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dusya [7]
3 years ago
12

A large room in a house holds 947 kg of dry air at 33.2°C. A woman opens a window briefly and a cool breeze brings in an additio

nal 62.4 kg of dry air at 13.2°C. At what temperature (in degrees Celsius) will the two air masses come into thermal equilibrium, assuming they form a closed system? (The specific heat of dry air is 1006 J/ kg * °C, although that value will cancel out of the calorimetry equation.)
°C
Physics
1 answer:
BartSMP [9]3 years ago
7 0

Answer:

31.96362 °C

Explanation:

m_1 = Mass of air in the room = 947 kg

m_2 = Mass of air entering the room = 62.4 kg

T_1 = Temperature in the room = 33.2°C

T_2 = Temperature air entering the room = 13.2°C

T = Equilibrium temperature

c = Specific heat of air = 1006 J/kg °C

In the case of thermal equilibrium we have the relation

m_1c(T_1-T)=m_2c(T-T_2)\\\Rightarrow T=\frac{m_2T_2+m_1T_1}{m_1+m_2}\\\Rightarrow T=\frac{62.4(13.2+273.15)+947(33.2+273.15)}{947+62.4}\\\Rightarrow T=305.11362\ K=305.11362-273.15=31.96362^{\circ}C

The temperature of thermal equilibrium is 31.96362 °C

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3 years ago
A planet of mass m moves around the Sun of mass M in an elliptical orbit. The maximum and minimum distance of the planet from th
zzz [600]

Answer:

the relation between the time period of the planet is

T = 2π √[( r1 + r2 )³ / 8GM ]

Explanation:

Given the data i  the question;

mass of sun = M

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Now, using Kepler's third law,

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T² ∝ R³

average distance a = ( r1 + r2 ) / 2

we know that

T² = 4π²a³ / GM

T² = 4π² [( ( r1 + r2 ) / 2 )³ / GM ]

T² = 4π² [( ( r1 + r2 )³ / 8 ) / GM ]

T² = 4π² [( r1 + r2 )³ / 8GM ]

T = √[ 4π² [( r1 + r2 )³ / 8GM ] ]

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5 0
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Answer:

Ist and second

Hope it helps

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Answer:

The moment (torque) is given by the following equation:

\vec{\tau} = \vec{r} \times \vec{F}\\\vec{r} \times \vec{F} = \left[\begin{array}{ccc}\^{i}&\^j&\^k\\r_x&r_y&r_z\\F_x&F_y&F_z\end{array}\right] = \left[\begin{array}{ccc}\^{i}&\^j&\3k\\0.23&0.04&0\\150&260&0\end{array}\right] = \^k((0.23*260) - (0.04*150)) = \^k (53.8~Nm)

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