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Study Figures 4-11 and 4-16 and describe the phases that Venus would have displayed to Galileo’s telescope if the Tychonic universe had been correct.
Figure 4-11 Tycho’s model of the Universe retained the first principles of classical astronomy; it was geocentric with the Sun and Moon revolving around Earth, but the planets revolved around the Sun. All motion was along circular paths.
Figure 4-16 (a) If Venus moved in an epicycle centered on the Earth-Sun line (see Figure 3b in the Chapter 2 Concept Art: An Ancient Model of the Universe), it would always appear as a crescent. (b) Galileo observed through his telescope that Venus goes through a full set of phases, proving that it must orbit the Sun.
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Chapter 4 Solutions
FOUNDATIONS OF ASTRON.-MINDTAP (2 TERM)
- please help me solve this questions. show all calculations and a good graph too :)arrow_forwardWhat is the force (in N) on the 2.0 μC charge placed at the center of the square shown below? (Express your answer in vector form.) 5.0 με 4.0 με 2.0 με + 1.0 m 1.0 m -40 με 2.0 μCarrow_forwardWhat is the force (in N) on the 5.4 µC charge shown below? (Express your answer in vector form.) −3.1 µC5.4 µC9.2 µC6.4 µCarrow_forward
- An ideal gas in a sealed container starts out at a pressure of 8900 N/m2 and a volume of 5.7 m3. If the gas expands to a volume of 6.3 m3 while the pressure is held constant (still at 8900 N/m2), how much work is done by the gas? Give your answer as the number of Joules.arrow_forwardThe outside temperature is 25 °C. A heat engine operates in the environment (Tc = 25 °C) at 50% efficiency. How hot does it need to get the high temperature up to in Celsius?arrow_forwardGas is compressed in a cylinder creating 31 Joules of work on the gas during the isothermal process. How much heat flows from the gas into the cylinder in Joules?arrow_forward
- The heat engine gives 1100 Joules of energy of high temperature from the burning gasoline by exhausting 750 Joules to low-temperature . What is the efficiency of this heat engine in a percentage?arrow_forwardL₁ D₁ L₂ D2 Aluminum has a resistivity of p = 2.65 × 10 8 2. m. An aluminum wire is L = 2.00 m long and has a circular cross section that is not constant. The diameter of the wire is D₁ = 0.17 mm for a length of L₁ = 0.500 m and a diameter of D2 = 0.24 mm for the rest of the length. a) What is the resistance of this wire? R = Hint A potential difference of AV = 1.40 V is applied across the wire. b) What is the magnitude of the current density in the thin part of the wire? Hint J1 = c) What is the magnitude of the current density in the thick part of the wire? J₂ = d) What is the magnitude of the electric field in the thin part of the wire? E1 = Hint e) What is the magnitude of the electric field in the thick part of the wire? E2 =arrow_forwardplease helparrow_forward
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