Fuerza entre do5 cargas 0e 20 N. Si duplicamos distanciz anina Ias catgas enoncesIrcerzaSeleccione Una:...

Consider a pair of forces, one having a magnitude of 20 $\mathrm{N}$ and the other a magnitude of 12 $\mathrm{N}$ . What is the strongest possible net force for these two forces? What is the weakest possible net force?

Force is proportional to one of our square. So if one over F two equals our do over r one square so we would have f two equals f one times I want over our two squared. Now if one equals 20 Newtons are due big guns. Our long now our do becomes our one over to half the distance square of that, so that makes it 80 commuter.

Hello students in this question we have to similar equal magnetic poles to suppose M one M two, which is equal to em when they are separated by a distance R equals to one m. Repel with a force F equals to 10 to the power minus three newton. Okay, so we have to determine this polish strength and know the force between the polish strength is given by noon or two, divided by 45 M one M two developed by our square. So we can substitute the value. So F. Which is equal to 10 to the power minus three and this is equal to the and to the power minus seven and M. One M. Two, which is M. So this Emmy square divers are which is one m and this holy spirit. So after solving we get em square equals to 10 to the power four. Or we get from here the polish 10th M. Is equals 200 ampere meter. Okay, so from the given options option D. Is the correct answer for the problem. Okay, thank you.

So over on the left, I have the values we need to solve this problem. We have the constant K, which is equal to nine times into the nine Newton meters squared over Coolum squared. The charge of one of the particles is eight times 10 to the negative five cool homes and the charge. The other particle is three times 10 to the negative five cool ums. And then the force between the two particles is 2.4 times 10 to the to Newton's. And so we're looking to find the separation between these two particles. So the equation we want to use here is electric force is equal to k times the two charges over the separation squared. But we wanna multiply both sides by r squared and then take the square root to sell for our And if we do that, we find ours equal to square. Okay, times the two charges over the force. And if we plug in our values on the left into that equation, we find the separation is 0.3 meters

So we're told that we can assume that bone breaks if the stress is greater than 0.15 Giga Pascal's If you have the stress greater than that, it's gonna break intentional. And the Youngs, ma journalists were told to assume, is 15 Giger past the diameter of the bone and the smallest cross section is 2.5 centimeters and the bone is 25 centimeters long. So we want to figure out what is the maximum force that can be exerted on this bone. Well, we know that the force is the stress times, the cross sectional area. And so the maximum force is gonna be the maximum stress times a cross sectional area four times again, the minimum cross sectional area. So we have a cross section of areas pile before d squared and we have everything here and we can plug in our get our number 70 73.6 killing. Could the bone could handle that much in intention? Anyway, Bones can handle a lot more stress and compression just like any kind of ceramic. But we were told that within asked that how much force is applied, how much does the if it's a fight and compression, if this maximum force were complied, applied and compression, how much does the 25 centimeters long bone shortened? Well, we know that the stress well, we're gonna We're gonna assume one of these things. We have to make an assumption here that this that the bone is uniform. Okay, so if the ball is in uniform, if it's like this, then we can't say that this this strain and the stress are not gonna be uniform through the bone. So we can't really get this. We can't really average over this whole way. But if the bone were uniform and he, um um, it was just 2.5 centimeters has a 2.5 centimeter diameter. Then we could get a value here. Um, this would be an overestimate, because the stress will be lower when the at the cross sections where the bone is wider. So there it would deformed less there. Anyway, if we have a if their bonus uniform, and then we have our maximum stress acting throughout that bone every wearing in here on the Youngs module is this. We know that the strain is then 0.1 and then the change in length is, and again, I suppose I should indicate clearly that these are initial links. Change in length is the strain time, the initial length and that's 2.5 millimeters.