College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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- What force holds the Earth in its place in the solar system? a. The gravitational pull caused by the moon within its orbit around the Earth b. The gravitational pull caused by the sun within the solar system c. The gravitational pull caused by the black hole in the center of the Milky Way d. The gravitational pull caused by Venus and Mars within the solar systemarrow_forwardA number of gas giant planets orbiting other stars at distances less than 1 A.U. have been discovered. Because of their proximity to their parent stars, and their compositional similarity to Jupiter, they have been labeled “Hot Jupiters”. The orbital radius of one of these planets is 0.06 A.U. with average orbital speed 600 km/sec. What is the length of this planet’s year in Earth (solar) days? Estimate the mass, M, of its parent star in terms of the mass of the sun (M) using Newton’s first form of Kepler’s 3rd Law. Calculate the star’s luminosity, L, in terms of the luminosity of the sun (L☉), Note: (LL=MM4where L ~ 4 × 1026 W ). The radius of this planet is 1.5 times the radius of Jupiter. Assuming its equilibrium temperature is the temperature at which the planet radiates as much energy as it receives from its star, estimate the temperature of the planet. The value of the planet’s albedo is 0.8. (NOTE: The intensity of the star’s radiant power at a distance d from the star is…arrow_forwardA distance R is measured to be 3.400±0.005m. What is the relative uncertainty in R? b) What is the relative uncertainty in R3? c) What is the relative uncertainty in R−2? d) What is the absolute uncertainty in R−2? e) What is the relative uncertainty in √R?arrow_forward
- 18 \arrow_forward2. Suppose you are standing on a field, which is at a latitude of 40 5/6°, and that you are 2.0 m tall. (a) Analytically find the speed of your feet, v, due to the rotation of the Earth. (Assume that the Earth is a perfect sphere, which is not, of radius R = 6378 km and that a day is exactly 24 hours long.) (b) Using your result from (a), determine the centripetal acceleration of your feet. 1 (c) Repeat parts (a) and (b) for your head and compute the difference between the acceleration at your head and feet. Show how round-off can corrupt your calculation and offer a way to fix the problem.arrow_forwardThe earth orbits the sun once a year (3.16 107 seconds) in a nearly circular orbit of radius 1.5-10¹¹m. a) Assuming the earth travels at a constant speed, determine that speed. (Hint: Average speed distance / time. Distance around a circle is called the circumference of the circle.) VEarth = m/s b) Determine the size of the centripetal acceleration of the earth. m ac = c) Determine the size of the net centripetal force needed to keep the earth orbiting the sun. What is its direction? Fret Narrow_forward
- According to Lunar Laser Ranging experiments the average distance LM from the Earth to the Moon is approximately 3.85 x 105 km. The Moon orbits the Earth and completes one revolution in approximately 27.5 days (a sidereal month). a) Calculate the orbital velocity of the Moon. b) Calculate mass of the Earth.arrow_forward6. For vectors B = -î – 4ĵ and Å = -31 – 2j, calculate: a. Ā +B and its magnitude and direction angle b. A-B and its magnitude and direction angle.arrow_forward8. Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? 13 y R N Ꮎ 1A W N S Earrow_forward
- Please I need help with the second part of the question, It wasn’t 1.003x10^6 or 1003000 please helparrow_forwardThe solution differs so I must ask the teacher to explain. The book states that the solution is 11mCarrow_forwardNeptune orbits the Sun with an orbital radius of 4.495 x 10^12 m. If the earth to sun distance 1A.U. = 1.5 x 10^11 m, a) Determine how many A.U.'s is Neptune's orbital radius (Round to the nearest tenth). b) Given the Sun's mass is 1.99 x10^30 kg, use Newton's modified version of Kepler's formula T^2 = (4pi^2/Gm(star)) x d^3 to find the period in seconds using scientific notation. (Round to the nearest thousandth). C) Convert the period in part b) to years (Round to the nearest tenth)arrow_forward
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