a-ball-is-rolling-such-that-its-velocity-v-in-mathrmcm-mathrms-as-a-function-of-time-t-in-s-is-vfrac

Question

A ball is rolling such that its velocity $v$ (in $\mathrm{cm} / \mathrm{s}$ ) as a function of time $t$ (in s) is $v=\frac{0.45}{0.25 t^{2}+1} .$ How far does it move in $10.0 \mathrm{s} ?$To find the electric field $E$ from an electric charge distributed uniformly over the entire $x y$ -plane at a distance $d$ from the plane, it is necessary to evaluate the integral $k d \int \frac{d x}{d^{2}+x^{2}} .$ Here, $x$ is the distance from the origin to the element of charge. Perform the indicated integration.

Darren Wilson · Accepted Answer

okay in this problem. We&#x27;ve got a table with a ball and the ball rolls off the table and we&#x27;re told that it&#x27;s in the air for a time of 0.3 80 seconds. We&#x27;re told that the velocity at any point the velocity vector is 1.2. I had direction minus 9.8 t in the J hat direction, where I is horizontal and Jay is vertical unnoticed. By the way, this is just an acceleration of 9.8 meters per second. This just means that the starting velocity, if you care, was 1.2 meters per second. And we&#x27;re told that the displacement vector are is equal to the definite integral from zero to 0.38 seconds of that velocity vector with respect to time. And so, if we solve that, we can say that D R is equal to the integral. From 2.38 of 1.2, I had minus 9.8 t j hat and Weaken treat 1.2 I had as a constant, and we can treat the 9.8 and the J hat as a constant so we can write this as two separate intervals. Delta R is equal to the inter goal of 1.2 I hat from 2.3 eight seconds minus the interval of 9.8 T J hat from zero 2.38 seconds. And so, when we do those intervals, we&#x27;re gonna go Delta are is equal to this is just a constant and so we&#x27;re going to multiply it by t should be a hat over that. So it&#x27;s gonna be 1.2. I had times t evaluated from zero 2.38 and the therapies will just four away minus that t becomes a t squared. So it&#x27;ll be nine point eight t squared, and it should have put a 1/2 in front. Because as that T square goes, we need to put it. Why half many for the 1/2 and J hat And that also is evaluated from zero 2.38 Now notice that both zeros. When I multiply those three, those will go away. And so my delta are Victor is going to be equal to 1.2 when I plug 2.38 in for tea. That&#x27;s just going to be 0.38 and that&#x27;s in the I had direction minus. We&#x27;ve got the 1/2 times 9.8 times 0.38 squared. Let me write those all down 1/2 9.8 0.3 eight squared and that&#x27;s in the J hat direction. And when we go through and do all the math for those we get, our Delta are is equal to point 456 I had and minus 0.70 eight j hat and that&#x27;s displacement vector as our object falls off the table, and that&#x27;s how you saw that from.

Guilherme Barros · Answer

in this problem, we&#x27;re going to talk about Gaza&#x27;s law and the electric potential. So first about gas is long. We need to remember, uh, that Let&#x27;s say that we have charge distribution whose total charge is equal to queue and that we draw. Let&#x27;s say that we draw a surface caution surface around discharge. Uh, then God&#x27;s law tells us that the electric flux defined as the surface integral off the electric field times The element of area is equal to the total charge enclosed by the surface divided by absolute zero. Okay, um, also we have the electric potential V is equal to minus the integral off the electric field dot d r from the point. Actually, let me write this as V minus zero. This is the potential difference between our zero and are okay, so now we can go on to our exercise. What we need to do is to consider, uh, a sphere and non conducting sphere that has a radius r zero. Let&#x27;s say that this here is our sphere and this year has a uniformed densities such that the total charge off the sphere is cute. And our goal is to find what are. What is the value off the potential energy? First in question A when R is greater than our zero. Well, in that case, the, uh, we can use God&#x27;s law in order to find the electric field. And then with that, we can find the electric potential are the first thing we need to notice is that when V goes to when I&#x27;m sorry, when our goes to infinity, we want V to go to zero. Okay, so that&#x27;s something that we&#x27;re gonna impose over our occasion. So first, uh, notice that since there is ah spherical symmetry in our system, then the electric flux is just let&#x27;s say that we draw the surface at a distance are from the center. The electric flux is just the electric field at that point. Notice that all throughout the surface the electric field is the same times the area, which is four pi r square. And this is cute. That&#x27;s the charge enclosed by the surface, divided by absolute zero. The electric feud is Q over four pipes, 10 R squared. Now V is minus the integral off. Uh, the electric feud, the are. And what I&#x27;m gonna do is to integrate from, um infinity 20 because the infinity visit, which is you so V is equal to minus que over four pi absent. Zero are and there is a minus here. So there&#x27;s a tree. Yeah, So there&#x27;s a no overall positive sign. So v is Q over four pi up soon a zero r notice that when Argos infinity v goes to infinity as we want to, then we want to find what is the value in our is smaller than are zero Here, we&#x27;re gonna need to impose that the electric potential must always be our continuous. So the electric potential at our the electric potential are is whether greater than are zero when calculated A Let&#x27;s say our is it what you are zero Plus, This means that it&#x27;s equal to eyes to our zero from outside of the conducting is here. This is able to v of r is equal to or zero minus. Okay, eso first, uh, I&#x27;m gonna draw. Here are surface. So now we have that are zero This thesis Syria&#x27;s are zero Our surface. The distance between the center and our surface is our eso. What we have is that the electric field. I&#x27;m sorry. The electric field times. Uh, the area of all the area element is equal again to yeah, times four pi r squared. And this is equal to the enclosed charge. I&#x27;m gonna call it Lorca&#x27;s que Invited by absence, you notice that lower case Q is equal to the density times the volume of four pi r square R cubed over three and, um, roll The density is the total charge divided by four pi r zero cube over three, then multiplied by four pi over three R Q. So this is just Hugh R cubed over our zero cube. So e times four pi r squared is it? What you Q r cubed over our zero cube. So e is equal to queue our cube over four. And actually, there is a an absolute zero here, So four pi absolute zero um R zero cube and we have to divide everything by our square. So I&#x27;m just gonna cancel the three year. So this is the electric field Now. The different the man is zero is equal to, uh, actually let me just right v when I write fee as minus the integral of e the are And then I&#x27;m gonna leave it as on indefinite Integral. So I&#x27;m going to some a term in the end, so V is equal to minus the integral of Q r over four pi at zero r, zero square d r So this is minus. Q R squared over eight pi absolute zero r zero squared. Okay, there is a plot, Bleus Certain constant here. I&#x27;m gonna call it K. Okay. And now we have to impose that v is continuous, so v, for our slightly greater than are zero are is, let&#x27;s say V at our zero plus that is equal to que over four pi aplomb. Zero R zero squared and this is minus que r zero squared. Ah, actually, there is a small mistake here. It should be our zero cube here. I&#x27;m sorry for that in here as well. So cue or zero cute over eight. Hi. Upon zero R zero Q plus K. So these terms recon swelled and your last with que I noticed that this is too eight. So K is equal to three. Hugh over eight. Hi. Absolute zero r zero squared. Um, actually, there&#x27;s some where just are zero. Yeah, s o V is equal to minus. Q R squared over eight pi absolute zero R zero cube plus three Q over eight high F zero are zero minimal. Gonna write it as, um All right. This as que over eight. Hi. Absolute zero R zero times three minus r squared over r zero squared. Then finally, in question D we have to plot V and E as a function of our, um so see, let&#x27;s start with V notice that V or are smaller than are zero eyes this value here. So it&#x27;s a problem, but decreasing one until it reaches the value. So it starts with three Q over eight pi eyes e o. I eat interruption zero R zero in this and finally becomes que rape I I zero or zero and it decays like this and actually has a problem. Really, breaches are zero. Let me write this a little better, and then it starts to Kane as one overall. Okay, this is easy tomorrow and then we can can draw. He is a function war as well. Okay, so notice that e from here starts at zero and increases linearly until it reaches the value of K of Q over four pi absalon zero r zero square. Okay, so it starts. Sorry. Started this point. There&#x27;s something like this. It reaches are zero. So its able to cue over four pi absolute zero r zero squared. And then, in the case, as one over R squared. Okay, which is the answer to our exercise?

Paul A. · Answer

once again. Welcome to new problem. This time we are given us for Rickles office with the radius and a total charge of Q. So we have a total charge of Q. And uh the electric the electric field strength. The electric field electric feel strength. Uh huh. Inside the ball, R. Is less than our is given by is given by mm We are equals two. Uh imax hard to the fourth power of capital are default. So this is the variable radius and this is the maximum radius. So there are three things we want to figure out in this problem, determined timing. Uh huh. The maximum electric field strength becoming the imax based on total charge and maximum radius. Uh huh. Sure. Or rather or rather determined determined on expression from the um volume charge density determinant expression for the volume charge density sigma are inside the ball. And this simply means that sigma is function is a function of our sigma as a function of art. And then uh this is but being that we&#x27;re looking at what C. Is saying mm Sure that the um the charge density sigma arm views a total charge of cube. Show that the charge densities sigma are, gives a total charge. Uh Off the key off. So we&#x27;re just gonna jump into it. We know that E. R. equals two. Imacs are full of capital are and then by ghosts is gross&#x27;s law. He lost his law sigma mm The electric Phil strength times the infinitesimal distance is human caused. All of you know you notice the connectivity. Um So at a distance at a distance R. Equals two R. Mhm. The electric field, the electric field here&#x27;s marksman and so is the Q enclosed. So we want to do this formula. Yes, we have human caused uh you know, but then also imax times four by R squared equals two Q. Of you know what? This is human closed his maximum. So that&#x27;s why we&#x27;re gonna use Q at that point. So then imax is the same as a few, you note times four pi R squared since we divided both sides by four pi R squared like that. And so the max is the same as wonderful full pipe. You know what time skew over R squared. That&#x27;s our imax. And then in the body mm we know that the field strength, the electric field strength has given us the variable radius in terms of the marks. But we&#x27;ve already solved for remarks. So we&#x27;re just gonna go ahead and plug in the values that we got for imax from the previous step. And so now we do have a new relationship for the electric Uh field strength, which is just Q. four pi We&#x27;re not we&#x27;re combining these two. So we have our six And then outside we have are to the 4th power. So electric field strength, mm hmm Inside the ball, the electric field strength. Inside the ball simply mm times the S. Is Q. And closed over the note. And the volume charge density. Remember we were finding a want to find an expression for the volume charge density. The expression for the ruling charge density. We know that Cuban closed. It&#x27;s gonna be the same as Going from 0 to buy sigma are four pi R. Square our that&#x27;s our cue. And close. So we go back and connect the two formulas the previous one yard deeds which was um Cuban caused a in art then Q. And closed one of our in our sigma. We bring this one down because this is the same as Cuban closed. Four pi R squared the R. With that mm. And so E. D. S. His E. R. Times four pi squared equals to one of you know, charge density. And this one goes from zero to are we don&#x27;t want to forget that for by R squared D. R. I&#x27;m all right. And we already have a formula. We already have a formula for E. R. If you go back you see that E. R. Is this formula. So we&#x27;re going to plug it in right here. I want to plug in that formula from E. R. Right there. And this is in the process is going to help us simplify the process. Zero to our sigma. Are four by us. Where are these two cancel out? And then you combine those two. So we have a Q. r. six of six equals two. Don&#x27;t forget to cancel out the notes. So we just have a simplified version where we can isolate the volume charge density. So we&#x27;ll go back to this section and we um want to isolate so we have CDMA from zero to our, I just want to make sure we have zero Rds so flip them and we have sigma are for pie our square the R. And then on the right side we have a previous equation. Um Then we do differentials with respect to our that&#x27;s going to cancel out this. So we have sigma are for parts past where they are when we do the differential for that we get six Q. Are to the 5th power all over mm six to our over our six because our six is a constant. We only do differential with respect to our so sigma are signals are Becomes equal to six q. Are to the 5th power. six Q. R. to the 5th power. Uh Let&#x27;s see. We want to simplify the six. Cute six Q arch. The fifth bottle, all of four pi R squared on six. So we divided both sides by four pi R squared. All these D. R. Disappeared. So this one after we did the into into uh differential. It disappeared. So we don&#x27;t really need it normal. These two are going to cancel out. So we have our three. So our final final part as an answer is six cube R. to the power of four pi r. six. This is an expression for volume charge density. And then the final part, part C. We want to verify that um the total charge is equivalent to our based on the charge density. So the charge density uh he&#x27;s just total charge equal to Q. And so we&#x27;re gonna take the integral on both sides. Remember we have the charge density right there. This is the charge density. So I want to do that replacement. So from zero to our And we have 6q. Three fought by are to the sixth. Power times four pi R squared the R. I want to see if that&#x27;s gonna give us Q. So let&#x27;s pull out the constance uh 60 to our who left with Are the 5th power times they are because these two combined these two council. So it&#x27;s hard to the feast Powell. Then we go ahead and have 6q. Our six. When you do the integral for that. We have a large the sixth power of six because you&#x27;re adding one and adding like that to give you the integral From 0 to our. So this is six Q over arch. The six. We plug in the are we have our six. All over six. These two council these to cancel and we have a total charge of cute. So hope you enjoy the problem. Feel free to send any questions or comments and have a wonderful day

Nathan Silvano · Answer

Okay, so we have ice particle e symmetric charge distribution, and we have to calculate the electric flux in this charge distribution. We know that using the electric flux, we can say that the clothes integral off the fuel to the electric feud D A is going to be cool. The charge inside big ocean surface divided by absolute zero. Okay, so since the electric field is Postant, we can see that this is going to be, uh, e They&#x27;re multiplies doing to grow off the area, which is the surface area off these off this fear. It&#x27;s always four pi are square. She&#x27;s the Goshen surface, remember? Ah, this is going to be equal. The charge inside big ocean surface and the charge inside the ocean surface can be described us in that city. Devi Actually the integral off the density DV divided by absolute zero. So, uh, what iss this integral? This is the integral of volume. Therefore, we can say this is going to be one moment that sport in here we have e there multiplies for pie are square. That is equal one divided by absolute zero closed The need to grow, not closing to grow off the density. And the problem say&#x27;s that the density is just a divided by R and the volume The volume is just the area That&#x27;s Putin here. The volume DV is just the area multiply by g r. Therefore, this surface area this is going to be just for pie are square d r and their radios goes from zero True are Therefore we just need to calculate this. So let&#x27;s continue here. We have that the electric field is for by our square is going to be equal one divided by absolute zero Uh, a goes out of the integral So this is just a divided absolute zero. Uh, we have for buying here Integral off. Oh, are the are from zero to our Okay, therefore, let me see. This is going to be Let&#x27;s continue here. It is going to be e four pi r square. It&#x27;s going to be equal for pie. Hey, absolute zero. Our square divided by two from our 20 So this is just for by a divided by to absolute zero are okay now we can finally calculate the electric food and electric huge is going to be, um it&#x27;s cross for by for by You know that refuge is going to be We can cross one of the radios here. Yeah, divided by, uh, to absolute zero. Actually, you have Regis Square Sarti. So this is radios a square, therefore going to cross both their radius. And the final answer is just a divided by two absolute Europe in the direction off the electric foods. It&#x27;s going to be readily outer based fear. Okay, so that&#x27;s your final answer. Thanks for watching.

Sri Datta Vikas Buchemmavari · Answer

So we have a spectacle Balloon. It starts off with radius Armand radius to our a time which means the radius must go with respect to time has are not less are not keep I t which is our not one plus Steve, I t If you a cheese course zero you can see that the latest we are not a teak or do captain d we&#x27;ll see that the regis is to know you are not so Now that we have this, the elephant field just outside the bedroom will be just as if the total charge which is spread uniformly across these surface, is at the center off the disappeared. So hence the electric field will be simply one by four by absolutely not you buy our square where art is the radius. So that is one by four by its it or not, Cuba are not scared into one place. T v I t square. No, but a distance of 3.2 are not. If you consider a Gaussian surface, the amount of charge enclosed will never change. Hence, using office law, we can see that the flux is equal. Do the electric bill Times area which is Judge in closed. I s it or not? We&#x27;re charging. Was this constant on has E is he&#x27;ll fight 45 are square exit on Lord are Q by four Fight. You know that this distance are is 3.2 are not squared. It&#x27;s almost. This will be the electric bill at a distance off three point you are not. That&#x27;s it. Notice that the in close charge does not change you and in the first case and the electric field changes only because the radius changes.

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

Grace H.

Anna Marie V.

Kristen K.

Samuel H.

Precalculus Review - Intro

Functions on the Real Line - Overview

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Basic Technical Mathematics with Calculus

Grace H.

Anna Marie V.

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Precalculus Review - Intro

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Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Grace H.

Anna Marie V.

Kristen K.

Samuel H.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus