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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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A jambo jet mass 4×10^2 kg travelling at a speed pf 5000 m/s lands on the airport . It takes 2 min to come to rest .calculate th
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The average force applied by the ground on the Aeroplane is  16.664 kN.

<h3>What is Newton's second law of motion?</h3>

It states that the force F is directly proportional to the acceleration a of the body and its mass.

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2 years ago
A car slows down from 27.7 m/s to 10.9 m/s in 2.37 s, what is the acceleration?
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Answer:

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A man pushes on a piano with mass 180 kg; it slides at constant velocity down a ramp that is inclined at 19.0° above the horizon
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Answer:

(a) F = 574.3 N

(b) F = 607.4 N

Explanation:

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data

β= 19° : Angle of inclination of the ramp

μk = 0 : Coefficient of kinetic friction

m =180 kg  :piano mass

g = 9.8 m/s² : acceleration due to gravity

W= m*g :  Piano Weight

W= 180*9.8= 1764 N

X-Y axes in the inclined plane

We define the x-axis in the direction of the inclined plane ,19° to the horizontal.

We define the y-axis and in the direction of the plane perpendicular to the inclined plane.

We calculate the weight component parallel to the displacement of the piano:

Wx= W*sin19°= 1764*sin19°= 574.3 N

Problem development

a) The man pushes the piano with a force (F) parallel to the inclined plane.

We apply formula (1):

∑F = m*a , a=0 : Because the velocity is constant

F-Wx = 0

F = Wx

F = 574.3 N

b) The man pushes the piano with a force (F) parallel to the floor.

We apply formula (1):

We define the F force component parallel to the displacement of the piano (Fx):

Fx= F*cos19°

∑F = m*a , a=0 : Because the velocity is constant

Fx-Wx = 0

F*cos 19°- 574.3 N = 0

F = (( 574.3 ) / (cos 19°) )N

F = 607.4 N

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