A 9.88 kg object speeds up from 15.91 m/s to 47.25 m/s over a distance of 122.75m. What is the acceleration of the object? If the frictional force is 19.05N, what is the net acceleration acting on the object?

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11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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1. A 9.88 kg object speeds up from 15.91 m/s to 47.25 m/s over a distance of 122.75m. What is the acceleration of the object? If the frictional force is 19.05N, what is the net acceleration acting on the object? (10 points)

2. A 3177.06 kg car crashed to a 534.95 lbs motorcycle which is at rest, throwing it to accelerate at 8.24m/s2. If the car was pushed back with a distance of 21 m, how much farther is the motorcycle from the crash site? What is the final velocity of the car upon being pushed back that is caused by the crash? use 1kg=2.20lbs (10 points)

3. An unknown planet (mass is equal to 1.29 x 10^12 kg) is orbiting around the sun (mass of the sun is equal to 6.623703×1035kg). Knowing that the gravitational force between the unknown planet and the sun is (5.61x 10^20)N, is the unknown planet nearer or farther to the sun compared to earth? (earth's distance to sun is 1 AU) Express your answer in terms of astronomical units or AU (1AU = 1.496 x 108 km). Assuming that the sun is fixated at a center point, calculate the area (in of the unknown planet's orbit (assume that the orbit is circle). Express your answer using proper scientific notation convention (Refer to the Review portion of introduction to General Physics 1). (10 points)

4.  the difference in gravitational force when an object is located at sea level (1m) vs located at a height of 10289m. Use mEarth as a a representation of the mass of the earth on your solution. The percentage of course should not exceed 100%. Write in proper scientific notation. Answer in complete sentence .Express, in percent

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