a-doctor-preparing-to-give-a-patient-an-injection-squirts-a-small-amount-of-liquid-straight-upward-f

Question

A doctor, preparing to give a patient an injection, squirts a small amount of liquid straight upward from a syringe. If the liquid emerges with a speed of $1.5 \mathrm{m} / \mathrm{s},$ (a) how much time does it take for it to return to the level of the syringe? (b) What is the
maximum height of the liquid above the syringe?

Nathan Silvano · Accepted Answer

Okay, so in this problem, we are squirting a small amount of liquid. It&#x27;s straight upwards from a syringe. I cannot draw a syringe. It&#x27;s that&#x27;s from the standard problem. First, we know that this liquid reaches a maximum height, then begins to fall again until he reaches the same height off this strange. Okay, so let&#x27;s see your problem. It&#x27;s the problem, say&#x27;s that the initial speed off the liquid was 1.5 meters per second. We know that just the liquid is under the gravity. The separation off nine points. You&#x27;re one meters per second square. And the first question is, how much time does it take for the liquid returned to the level? Simply how much time this hope process took? We know that the time to the liquid go up is equal to the time to the liquid to go down. So we free discover their time to the secret go up. We just need to multiply by two. And we have d&#x27;oh whole time of the process. So let&#x27;s think. Let&#x27;s look on Lee. So the first part off this movement and the first part of the movement we know we have, uh, initial velocity off 1.5 meters per second. We have a gravity, the separation negative because the gravity&#x27;s points in the opposite direction. Off the movement, off nine points. You are meters per second square and we want to discover death time. So simply years velocity as a function off time putting minus g here because we know we have to use minus G. Well, that&#x27;s put a positive sign and change later. So you guys get not confuse. It&#x27;s always positive, but the value off the gravity for this problem is going to be minus. Okay, so let&#x27;s see. Final velocity zero because he reaches his maximum height. Initial velocity 1.5 miners, nine point 81. So the time to go up it&#x27;s going to be 1.5 divided by 9.81. And this and this is equals two year of point 15 seconds. So the entire time off the movement we need to answer. It&#x27;s just multiply this by two. I&#x27;m going to be zero point tree seconds. This is the answer for the first item. This second item I can be what we want to discover. We want to discover the maximum height. That means the maximum height off the liquid. Okay, so we already know the time. You already know velocity can choose. Any equation to calculate is I&#x27;m going to choose the velocity as a function off position because I think is the easier one. But you can also use a position as a function of time and you get the same result. So let&#x27;s see. Final velocity zero initial velocity, 1.5 square minus two times 9.81 doubt x so that maximum height. It&#x27;s just going to be two points. 25 divided by two times 9.81. This is going to be equal 20.11. Yes, that&#x27;s right. Zero point 11 5 meters. This is the maximum height off the liquid after after leaving this revenge. So this is the final round

Surjit Tewari · Answer

question was 63 for party. We know that the pressure is equal to the hydrostatic pressure row G etch for party. The pressure is evenness. 55 m m a. G. So 55 into 1. 33 Paschal&#x27;s equal toe density off water is 1000 in 29.8 in tow. Etch A From this, we will get Jake will too. Zero point 746 meter for Part B. The pressure is given us 6 50 as tool, so we will convert it into Paschal&#x27;s 6 15 to 9.81 Rascals if we&#x27;re to density 1000 in 29.8 in tow edge. We from this we will get actually is approximately equal to zero point 65 meter. And for policy, it is given that 18 m m a. G into 1 33 Prospal&#x27;s Equal toe 1000 density into nine pointed into etc. From this, we will get at sequel to zero point 244 meter

Averell Hause · Answer

so intravenous transfusions are usually done through gravity. Um, we need to find the heights of the bottle such that it gives us certain pressures. So we have The density is going to be, uh, 1.0 g per cubic centimeter. Blood is mostly just water, and it&#x27;s gonna be 1000 kg per cubic meter. We also have the change in pressure is going to be equal to the density times and gravity times H where the rather delta h. So it&#x27;s redefine this Delta h the change in height. So the change in height will simply be equal to the change in pressure divided by the density times gravity. And so we can do each of these parts directly. Delta H for part A is going to be equal to 52 millimeters of mercury, and then we&#x27;re going to times this by 133 Pascal&#x27;s for every millimeter of mercury, all divided by 1000 times 9.8, and we get that Delta H for part A is going to be 0.706 m for Part B. We have Delta H equals, and then here we&#x27;ll have 680 680 millimeters of H 20 water and we get that we have 9.81 Pascal&#x27;s for every millimeter of water, all divided by 1000 times 9.8. And we&#x27;re getting that The the change in height for this would be 0.681 m again, this Delta ace just represents how how high the bottle must be in order to complete the transfusion. So Delta H for part B will be 0.681 m. The bottle must be placed 0.8681 m high. Quite see, we have, uh, again set. We have millimeters of mercury, 75 millimeters of mercury apologized and we know that we have again 133 Pascal&#x27;s for every millimeter of mercury. So let&#x27;s divide this by again 1000 times 9.8. So Delta H is going to give us 1.2 m and this would be our answer for C. That&#x27;s the end of the solution. Thank you for watching

Dading Chen · Answer

because we know the total pressure and can vehicles with the pressure at the most year lost the pressure funding fluid It was an arrangement here. Eventually have the hide off fluid. If you go to a P mines P at the most were over Rocchi. So we know the total pressure is given 100 and I still Pascal which is 1.9 times 10 to the power of five. Pascal and um what as the most feared pressure is about 1.13 times 10 to the power of fire Pesko And when on the destiny of fluid is given as 10 20 kilogram from Munich, you and the acceleration will grab these little 9.8 meters per second squared. So not it&#x27;s fucking better equation to determine the heights which is you go to one boy zero and I guys tend to the poor. If I pass cow minus 1.13 times tend to the horrify Pasco over 10 20 Kill. Where? For me to you? What about by 9.8 meter for a second square and it will give us he hide over fluid is about zero boy 770 meter. So now if we use the last stance fluid instead. So if we use the last stance flu it, let me see testy. Well, degrees, as you can tell, the heights, is a go to PG minus. P a t M over. Oh, gee, so Israel decrease. That means the heights will be increased if the other variables didn&#x27;t change. Okay, so their body hide of a suspension will be increased if the last dance flu is is using staff in. These are the answers for this question.

John Barone · Answer

So here we&#x27;re being told that this patient needs room leader of ivy solution over the course of 12 hours. Now the first thing I want to do is convert the leader into Miller leaders because the other rates during milliliters on this just makes it easier. So remember that one leader is the same thing as 1000 milliliters, and then the ivy can drop. Ah, For every Miller leader, it&#x27;s set up to give 15 drops. And then if we do our crossing out here, uh, there and we do the math, we end up getting 15,000 Trump&#x27;s for 12 hours. All right, that&#x27;s really the first step. And then then we need to figure out how many drops per minute were expected to. Dio and I have made this into two equations. You considerably just do it is one. But, um, I should have run out of space, will on the radio inside there. So we have 15,000 drops for 12 hours, so we will convert hours into minutes cross those, huh? And when you do that, you end up getting a REIT. That&#x27;s about 20 one drops. What? We&#x27;re G t ts per minute when you know how you rounded things that might come out to be a little bit closer to 20 drops per minute. All right, So again, this could be all pushed together into one big long equation, and it&#x27;s often easier for me on this white board, just her right out to equations, and that&#x27;s made in here.

Jamie Luskin · Answer

soon this problem, we&#x27;re gonna be taking a tube. And on one end it&#x27;s connecting into a vein. And on the other end, it&#x27;s gonna connect to Reservoir. This is gonna be open to the air, has fluid in it and were given the density. Okay, The height of the reservoir above the vein is gonna be to notify H and were given a gauge pressure of the vein. So I&#x27;m gonna write. That is PG. We&#x27;ll talk a bit about what that means. So, first, what are we trying to find in this problem? Well, we&#x27;re looking for the minimum height so that water will flow from the reservoir into the vein. And so just to make sense of all the values were given, let&#x27;s talk about what engage pressure is so engaged. Pressure is not an absolute total pressure. So here we&#x27;re not talking about the total pressure in the vein. We&#x27;re talking about the relative pressure inside the vein toe outside the bain. So inside the vein right here outside the vein, that&#x27;s gonna be at our reservoir. So it&#x27;s gonna be really important to keep that in mind as we make sense of things. And so Let&#x27;s really quickly just go over the conditions that we have, so notice that the density is constant. We&#x27;re also told in the problem that we don&#x27;t have to deal with any viscosity effects. We can ignore that. And so when we&#x27;re first considering the problem, nothing in the fluid is moving first. So it&#x27;s a static, fluid scenario. It&#x27;s nothing too complicated is going on here. Ultimately, we&#x27;re gonna need to relate the pressure differences and the fluid to the height. Difference is, it&#x27;s a sort of using that as intuition equation 12.6 from the text is gonna be pretty handy. So I&#x27;m just gonna go ahead and write that down, and they will make sense of the terms. Okay, So sorry if this is a bit hard to read, but this P and that&#x27;s gonna be the pressure at any given death. And so here with depth, we want to think about, you know, height below the reservoir. He not is the pressure at the service. So again, that&#x27;s the circus of the fluid, which is the reservoir row. Oh, sorry. So this is supposed to be a plus. Okay, looking much better. So anyway, moving on, um, row is the density which were given G is the acceleration due to gravity. Just go ahead and write that value down and h is air height. And so we&#x27;re looking to sell for height. So let&#x27;s go ahead and just rearrange this equation to get it in terms of h. So we subtract he not from both sides. First, divide out the road g both sides. Okay, so now we have this equation rearranged in terms of height, which is what we&#x27;re solving for. And let&#x27;s just really quickly take a look at the equation and see what we have and what we don&#x27;t have. We&#x27;re looking to sell for each. We are given me densities we have. But we know the acceleration due to gravity. And here we have p minus peanut. That&#x27;s a pressure difference or relative pressure. So keep in mind what we&#x27;re given. And the problem is a relative pressure. So let&#x27;s just really quickly make sure it makes sense. So h is the height of the reservoir above the vein than pee here is gonna be the pressure at the depth of the vein, and peanut is gonna be the surface pressure so it&#x27;s gonna be at the reservoir. So looking into these terms, this is actually the engage pressure. So what we&#x27;re saying is another way to write this in terms of our very beings that we have his team artist, you know, So we have everything we need. So go ahead and like these values in and sell for H through in gravity and okay. And so if you evaluate this on your calculator, she get approximately 0.58 and we&#x27;re talking about height within the units of meters. And so that&#x27;s the minimum height, where you&#x27;ll start to see fluid flowing from the reservoir into a vein.

Problem

A hot-air balloon has just lifted off and is risi…

Problem 100 Hard Difficulty

Answer

(a) 0.31 $\mathrm{s}$
(b) 0.115 $\mathrm{m}$

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(a) 0.31 $\mathrm{s}$(b) 0.115 $\mathrm{m}$

Physics

Liev B.

Zachary M.

Aspen F.

Meghan M.

Math Review - Intro

Algebra - Example 1

James S. WalkerPhysics

Liev B.

Zachary M.

Aspen F.

Meghan M.

(a) 0.31 $\mathrm{s}$
(b) 0.115 $\mathrm{m}$

James S. Walker

Physics