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choli [55]
2 years ago
14

g HW 5.6.The Moon and the Earth both orbit the center of mass of the Earth-Moon system.Earth is 81 times more massive than the M

oon. Assume that the distance between the centerof the Earth and the Moon is 384,400.0 km. Determine the distance of the center of mass of theEarth-Moon system from the center of the Earth. Does it lie inside or outside the Earth with theradius of 6,371.0 km
Physics
1 answer:
Law Incorporation [45]2 years ago
6 0

Answer:

R_{cm}  = 4.688 106 m

Explanation:

The expression for the center of mass is

       R cm = 1 / M ∑ r_{i} m_{i}

Where M is the total mass of the system, ri and m1 are the distance and mass of each particle from a defined origin

Let's look for the total mass, the mass of the moon is m and that of the earth Me

    M = M_{e} + m

    M = 81m + m

    M = 82 m

The distance from the earth to the moon is R = 384.4 106 m

    R_{cm} = 1 / M (M R1 + m R)

   R_{cm}  = 1 / 82m (0 + m R)

    R_{cm}  = R / 82

   R_{cm}  = 384.4 106/82

   R_{cm}  = 4.688 106 m

The radius of the earth is

    R_{e} = 6371 103 m

    R_{e}  = 6.371 106 m

We can see that the center of mass of the system is within the beating radius

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P_u =  \frac{8.9}{100} P =  \frac{8.9}{100}  (15 W)=1.34 W

The rest of the power instead is wasted as heat, and this amount of power wasted is given by the difference between the total power and the useful power:
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2 years ago
A cup containing 200 g of hot water is taken off the stove placed on the kitchen table. Initially the water is at 75Degree C. bu
8090 [49]

Answer:

ΔH = -45.1872 kJ , where negative sign signifies heat loss.

Q = ΔH = -45.1872 kJ  , where negative sign signifies heat loss.

S system = -0.141 kJ/K

S surroundings = 0.1536 kJ/K

S universe = -0.141 kJ/K + 0.1536 kJ/K  = 0.0126 kJ/K

Explanation:

Given:

Cp = 4. 184 J/(mole. K)

T₁ = 75 ⁰C

T₂ = 21 ⁰C

Mass of water = 200 g = 0.2 kg

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ΔH = 0.2*4.184*(21-75)  kJ

<u> ΔH = -45.1872 kJ , where negative sign signifies heat loss.</u>

Since the process is at constant pressure

<u> Q = ΔH = -45.1872 kJ  , where negative sign signifies heat loss.</u>

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

T₁ = 75 ⁰C = 348 .15 K

T₂ = 21 ⁰C = 294.15 K

The entropy of the water is given by:

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S = 0.2*4.184*ln(294.15/348.15)

<u> S system = -0.141 kJ/K</u>

The heat gain by surroundings

<u> dQ = -Qreaction =  45.1872 kJ </u>

The entropy change of surroundings is

S surr = dQ/T₂ = 45.1872/294 .15

<u> S surr = 0.1536 kJ/K </u>

The entropy of universe  is the sum total of the entropy of the system and the surroundings and thus,

S universe = Ssys + Ssurr

<u> S universe = -0.141 kJ/K + 0.1536 kJ/K  = 0.0126 kJ/K</u>

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