Physics Laboratory Experiments
8th Edition
ISBN: 9781285738567
Author: Jerry D. Wilson, Cecilia A. Hernández-Hall
Publisher: Cengage Learning
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Chapter 10, Problem 7ASA
Explain how the
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G = 6.67 * 10 ^-11 N*m^2/kg^2
M earth = 5.97 * 10^24 kg
Chapter 10 Solutions
Physics Laboratory Experiments
Ch. 10 - Prob. 1ASACh. 10 - Prob. 2ASACh. 10 - Prob. 3ASACh. 10 - Prob. 4ASACh. 10 - If the centripetal force acting on an object in...Ch. 10 - Suppose that the centripetal force acting on an...Ch. 10 - Explain how the centripetal force is directly...Ch. 10 - How does the centripetal force vary with the...Ch. 10 - If the centripetal force on an object in uniform...Ch. 10 - Does the centripetal force acting on an object in...
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- Needs Complete typed solution with 100 % accuracy.arrow_forwardplease prove this sample problem's answer using GRESA method Determine the force of gravitational attraction between the earth (m = 5.98 x 1024 kg) and a 70-kg physics student if the student is standing at sea level, a distance of 6.38 x 106 m from earth's center. The solution of the problem involves substituting known values of G (6.673 x 10-11 N m2/kg2), m1 (5.98 x 1024 kg), m2 (70 kg) and d (6.38 x 106 m) into the universal gravitation equation and solving for Fgrav. The solution is as follows:arrow_forward3. Suppose a ball is dropped from 2 m above the floor. If it took 0.45 seconds to hit the ground, what is the gravitational acceleration that can be determined from the experiment? If the error in the height measurement is 0.050 m and the error in time measurement is 0.10 s, what is the total error value of the gravitational acceleration? Is this value consistent (or inconsistent) with the literature value?arrow_forward
- How would you solve this: In the search for the Higgs Boson, it was posited that the potnetial energy storedbetween two higgs particles is given by a 1 dimensional function of the distancebetween the two particles (x) as U(x) = c + ax2 − bx4 where a, b, c are constants.a) Find a function for the force as a function of position.b) what value of position (x) is the equilibrium position in terms of a and b.arrow_forwardIn the law of universal gravitation, Newton assumed that the force was proportional to the product of the two masses (m1m2) . While all scientific conjectures must be experimentally verified, can you provided arguments as to why this must be? (You may wish to consider simple examples in which any other form would lead to contradictory results.)arrow_forwardPlease solve this problem and show the written solution. Thank you very mucharrow_forward
- Solve and explain. What is the mass of the sun using the period of the Earth which is 365 days and its distance from the sun of 147.38 million km?arrow_forwardEx. 94 : In Melde's experiment when 6 gm weights are added to the pan the number of loops is 5 and when 10.5 gm weights are added to the pan the number of loops is 4. Find the number of loops when there are no weights in the pan.arrow_forwardA body with a mass of 100kg is pulled towards the center of the moon with a gravitational acceleration equivalent to 3 less than half the earth's gravitational acceleration. Compute the force acting on the body in N.arrow_forward
- We will use differential equations to model the orbits and locations of Earth, Mars, and the spacecraft using Newton’s two laws mentioned above. Newton’s second law of motion in vector form is: F^→=ma^→ (1) where F^→ is the force vector in N (Newtons), and a^→ is the acceleration vector in m/s^2,and m is the mass in kg. Newton’s law of gravitation in vector form is: F^→=GMm/lr^→l*r^→/lr^→l where G=6.67x10^-11 m^3/s^2*kg is the universal gravitational constant, M is the mass of the larger object (the Sun), and is 2x10^30 kg, and m is the mass the smaller one (the planets or the spacecraft). The vector r^→ is the vector connecting the Sun to the orbiting objects. Step one ) The motion force in Equation(1), and the gravitational force in Equation(2) are equal. Equate the right hand sides of equations (1) and (2), and cancel the common factor on the left and right sides. Answer: f^→=ma^→ f=Gmm/lr^→l^2 a^→=Gmm/lr^→l^2 x r^→/lr^→l r^→=r^→/lr^→l * Gmm Could you please…arrow_forwardLet's consider an astronaut with a mass of 76.0 kg standing on the Earth. What is the gravitational force (in N) on the astronaut? (Review your work from the previous part. Remember that you can use the full gravitational force formula, or w = mg.) weight on the Earth's surface = 744.8 N Now let's imagine that same astronaut in orbit aboard the space station, orbiting at an altitude (height above the Earth's surface) of 380 km. We first need to find the astronaut's distance to the center of the Earth. What is the altitude expressed in units of meters? 380 km = 380000 The radius of the Earth is 6.37 ✕ 106 m. What is the total distance (in m) from the center of the Earth to the astronaut? r =_________m Finally, let's put it all together. Use the universal gravitational force formula to find the force (in N) on the astronaut, where m = 76.0 kg, the Earth's mass ME = 5.97 ✕ 1024 kg, and the distance r is the result you just found. F = ______N solve for the two bolded…arrow_forwardPlease help with P-3.1arrow_forward
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