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Molodets [167]
3 years ago
15

You leave a pastry in the refrigerator on a plate and ask your roommate to take it out before youget home so you can eat it at r

oom temperature, the way you like it. Instead, your roommateplays video games for hours. When you return, you notice that the pastry is still cold, but thegame console has become hot. Annoyed, and knowing that the pastry will not be good if it ismicrowaved, you warm up the pastry by unplugging the console and putting it in a clean trashbag (which acts as a perfect calorimeter) with the pastry on the plate. After a while, you find thatthe equilibrium temperature is a nice, warmTeq.. You know that the game console has a mass ofm1. Approximate it as having a uniform initial temperature ofT1. The pastry has a mass ofm2and a specific heat ofc2, and is at a uniform initial temperature ofT2. The plate is at the sameinitial temperature and has a mass ofm3and a specific heat ofc3. What is the specific heat ofthe console
Physics
1 answer:
Sidana [21]3 years ago
3 0

Answer:

   c_{e1} =  \frac{(m_2 c_{e2} \ + m_3 c_{e3} ) \  (T_{Teq} - T_2)     }{m_1 (T_1 - T_{eq}) }

Explanation:

This is a calorimeter problem where the heat released by the console is equal to the heat absorbed by the cupcake and the plate.

           Q_c = Q_{abs}

where the heat is given by the expression

           Q = m c_e ΔT

            m₁ c_{e1) (T₁-T_{eq}) = m₂ c_{e2} (T_{eq} -T₂) + m₃ c_{e3} (T_{eq}- T₁)

note that the temperature variations have been placed so that they have been positive

They ask us for the specific heat of the console

           c_{e1} =  \frac{(m_2 c_{e2} \ + m_3 c_{e3} ) \  (T_{Teq} - T_2)     }{m_1 (T_1 - T_{eq}) }

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Answer:

r =  255.68 m

Explanation:

When a body moves in a circular path, an acceleration, due to constant change in its direction, is developed, known as centripetal acceleration. The centripetal acceleration acts towards the center of the circular path. The formula to calculate the centripetal acceleration is given as follows:

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On the way to the moon, the Apollo astronauts reach a point where the Moon’s gravitational pull is stronger than that of Earth’s
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Answer:

rm = 38280860.6[m]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2}  } \\

F_{m} =G*\frac{m_{m}*m_{a}  }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]

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G*\frac{m_{e} *m_{a} }{r_{e}^{2}  } = G*\frac{m_{m} *m_{a} }{r_{m}^{2}  }\\\frac{m_{e} }{r_{e}^{2}  } = \frac{m_{m} }{r_{m}^{2}  }

To solve this equation we have to replace the first equation of related with the distances.

\frac{m_{e} }{r_{e}^{2}  } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m}  )^{2}  } = \frac{7.36*10^{22}  }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17}  \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c }  }{2*a}\\  where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) }  }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

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