice slides across a rough floor. Its initial velocity is 2.5m/s and its final velocity is 0.50m/s. How much of the ice block melted as a result of the work done by friction? (Latent Heat of water is 3.3*10^5J/kg)

Explanation:

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You have a 3.00-liter container filled with N₂ at 25°C and 4.45 atm pressure connected to a 2.00-liter container filled with Ar
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Answer : The final pressure in the two containers is, 2.62 atm

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

Thus, the expression for final pressure in the two containers will be:

PV=P_1V_1+P_2V_2

P=\frac{P_1V_1+P_2V_2}{V}

where,

P_1 = pressure of N₂ gas = 4.45 atm

P_2 = pressure of Ar gas = 2.75 atm

V_1 = volume of N₂ gas = 3.00 L

V_2 = volume of Ar gas = 2.00 L

P = final pressure of gas = ?

V = final volume of gas = (4.45 + 2.75) L = 7.2 L

Now put all the given values in the above equation, we get:

P=\frac{(4.45atm)\times (3.00L)+(2.75atm)\times (2.00L)}{7.2L}

P=2.62atm

Thus, the final pressure in the two containers is, 2.62 atm

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