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yarga [219]
3 years ago
12

At its closest point, Mercury is approximately 46 million kilometers from the sun. What is this distance in AU?

Physics
2 answers:
alina1380 [7]3 years ago
5 0
1 astronomical unit 1 AU = 1.4960 * 10^11 meters
it is the average distance between earth and sun
mercury to sun distance is = 46,000,000 * 1000 meters
               = 4.6 * 10^9 meters = 4.6 * 10^9 / 1.4960 * 10^11  AU
               =  3.0.74 / 100  = 0.0374 AU

Radda [10]3 years ago
5 0

Explanation :

Astronomical unit is the unit to find the average distance between the earth and the sun.

1\ AU=1.496\times 10^8\ Km

So, 1\ Km=\dfrac{1}{1.496\times 10^8}\ AU

46 million kilometers means, 46\times 10^6\ Km

46\times 10^6\ Km=\dfrac{46\times 10^6}{1.496\times 10^8}

46\ million\ Km=0.3074\ AU

Hence, this is the required solution.      

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Two bricks are released at the same time from the same point ten feet above the ground. One of the bricks is falling straight do
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A piston of volume 0.1 m3 contains two moles of a monatomic ideal gas at 300K. If it undergoes an isothermal process and expands
seropon [69]

Answer:

the work is done by the gas on the environment -is W= - 3534.94 J (since the initial pressure is lower than the atmospheric pressure , it needs external work to expand)

Explanation:

assuming ideal gas behaviour of the gas , the equation for ideal gas is

P*V=n*R*T

where

P = absolute pressure

V= volume

T= absolute temperature

n= number of moles of gas

R= ideal gas constant = 8.314 J/mol K

P=n*R*T/V

the work that is done by the gas is calculated through

W=∫pdV=  ∫ (n*R*T/V) dV

for an isothermal process T=constant and since the piston is closed vessel also n=constant during the process then denoting 1 and 2 for initial and final state respectively:

W=∫pdV=  ∫ (n*R*T/V) dV =  n*R*T  ∫(1/V) dV = n*R*T * ln (V₂/V₁)

since

P₁=n*R*T/V₁

P₂=n*R*T/V₂

dividing both equations

V₂/V₁ = P₁/P₂

W= n*R*T * ln (V₂/V₁)  = n*R*T * ln (P₁/P₂ )

replacing values

P₁=n*R*T/V₁ = 2 moles* 8.314 J/mol K* 300K / 0.1 m3= 49884 Pa

since P₂ = 1 atm = 101325 Pa

W= n*R*T * ln (P₁/P₂ ) = 2 mol * 8.314 J/mol K * 300K * (49884 Pa/101325 Pa) = -3534.94 J

5 0
3 years ago
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