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I) How much tension must a rope withstand if it is used to accelerate a 1210-kg car horizontally along a frictionless surface at 1.20 m/s$^2$ ?
(II) According to a simplified model of a mammalian heart, at each pulse approximately 20 $g$ of blood is accelerated from 0.25 m/s to 0.35 m/s during a period of 0.10 s. What is the magnitude of the force exerted by the heart muscle?
(I) A 7150-kg railroad car travels alone on a level frictionless track with a constant speed of 15.0 m/s. A 3350-kg load, initially at rest, is dropped onto the car. What will be the car's new speed?
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here we calculate the electric potential due to a uniformly charged insulating sphere. A sphere has a radius R, as I've drawn here and a total charge of Q. And as we mentioned, it's uniformly charge. So each unit of volume has the exact same charge. We've previously calculated the electric field for such a insulating sphere, which I've drawn here. Now we want to calculate the electric potential because we have two different regions of electric field or the electric field behaves differently in two different regions. We need to calculate potential in both regions. So first we consider are greater than our. And as we see, this is simply the electric field of a point charge. And so we'll expect correctly that be of our is equal to take over our now let you verify the math on that. But you'll go through the same routine we've done before and see that this is what we're gonna cover in our second region is when our radius of our Goshen surface is less than capital are on. What's important is when you are finding the potential in different regions or regions where the electric field behaves differently. It's very important that you always maintain the same zero potential reference. So here, when we calculate RV Abe or the potential point with respect to that point B, this will again be zero when b equals infinity, as we did for the point charge. Eso No So now we integrate e dot d l from A to infinity. And of course, because the electric field in all regions is radio lee outward this becomes e times d r now e varies by region. And so we break this integral into each region And for the first one inside the sphere we go from A to capital are our electric field is que que little are over capital r cubed d r and our second integral goes from capital are to infinity of cake over r squared d r And so for the first integral we get cake r squared over to Capitol Are cube evaluated from eight a capital R and the second is simply the potential when we're distance capital are away from a point charge or cake over capital are now we simplify this by evaluating our limits of integration and we see that we can simplify this even further by writing it ISS three k capital que over to our minus que que a squared over to capitol r cubed. Now, of course, A is just a placeholder. And really, we could write the potential as a function of our distance from the center of the sphere. Are his pulling out? Constant terms of cake over to Capitol are three minus times three minus little r squared over capital R squared. And so here's for the region. When mhm we're inside the sphere. Well, no, This is different from the potential of a conducting charge sphere because there is no electric field inside a conductor. Eso weaken graph these against each other potential versus our little are. And of course, our point of interest is here at large. Are so our insulate er which I'll put in blue falls off outside as one over r where this value here his cake over capital are and we see again inside that it will fall off as one over r squared. But this redraw that a little more quadratic Lee were this value here, just three k que over to Capitol are we'll call for a conductor which I'll put it read that outside. It's the exact same thing in both cases when we're outside of the spirit we treated as a point charge. Q. But inside because there's no electric field, the potential remains constant. And so there's a visual comparison of the electric field electric potential due toa conducting sphere and a uniformly charged insulating sphere. One last thing to check is what happens to our potential within the sphere as our goes toe are. Well, then we get cake we kept to capital are and it's three r squared over r squared or que que over our which is our limit. I'll tell you this fear as little Argos, the big are. So we've verified that are potentials are logically sensible in that different regions. As you approach the boundary from either side, you get the exact same potential. Next, we calculate the potential associate ID with the infinitely large charge conducting slab and by a slab. I merely mean a sheet that has some thickness in case in this case, the thickness of D. We're gonna place this sheet slab so that it's parallel to the Y Z plane and such that the front faces defined by the plain X equals D over two, and the back face is defined by the plane. X equals negative d over to now. Whatever the charge in the slab is, we know two things. One, all of the charge will go to the surface is as it always does in a conductor. And since it's infinitely big, it's got to go to the front or back surface that facing the positive X direction or negative extraction. Also by symmetry, the charges want to get us far away from each other as possible in order to minimize the forces between them. And so the front face of the slab will have the same total charge or charge per unit area as the back slab. So, really, what we have is two parallel infinite conducting sheets with same sigma or surface charge density, and will recall that for a single sheet we found this to be our electric field. So for two sheets, we'll be working with superposition of this. And so when we're in the positive X direction from this slab, our electric field is sigma over Epsilon, not in the extraction, and on the other side we get the exact same magnitude Onley in the opposite direction. So we'll use a negative sign. And we know when. A conductor The electric field is also zero, but also works by superposition. The electric field from the front face moves in that direction. The electric field in, uh from the back face points in that direction, and since they're of equal magnitude, they cancel out. So now let's calculate what are potential is and we'll start with the potential in the positive X area. So here we have three regions to the right, in the positive extraction to the left, the negative extraction and within our slab. So now our baby and for this is going to be for our greater or X greater than D over to V A minus. B B course is the integral from A to B of e d l. But the electric field on Lee points in the extraction, and so we can rewrite this as e. D. X. Now for a point charge, we put our zero potential over at infinity. We tried that with the infinite connecting sheet and saw that it did not work. We ended up getting an infinite potential everywhere So what we did and said is we defined zero as at at the sheet itself. So I'm going to do the same thing here. But I'm gonna do it from the front sheet. This will be our zero potential. In other words, this is zero when b equals d over two and we change our limit of integration. So now we integrate from a to D over to of Sigma over Epsilon, not DX, and we'll see that we get Sigma over Epsilon not de over to minus a And of course, that's are zero. So, more generally, we can write this in terms of x, a Sigma X over Epsilon. Not with the negative sign. Does this make sense? Well, at the face, we are at zero potential and we expect that assuming signals positive as we move away from a positive charge, potential should drop Well, that happens here as X gets larger and larger, this number becomes more and more negative. So we're at a lower potential. So next let's evaluate some region in here Now there's no electric field in there, so we don't expect any electric field to contribute to the potential. So we expect the potential to remain zero and let's verify that. So now this is actually or are zeros via de over to we take this integral from a to D over to of e yeah, e d x. And of course, that is zero. So we see that this is zero Finally, So that was the region of negatively over two is less than X has less than d over to finally will calculate the potential in the region where X is less than negative d over to again. We're going to use the same reference v f d over to value from A to B over to e. D. X. Of course, we need to break this up into each region and so we'll have one region from a to negative d over to he d x and another region from negative d over to d over to e DX. Of course, he is zero in that region, so we just we only worry about the first integral and so this gives us a to negative d over to integrating negative sigma over epsilon, not D X, and so we get negative Sigma X over epsilon not evaluated at those positions and we get Sorry, I should remove this X and we get Sigma over Absalon not de over two plus a or more generally, it will move this aside a little bit R V of X, a sigma over Epsilon not de over to plus X. So now we see that as ex approaches negative d over to, this becomes zero Great. We've seen that it should be consistently. The potential should be consistently zero through the conducting service and through the entire conductor. And as we get more and more negative are potential drop. So as we get away from this presumably positive sigma, we get a lower and lower potential, which is what we want. And so now we have our regions of interest and we have proven what the electric potential is in all of these regions.
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