1. A spaceship with mass of 450 kg, orbits the Earth in 100 minutes in a circula
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1. A spaceship with mass of 450 kg, orbits the Earth in 100 minutes in a circular orbit. At one point in its orbit it fires an instantaneous burst in the forward direction, reducing its speed by 2.00 %. After this burst, the ship follows an elliptical orbit. What will be its new orbital period? Cole weighs 145 lb on the surface of the Earth. Cole travels to an exoplanet discovered recently. This exoplanet orbits a star that is 2.5 times as massive as our Sun. The planet itself is 0.40 times as massive as the Earth, and its diameter is half the diameter of the Earth. How many pounds will Cole weigh on the surface of this planet? 2. The Earth orbits around the Sun in a nearly circular orbit with a period of about 365 days. Which of the following changes will absolutely result in a change in the orbital period? 3. a. b. c. d. e. Doubling the speed of the Earth Doubling the magnitude of the momentum of the Earth Doubling the distance between the Earth and the Sun Doubling of the mass of the Sun Doubling the speed of the Earth Comets are known to have very eccentric (far from circular) orbits around the Stars. Imagine a comet with a mass of 4000 kg that orbits the Sun in an eccentric elliptical orbit. We know that this comet's perihelion (the closest distance to the Sun) is 0.10 AU, and its aphelion (the farthest distance from the Sun) is 22.0 AU. 1 AU is the mean distance between the Earth and the Sun and it is equal to 150 million km. If this comet has been to its perihelion in 2017, how many years later should we expect it go farthest from the Sun, that is, how many years later will it visit its aphelion? Enter with a precision down to 0.1 year or better. (The mass of the Sun is 1.99x10430 kg.) 4. 5. Which of the following statements is (are) correct? Please check all that apply. a. We measure the weight of an object 10 km above the surface of the Earth, and call this weight W. We then measure the weight of the object 10 km below the surface of the Earth, and call this weight W, If the weight of the object on the surface of the Earth is Wo then we can always write 2W0=w,+W b. If we plot the weight of an object as a function of its distance from the center of the Earth, the plot will look like the following:Explanation / Answer
6) the water will stop, as in free fall, water and container both will travel with equal speed and hence no relative motion.
7) fraction above water surface = 1 - dwood/dwater
=> 0.6 = 1 - dwood/1000
=> dwood = 400 kg/m3
8) ALvL = ATvT
=> vT = (36/16)x 5 = 11.25 m/s
now, (P/d + 0.5v2 + gh)L = (P/d + 0.5v2 + gh)T
=> ((101300 + 85000)/1000 + 0.5 x 52 + 9.8 x 0) = ((101300 + P2)/1000 + 0.5(11.25)2 + 9.8x2)
=> 186.3 + 12.5 = 101.3 + P2/1000 + 63.28125 + 19.6
=> P2 = 14618.75 Pa = 14.61875 kPa
9) pressure at same level will be same
=> F/A1 + dgh = mg/A2
=> F/0.05 + 1000 x 9.8 x 0.3 = 500 x 9.8/(1.2x0.8)
=> F = 108.21 N
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