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julsineya [31]
3 years ago
10

Which of the following is a non renewable resource?

Physics
1 answer:
Karo-lina-s [1.5K]3 years ago
3 0
Carbon isn’t because all the others on the list are naturally made apart from carbon which is man made because of all the pollution . X
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The gravitational force between two asteroids is 6. 2 × 108 N. Asteroid Y has three times the mass of asteroid Z. If the distanc
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Answer:Let m =  mass of asteroid y.Because asteroid y has three times the mass of asteroid z, the mass of asteroid z is m/3.Given:F = 6.2x10⁸ Nd = 2100 km = 2.1x10⁶ mNote thatG = 6.67408x10⁻¹¹ m³/(kg-s²)The gravitational force between the asteroids isF = (G*m*(m/3))/d² = (Gm²)/(3d²)orm² = (3Fd²)/G     = [(3*(6.2x10⁸ N)*(2.1x10⁶ m)²]/(6.67408x10⁻¹¹ m³/(kg-s²))    = 1.229x10³² kg²m = 1.1086x10¹⁶ kg = 1.1x10¹⁶ kg (approx)Answer: 1.1x10¹⁶ kg

Explanation:

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2 years ago
What is the average velocity of the particle from rest to 9 seconds?
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2 years ago
How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe
Rus_ich [418]

Explanation:

(a)  For an isothermal process, work done is represented as follows.

             W = -nRT ln(\frac{V_{2}}{V_{1}})

Putting the given values into the above formula as follows.

        W = -nRT ln(\frac{V_{2}}{V_{1}})

             = - 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})

             = -12280.82 \times ln (0.09)

             = -12280.82 \times -2.41

             = 29596.78 J

or,         = 29.596 kJ       (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

         W = \frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}

              = \frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}

              = \frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}

              = 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c)   We know that for an isothermal process,

               P_{1}V_{1} = P_{2}V_{2}

or,       P_{2} = \frac{P_{1}V_{1}}{V_{2}}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})

                    = 11 atm

Hence, the required pressure is 11 atm.

(d)   For adiabatic process,  

          P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}

or,       P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}

                    = 28.7 atm

Therefore, required pressure is 28.7 atm.

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Marie Curie and her husband Pierre were Henri Becquerel's graduate students when Becquerel
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Answer:

It is C on edge.

Explanation:

Because I just figured it out and got it right and because it says so in the link provided from the question.

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