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Alina [70]
3 years ago
6

(b) An example of an inverse square field is the gravitational field F = −(mMG)r/|r|3. Use part (a) to find the work done by the

gravitational field when the earth moves from aphelion (at a maximum distance of 1.52 108 km from the sun) to perihelion (at a minimum distance of 1.47 108 km). Use the values m = 5.97 1024 kg, M = 1.99 1030 kg, and G = 6.67 ✕ 10−11 N·m2/kg2. (Round your answer to two decimal places.)
Physics
1 answer:
enyata [817]3 years ago
4 0

Answer:

To solve this, we need to recall that Work done W,

W = F x dₐ                         Eqn 1

where F is the gravitational field force and d is the distance travelled by the earth.

But F = −(mMG)r/|r|³                    Eqn 2

then, W = −∫ (mMG)r/|r|³ x dₐ

Integrating this expression gives us

W = -4mMG ((1/d1) -(1/d2))

m = 5.971024 kg, M = 1.991030 kg G = 6.67 ✕ 10⁻¹¹ N·m²/kg²

d2 = 1.52x10¹¹m, d1 = 1.47x10¹¹m  make the units in S.I units and similar

Substituting these value in to the eqn above gives

Now , W = 1.77 x 10³²J  (2 Decimal Places)

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We can use the work-energy theorem to solve.

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\large\boxed{W = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2}

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W = \frac{1}{2}(50)(5.61^2) = 786.8025 J

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Part A What will be the equilibrium temperature when a 227 g block of copper at 283 °C is placed in a 155 g aluminum calorimeter
stellarik [79]

Answer:

T = 20.84°C

Explanation:

From the law of conservation of energy:

Heat Lost by Copper Block = Heat Gained by Aluminum Calorimeter + Heat Gained by Water

m_cC_c\Delta T_c = m_wC_w\Delta T_w + m_aC_a\Delta T_a

where,

m_c = mass of copper = 227 g

m_w = mass of water = 844 g

m_a = mass of aluminum = 155 g

C_c = specific heat capacity of calorimeter = 385 J/kg.°C

C_w = specific heat capacity of water = 4200 J/kg.°C

C_a = specific heat capacity of aluminum = 890 J/kg.°C

\Delta T_c = change in temperature of copper = 283°C - T

\Delta T_w = change in temperature of water = T - 14.6°C

\Delta T_a = change in temperature of aluminum = T - 14.6°C

T = equilibrium temperature = ?

Therefore,

(227\ g)(385\ J/kg.^oC)(283^oC-T)=(844\ g)(4200\ J/kg.^oC)(T-14.6^oC)+(155\ g)(890\ J/kg.^oC)(T-14.6^oC)\\\\24732785\ J - (87395\ J/^oC) T = (3544800\ J/^oC) T - 51754080\ J+ (137950\ J/^oC) T-2014070\ J\\\\24732785\ J +51754080\ J+2014070\ J = (3544800\ J/^oC) T+(137950\ J/^oC+(87395\ J/^oC) T\\\\78560935\ J = (3770145\ J/^oC) T\\\\T = \frac{78560935\ J}{3770145\ J/^oC}

<u>T = 20.84°C</u>

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2 years ago
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