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zepelin [54]
2 years ago
13

It is observed that the number of asteroids (or meteoroids) of a given diameter is roughly inversely proportional to the square

of the diameter.
Part A

Assuming a density of 3000 kg/m3 and approximating the actual distribution of asteroids and meteoroids as a single 1000-km body (Ceres), one hundred 100-km bodies, 10 thousand 10-kmbodies, and so on, calculate the total mass in the form of 1000-km bodies.

Part B

Calculate the total mass in the form of 100-km bodies.

Part C

Calculate the total mass in the form of 10-km bodies.

Part D

Calculate the total mass in the form of 1-km bodies.
Physics
1 answer:
MaRussiya [10]2 years ago
4 0

Answer:

(a) M =  =1.57 *10²¹kg

(b) M =  =1.57 *10²⁰kg

(c) M =  =1.57 *10¹⁹kg

(d) M =  =1.57 *10¹⁸kg

Explanation:

(a) For 1000km body, diameter D = 1000km = 1,000,000 m

and N = (1000/D)² = (1000/1000km)² =  1

substituting into the formula, we have

M = 1 * 4/3 π*3000*[1000000/2]³

   =1.57 *10²¹kg

(b) For 100km body, diameter D = 100km = 100,000 m

and N = (1000/D)² = (1000/100km)² =  100

substituting into the formula, we have

M = 1 * 4/3 π*3000*[100000/2]³

   =1.57 *10²⁰kg

(C) For 10km body, diameter D = 10km = 10,000 m

and N = (1000/D)² = (1000/10km)² =  100

substituting into the formula, we have

M = 1 * 4/3 π*3000*[10000/2]³

   =1.57 *10¹⁹kg

(C) For 1km body, diameter D = 1km = 1,000 m

and N = (1000/D)² = (1000/1km)² =  100

substituting into the formula, we have

M = 1 * 4/3 π*3000*[1000/2]³

   =1.57 *10¹⁸kg

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Explanation:

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A tank having a volume of 0.85 m 3 initially contains water as a two-phase liquid-vapor mixture at 260 o C and a quality of 0.7.
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Answer:=14,160 kJ

Explanation: Let m1 and m2 be the initial and final amounts of mass within the tank, respectively. The steam properties are listed in the table below

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C MPa m^3/kg kJ/kg kJ/kg

1 260 4.689 0.02993 2158 2298 0.7 Liquid Vapor Mixture

2 260 4.689 0.0422 2599 2797 1 Saturated Vapor

The mass initially contained in the tank is m1 = V/v1

m1 =0.85 m^3 /0.02993 m^3 /kg

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The mass finally contained in the tank is

m2 =V2/v

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The heat transfer is then

Qcv = m2u2 − m1u1 − he(m2 − m1)

Qcv = (20.14)(2599) − (28.4)(2158) − (2797)(20.14 − 28.4) = 14,160 kJ

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The unseen force of gravity acts on the sled,
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