21st Century Astronomy
6th Edition
ISBN: 9780393428063
Author: Kay
Publisher: NORTON
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Chapter 14, Problem 10QP
To determine
The justification for difference in the time periods.
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Chapter 14 Solutions
21st Century Astronomy
Ch. 14.1 - Prob. 14.1ACYUCh. 14.1 - Prob. 14.1BCYUCh. 14.2 - Prob. 14.2CYUCh. 14.3 - Prob. 14.3CYUCh. 14.4 - Prob. 14.4CYUCh. 14 - Prob. 1QPCh. 14 - Prob. 2QPCh. 14 - Prob. 3QPCh. 14 - Prob. 4QPCh. 14 - Prob. 5QP
Ch. 14 - Prob. 6QPCh. 14 - Prob. 7QPCh. 14 - Prob. 8QPCh. 14 - Prob. 9QPCh. 14 - Prob. 10QPCh. 14 - Prob. 11QPCh. 14 - Prob. 12QPCh. 14 - Prob. 13QPCh. 14 - Prob. 14QPCh. 14 - Prob. 15QPCh. 14 - Prob. 16QPCh. 14 - Prob. 17QPCh. 14 - Prob. 18QPCh. 14 - Prob. 19QPCh. 14 - Prob. 20QPCh. 14 - Prob. 21QPCh. 14 - Prob. 22QPCh. 14 - Prob. 23QPCh. 14 - Prob. 24QPCh. 14 - Prob. 25QPCh. 14 - Prob. 26QPCh. 14 - Prob. 27QPCh. 14 - Prob. 28QPCh. 14 - Prob. 29QPCh. 14 - Prob. 30QPCh. 14 - Prob. 31QPCh. 14 - Prob. 34QPCh. 14 - Prob. 35QPCh. 14 - Prob. 36QPCh. 14 - Prob. 37QPCh. 14 - Prob. 38QPCh. 14 - Prob. 39QPCh. 14 - Prob. 40QPCh. 14 - Prob. 41QPCh. 14 - Prob. 42QPCh. 14 - Prob. 43QPCh. 14 - Prob. 44QPCh. 14 - Prob. 45QP
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- A bobsled starts at the top of a track as human runners sprint from rest and then jump into the sled. Assume they reach 40 km/h from rest after covering a distance of 50 m over flat ice. a. How much work do they do on themselves and the sled which they are pushing given the fact that there are two men of combined mass 185 kg and the sled with a mass of 200 kg? (If you haven't seen bobsledding, watch youtube to understand better what's going on.) b. After this start, the team races down the track and descends vertically by 200 m. At the finish line the sled crosses with a speed of 55 m/s. How much energy was lost to drag and friction along the way down after the men were in the sled?arrow_forwardFor what type of force is it not possible to define a potential energy expression?arrow_forward10. Imagine you have a system in which you have 54 grams of ice. You can melt this ice and then vaporize it all at 0 C. The melting and vaporization are done reversibly into a balloon held at a pressure of 0.250 bar. Here are some facts about water you may wish to know. The density of liquid water at 0 C is 1 g/cm³. The density of ice at 0 C is 0.917 g/cm³. The enthalpy of vaporization of liquid water is 2.496 kJ/gram and the enthalpy of fusion of solid water is 333.55 J/gram.arrow_forward
- Consider 1 mole of supercooled water at -10°C. Calculate the entropy change of the water when the supercooled water freezes at -10°C and 1 atm. Useful data: Cp (ice) = 38 J mol-1 K-1 Cp (water) 75J mol −1 K -1 Afus H (0°C) 6026 J mol −1 Assume Cp (ice) and Cp (water) to be independent of temperature.arrow_forwardThe molar enthalpy of vaporization of benzene at its normal boiling point (80.09°C) is 30.72 kJ/mol. Assuming that AvapH and AvapS stay constant at their values at 80.09°C, calculate the value of AvapG at 75.0°C, 80.09°C, and 85.0°C. Hint: Remember that the liquid and vapor phases will be in equilibrium at the normal boiling point.arrow_forward3. The entropy of an ideal gas is S = Nkg In V. Entropy is a state function rather than a path function, and in this problem, you will show an example of the entropy change for an ideal gas being the same when you go between the same two states by two different pathways. A. Express ASV = S2 (V2) - S₁(V1), the change in entropy upon changing the volume from V₁to V2, at fixed particle number N and energy, U. B. Express ASN = S₂(N₂) - S₁ (N₁), the change in entropy upon changing the particle number from N₁ to N2, at fixed volume V and energy U. C. Write an expression for the entropy change, AS, for a two-step process (V₁, N₁) → (V2, N₁) → (V2, N₂) in which the volume changes first at fixed particle number, then the particle number changes at fixed volume. Again, assume energy is constant.arrow_forward
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