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016 (part 1 of 2) 10.0 pointsObjects with Inasses Of 126 kg and 708 kg nre separated by 0.404 M. A 35 kg maSS is placed midway betwccn them: 708 kg 126 kg 35 kg0.40...

Question

016 (part 1 of 2) 10.0 pointsObjects with Inasses Of 126 kg and 708 kg nre separated by 0.404 M. A 35 kg maSS is placed midway betwccn them: 708 kg 126 kg 35 kg0.404Find the magnitude ofthe net gravitational [orce exerted by the Lwo larger [Iasses OI the 35 kg mass_ Thc value of the universal gravi- tational constant is 6.672 X 10-" m? /kg? . Answer in units of N.017 (part 2 of 2) 10.0 pointsLeaving the distance betwecn the 126 kg and the 708 kg masses fixed, at what distance from the 708 k

016 (part 1 of 2) 10.0 points Objects with Inasses Of 126 kg and 708 kg nre separated by 0.404 M. A 35 kg maSS is placed midway betwccn them: 708 kg 126 kg 35 kg 0.404 Find the magnitude ofthe net gravitational [orce exerted by the Lwo larger [Iasses OI the 35 kg mass_ Thc value of the universal gravi- tational constant is 6.672 X 10-" m? /kg? . Answer in units of N. 017 (part 2 of 2) 10.0 points Leaving the distance betwecn the 126 kg and the 708 kg masses fixed, at what distance from the 708 kg Inass (other than inlinitely remote



Answers

(a) The electrostatic force (not energy) of attraction between two oppositely charged objects is given by the equation $F=\kappa\left(Q_{1} Q_{2} / d^{2}\right)$ where $\kappa=8.99 \times 10^{9} \mathrm{~N}-\mathrm{m}^{2} / \mathrm{C}^{2}, Q_{1}$ and $Q_{2}$ are the charges of the two objects in Coulombs, and $d$ is the distance separating the two objects in meters. What is the electrostatic force of attraction (in Newtons) between an electron and a proton that are separated by $0.23 \mathrm{nm} ?$ (b) The force of gravity acting between two objects is given by the equation $F=G\left(m_{1} m_{2} / d^{2}\right)$ where $G$ is the gravitational constant, $G=6.674 \times 10^{-11} \mathrm{~N}-\mathrm{m}^{2} / \mathrm{kg}^{2}, m_{1}$ and $m_{2}$ are the masses of the two objects, and $d$ is the distance separating them. What is the gravitational force of attraction (in Newtons) between the electron and proton?
(c) How many times larger is the electrostatic force of attraction?

Hello friends. In this problem, we have to calculate the force on the moon. New toe Art Anderson given Moss off the moon is 7.35 10 to the power country to K g Moss off art 7135 10 to the power country to K G on March after Centobie 5.98 10 to the power 24 kg. The distance off Earth from the sun is 1. 46.6 in tow. 10 to the power six kilometers distance off Art from moon is 3 84 in tow 10 to the power three kilometers The restraints off moon from the Sun Bill B hardest square distance off earth from the sun minus the square of the distance from art from the moon No net force on the moon Due to our thins and bill be due to force on window to earth will be G and most of the moon Most of the art upon distance off the move from the art and force on one due to son G march off moon moss off son upon square off distance off one from doesn't the result enforceability f m e is square plus F m s C square at theater Toby 10 and worse off F m e upon F m s. So substituting develop, they will get after we equal toe g must off moon upon marks off art square upon radius Distance to the power for must distance off the moon from art Pless Moss off Sonny Square upon hard to the power the stains off moon from son substituting the value to simplify it 6.67 turned to the power 11 Most of the moon is given 7.35 10 to the power county too into Let me write it in the next line it will be better. Just a moment, please. So I am writing this slide No substituting the value 6.67 10 to the power 11 must after moon is given 7135 10 to the power county too in tow mass off the art Well 5.98 tend to the power 24 square upon distance is given three point hateful in tow 10 to the power eight with the powerful plus mass of the sun one point 99 tend to the power 30 square applied mass of distance of the earth from the sun 1.46 10 to the power 11th square minus 3.84 10 to the power 80 square. Hold to the powerful. So on solving it this forced to be 4.96 in tow, 10 to the power 22 Newton. Similarly, we can find the direction theater Toby 10 and worse. Off most off art in tow. Our square E s minus Artist square E m Upon Moss off, son in tow. Our square he substituting the value all parameters are given ever. You will get 23.7 degrees. That's all. Thanks for watching it.

For a let's calculate the electrostatic force between a proton and an electron at a distance of one times 10 of the two PICO meters, we have our constant Q one Q two over D squared Constant is 8.99 times 10 to the ninth Newton meters squared, whom squared and Q one 1.60 times 10 to the negative. 19 que lumps charged over Proton 1.60 times 10 to the 19. Colom's charge of an electron distance is 10 to the two times 10 to the negative 12 m squared and solving. Here we get 2.30 times 10 to the eighth Newton's and that would be the electrostatic force for Be well have to calculate the gravitational force. And to calculate this it'll be G m one m two over D squared G s air constant here 6.674 times 10 to the minus 11 Newtons meters squared kilograms squared mass of an electron is 9.11 times 10 to the negative. 31 kg mass of a proton is 167 times 10 to the negative 24 kg. The distance is 10 to the two times, 10 to the negative, 12 m squared and solving for the force of gravity. Here it's 101 times 10 to the negative. 44 Newton's for part C. How many times is F E eyes? He? How many times larger is the electrostatic force of attraction, So F over F G. This is equal to 2.30 times 10 to the eight Newton's over 1.1 times 10 to the negative 44 Newtons and we get 2.28 times 10 to the 36. So F E is equal to 2.28 times 10 to the 36 f g and therefore Thief Force. The electrostatic force of attraction is 2 to 8 times 10 to the 36 times greater then the force of gravity.

For a let's calculate the electrostatic force between a proton and an electron at a distance of one times 10 of the two PICO meters, we have our constant Q one Q two over D squared Constant is 8.99 times 10 to the ninth Newton meters squared, whom squared and Q one 1.60 times 10 to the negative. 19 que lumps charged over Proton 1.60 times 10 to the 19. Colom's charge of an electron distance is 10 to the two times 10 to the negative 12 m squared and solving. Here we get 2.30 times 10 to the eighth Newton's and that would be the electrostatic force for Be well have to calculate the gravitational force. And to calculate this it'll be G m one m two over D squared G s air constant here 6.674 times 10 to the minus 11 Newtons meters squared kilograms squared mass of an electron is 9.11 times 10 to the negative. 31 kg mass of a proton is 167 times 10 to the negative 24 kg. The distance is 10 to the two times, 10 to the negative, 12 m squared and solving for the force of gravity. Here it's 101 times 10 to the negative. 44 Newton's for part C. How many times is F E eyes? He? How many times larger is the electrostatic force of attraction, So F over F G. This is equal to 2.30 times 10 to the eight Newton's over 1.1 times 10 to the negative 44 Newtons and we get 2.28 times 10 to the 36. So F E is equal to 2.28 times 10 to the 36 f g and therefore Thief Force. The electrostatic force of attraction is 2 to 8 times 10 to the 36 times greater then the force of gravity.

Were given that this is the gravitational force between two objects of mass couple and lower case on when they're separated by a distance off far. So if we have two objects. So let's say this one has mess on, and this one has mass lower case on were asked how much work is required to move it from a distance of our one to another distance of our two. So what we can do is we can think of the distance between them, which is our as tiny segments. So each segment is going to be built. So then we know that work is equal to the force. Times best thanks. So if it was at a distance of Arwen now we need to move it to another distance, which is are too. Then what we're doing is, um, the force is going to be, Hey, it's not going to be a constant because it's equal to this function of here. So which means we need to take each Delta r and then some it up. So after we do our summation, it is going to, um converged on an integral so which tells us the work is going to be the integral of the force Times distance. So forces g, um over are great and in the distance that's traveling that each of those tiny little distances, which is Delta are. So when it when we take the integral that becomes d R. Now our limits integration is going to be from the original distance, which is our one to the new distance, which is our two. So if we do this, we're going to get G. This is our to carve negative, too. So the anti derivative is negative. Arch of negative one valid from are one or two. So this is going to be G um um, times it'll be negative one over R two minus negative zero plus one over R one so we can swap those. It'll be one over or one minus one over or two. Which is exactly what we're asked to show


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