Questions 1-4: A 2-kg mass is moving in a circularpath on the end of a 0.50-m string. The mass movesin a horizontal circle with a speed of 4.0 m/s.A. 0.50m12.0kga, 4. The speed of the mass is changed to 2.0 m/s.Compared to the centripetal acceleration of themass when moving at 4.0 m/s, its centripetalacceleration now would bea. Half as greatb. Twice as great. One-fourth as greatd. Four times as great

Given: Mass of the object = 2 Kg The radius of the circular path = 0.50 m Speed of the object in the…

Answered: Questions 1-4: A 2-kg mass is moving in…

Questions 1-4: A 2-kg mass is moving in a circular path on the end of a 0.50-m string. The mass moves in a horizontal circle with a speed of 4.0 m/s. A 0.50m 12.0kg

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

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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Every problem has a solution, Solve it! Instruction: Carefully read the word problem then identify the given, what is asked in the problem and provide solution. Write your answer with solution on a separate sheet of paper. 1. Four objects are situated along y-axis as follows; a 2.00-kg object is at + 3.00 meters, 5.00-kg object at 1.5 meters, a 4.5 – kg object at 3.0 meters and a 2.5 -kg object at 0.50 meters. Locate the center of mass of the system. 2. A billiard ball with a mass (m1) = 0.75-kg moves at a velocity vị of -2.0 m/s collides head-on with another billiard ball with a mass (m2) of =0.65 kg that is initially at rest (v2=0m/s) on the billiard table. No external forces act on the balls and collision is elastic. After collision, mi moves at the velocity of m2 velocity is equal to while the Note: To solve for vfi and vp we have these equations: where, Vf1 = mj-m2 : uy = final velocity/ velocity after collision Vi1 m2 2m, mị + m2 Vi = initial velocity/velocity before collision…
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In the problems below, “little g” plays a variety of roles. Near the earth’s surface, g has a value of 9.81 m/s2. At other locations g may have different values. Solve these word problems using the data provided below. MMoon = 7.35 × 1022 kg rMoon = 1.74 × 106 m rMoon’s orbit = 3.8 x 108 m mEarth = 5.97 × 1024 kg rEarth = 6.37 × 106 m rEarth’s orbit = 1.5 x 1011 m Question 1: An astronaut is standing on the moon, holding a feather. a) Calculate the acceleration due to gravity on the moon. b) The astronaut drops the feather from a height of 2.0 meters. How long does it take the feather to hit the ground? c) The feather has a weight of 0.020 N on Earth. What is its weight on the moon?
12. A physics student makes a device that prevents an egg from cracking when dropped. The mass of the device with the egg inside is 0.5 kg. The device is dropped and hits the ground with a velocity of 5 m/s. If 15 N of force is applied to the egg, how long did it take for the velocity to reach zero?
A Starting pitcher is able to pitch a .145 Kg baseball with a velocity of 40.0 m/s. What velocity must a 625 gram basketball achieve in order to have the same kinetic energy as the baseball?
How to solve?
A puck of mass m=2 kg moves on a circle of radius r=0.5 meters, on a frictionless table. The puck is attached to an object of mass M=10 kg by a cord that extends through a hole on the table. Find the speed of the puck if the object is at rest. You may take g=10m/s². Select one: O a. 4/3m/s O b.5/2m/s O c. 5m/s O d. 25m/s O e. 4m/s
The linear mass density of a straight rod of length L is given by a = a +?x (kg/m). Here x is the distance measured from A. The center of mass of the rod is xem = (-)L (m) away from the A end of the rod. What is the relationship between a and b? A x = 0 x = L A) a = 2b B) a = b C) a = b D) a = b E) a =b
1. a. A 100 kg astronaut on a spacewalk is 25m away from the 10000 kg space capsule and at the full length of her safety cord. To return to the capsule, she pulls herself along the cord. Where do the astronaut and the capsule meet from the capsule original position? b. A laser beam is directed at the International Space Station, 420 km above the Earth. The beam diverges at an angle of 1.43*10-5 rad. What diameter spot will it make on the ISS?
A meteoroid is moving towards a planet. It has mass m = 0.54×109 kg and speed v1 = 4.7×107 m/s at distance R1 = 1.6×107 m from the center of the planet. The radius of the planet is R = 0.78×107 m. The mass of the planet is M = 5.6×1025kg. There is no air around the planet. a)Enter an expression for the total energy E of the meteoroid at R, the surface of the planet, in terms of defined quantities and v, the meteoroid’s speed when it reaches the planet’s surface. Select from the variables below to write your expression. Note that all variables may not be required.α, β, θ, d, g, G, h, m, M, P, R, R1, t, v, v1 b)Enter an expression for v, the meteoroid’s speed at the planet’s surface, in terms of G, M, v1, R1, and R. c)Calculate the value of v in meters per second.