a-catapult-can-launch-a-plane-from-the-deck-of-an-aircraft-carrier-from-0-to-260-mathrmkm-mathrmh--2

Name: a catapult can launch a plane from the deck of an aircraft carrier from 0 to 260 mathrmkm mathrmh 2
Uploaded: 2021-06-08
Duration: 3 min 38 s
Channel: Numerade
Description: A catapult can launch a plane from the deck of an aircraft carrier from 0 to $260 \mathrm{km} / \mathrm{h}$ in $2.0 \mathrm{s}$. How many $\mathrm{g}$ 's is the average acceleration for such a launch? $\left(1 \mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}$.) \right.

Question

A catapult can launch a plane from the deck of an aircraft carrier from 0 to $260 \mathrm{km} / \mathrm{h}$ in $2.0 \mathrm{s}$. How many $\mathrm{g}$ 's is the average acceleration for such a launch? $\left(1 \mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}$.) \right.

Kian Manafi · Accepted Answer

So the average acceleration, I&#x27;m glad to note average a it&#x27;s equal to our change in velocity divided by our change in time. So here are change in velocity was equal to 260 and this was in km our minus zero kilometers per hour since we were told it went from 0 to 260 kilometers per hour in two seconds. And so what we want to do here is we want to actually convert this kilometers per hour into um meters per second. So Since this is equal to 260 km/h, we&#x27;re going to multiply by 1000 m per one kilometer and then by one our for 60 minutes And then by one minute For 60 seconds so that we can get this value into meters per second. So 260 times 1000 is 260,000. So this would be equal to 260,000 Divided by 3600. And this was this is in now meters per second. So if you plug this into calculator 260,000, Divided by 3600, it&#x27;s equal to 72 .22 feeding and this is in m per second. And so now that we have this average um velocity in meters per second, we can go back and look at our average acceleration equation. She is equal to our average velocity divided by our average or sorry, or change in velocity divided by our change in time. And so now we can put our change in velocity is 72.22 m/s and we can divide by two seconds since we&#x27;re going from were just we were only accelerating in those two seconds. So we have our average Acceleration is equal to 72 .22 repeating meters per second, divided by two seconds. And so this is equal to 36 .11 repeating meters per second squared. And so this is our average acceleration. Now to figure out how many jesus is we just multiply this. So RGs, it&#x27;s going to be equal to three, sorry, 36.11 m per second squared times one, divided by It was 9.8 meters per second square. Okay, so now our units are going to cancel and we&#x27;re just going to have Um, RGs is equal to 36.11 Repeating, divided by 9.8. And so if you plug this into a calculator, this comes out to be 3.68 five Gs. So our average acceleration was again 70 or sorry, 36.11 m per second squared. And then to convert that into Gs, we have 3.685 Gs.

Charles Carter · Answer

Okay, So this problem tells us that aircraft carrier a jet can be catapulted catapulted from 0 to 155 MPH. Uh um, two seconds. Okay. Um and then it says that the average force exerted by the catapult so they have force equals 9.35 times 10 to the fifth Newtons. It wants us to calculate the mass. So the thing we&#x27;re missing is the acceleration. Because we&#x27;re Nunes. I&#x27;m gonna change my MPH in two meters per second. So 1 55 MPH meters per second to change MPH. Two meters per second. This might be a helpful thing tohave. What do you like for the test? Um, the conversion is just 2.237 So you divide by 2.237 and that gives you 69.2912 meters percent out. Okay, that&#x27;s just a conversion. Get from MPH. Two meters per second. It&#x27;s a good thing to have memorized. So now that we have our final velocity, we can use a kinetic equation. We&#x27;re gonna have 69 points. 2912 That&#x27;s our final velocity. And that&#x27;s gonna equal our initial velocity, which was zero somebody plus our acceleration times to okay, said to find the acceleration, we just divide 69.2912 went to give us 34.6456 next meters per second squared. Okay. And then we knew our force. Eso forces 9.35 times 10 to the fifth that&#x27;s gonna equal are mass of the jet. Times are acceleration that we just got as 34.6456 So dividing 9.35 times 10 to the fifth Give us 26,987 0.55 So the mass of our jet, it&#x27;s gonna be 26,000 1987 0.55 kilograms. Thank you very much.

Jacob Adamczyk · Answer

here we have a being catapulted by a steam engine up to a velocity of 175 MPH. This process takes 2.5 seconds. And we would like to calculate from this data the average acceleration of the plane and the distance to plane moves within this time interval 2.5 seconds. The latter we will be assume the acceleration is constant. But in the former, we&#x27;re just going to find the average acceleration of the plate. So for part A, we&#x27;d like to find the average acceleration. To do that, we must know the initial velocity. And we&#x27;re assuming that the jet starts at rest. So V zero equals zero and then it goes to the final equals 175 MPH. Okay, remember, the acceleration is defined as the change in velocity divided by the change in time. This is the average for loss, the average acceleration. If we were like the instantaneous acceleration, then we would have to calculate the derivative with a velocity with respect to time. So to calculate the average, would you take the final minus two. The initial divided by the time interval, so in this case, we have 1 75 MPH in a zero MPH, divided by 2.5 seconds. Okay, so notice you have some units of MPH divided by seconds. That&#x27;s perfectly fine, because that&#x27;s the same units, Um, as any acceleration. Really, It just has to be distance divided by time, divided by time. Okay, so if we can get this we have 1 75 divided by 2.5, which is simply 70 70. And the units are miles for our troops period and MPH for a second. Or, in other words, 70 MPH per second. So for every second that the jet is being launched, it gains 70 MPH, has beaten. That&#x27;s what this means on average. So another, another way bring the average just as a heads up is like this. This is maybe what some textbooks will use. So this is just another way of writing. These are all saying the same thing. Okay, so we have 70 MPH per second on average, and now we&#x27;re going to assume that to be our constant acceleration throughout this process for part B to determine the amount of time. Okay, so this is part betting answer. So Part B, we would like to know the amount of distance Well, we know that the displacement is equal to the initial velocity plus 1/2 the acceleration inspired by the time squared. Here, the initial velocity we already mentioned that&#x27;s going to be zero recently. Have Delta X equals 1/2 70 MPH for a second multiplied by 2.5 seconds squared. Can&#x27;t, so don&#x27;t x equals one. Halftime. 70 is 35 35 times 2.5 times 2.5. That&#x27;s 218 0.75 But now watch the units here. We have MPH per second despite my second square. So we really have MPH times seconds. Okay, so what does that really mean? Well, we&#x27;re gonna have to convert either the hours with seconds. So if we convert, let&#x27;s go back up here and convert this MPH per second or convert that 2.5 seconds so that 2.5 seconds Let&#x27;s write that in terms of, uh let&#x27;s see, how should we do this? Well, but Let&#x27;s look down here. We have 200 say, 220 MPH time seconds. And we know that 60 seconds aren&#x27;t in one minute and 60 minutes or in one hour. Maybe this better. So we have one a minute for every 60 seconds, and we have 60 minutes in every hour. So we have one hour every 3600 seconds. Okay, so we&#x27;re going to have to, uh, multiply. This conversion factor by her Delta X is will have 218.75 We have miles and then we have seconds divide by hours. So we want some sort of unit that&#x27;s hours divided by seconds. So times one over 3600 hours, divided by seconds. Because this in and of itself, his unity, that&#x27;s equal to one. That&#x27;s just a conversion factor. So we can both play by toe by it all we want. So that&#x27;s our hands with this hour, the second cancers with this second. Okay, and then we divide by 3600 and we get a value for Delta X being equal to 0.6 07 miles. That&#x27;s not satisfactory. Then that&#x27;s multiply that by the conversion factor 50 to 80 feet per mile. Right time shifted to 80 and we get a value approximately 321 feet. So about 300 feet were able to reach a speed of 175 MPH with this jet.

Keshav Singh · Answer

for this problem on the topic of motion in one dimension, we&#x27;re told that a catapult launches a jet aircraft, giving it a speed of 175 miles an hour in 2.5 seconds. We want to find the average acceleration of the plane and assuming this acceleration is constant to find the distance that the plane will move. So the average acceleration of the plane a bar is equal to the rate of change of velocity delta V over delta T. This is 175 mph minus the initial velocity zero, since it starts from rest, Divided by a time of 2.5 seconds. This gives us an average velocity or average acceleration rather of 70 miles per hour seconds. So we need to convert this to a. C. Units. And so this is equal to 70 miles for our seconds times, 1609 meters for every mile times an hour For every 3000 600 seconds. And this gives us the average acceleration in si units to be 31 0.3 m/km2. That&#x27;s the average acceleration Of the plane in those 252.5 seconds next, you want to find the distance over which the plane moves using a kind of magic equation, delta X is equal to V, no t plus half a t squared. And assuming the acceleration is constant, We get this to be zero, since the initial velocity zero plus a half times the acceleration of 31.3 meters per square second times a time of two point five seconds squared. This gives us the displacement of the plane in this time To be 97 0.8 meters.

Meghan Miholics · Answer

So in this question, we are told that there is a 200 gram rockets and it has an acceleration of 10 meters per second squared and were asked, You know what thrust would needed to what would be needed on Earth as well as on the moon, where there is a different acceleration due to gravity. So thrust is basically just a fancy word for of the force that is acting on the rockets based on the interaction between the, um the air outside the rocket and, um, the rocket itself. So if you take a look at a free body diagram for this rocket, there&#x27;s gonna be a force of gravity pulling it down towards the Earth or the moon. And then there&#x27;s going to be the force upward provided by the thrust. So I&#x27;m just gonna label as f t h after we&#x27;ve done this free body diagram. Ah, it&#x27;s pretty simple to go ahead and salt this we&#x27;re just gonna use our f net equals m A. And I&#x27;m going to start off very general. So I&#x27;m gonna create equation we can use for both and be so the net force acting on the object. Let&#x27;s take up to be positive. So the net force acting on the object will be, ah, the thrust minus the gravitational force. And that will be equal Teoh m a. And then let&#x27;s substitute in mg for F G. And as you can see well, simply just solve for the thrust force. We can take the mg over to the other side and so we get an equation. If you factor out the M, it&#x27;s even easier to use. Ah, we get an equation that tells us that thief force, um, the thrust forces equal toe m times a plus t. So then we can just substitute in our numbers, um, for Earth and for the moon. So on earth, we have a value of G of 9.81 approximately and so we can plug that into the formula. So we get F T h. The mass is 0.2 kilograms because the rocket is 200 grams, so divide by 1000 to convert into kilograms. The acceleration is the 10 meters per second squared. That&#x27;s how much you know, we are looking for the object to be accelerating overall, and then we&#x27;ll do the plus team. And so if you plug right into the calculator and round Teoh one significant figure, which is all that were given in this question you get about four Newtons. So that&#x27;s our final answer for Part A. We need a thrust force of four Newtons and then for Part B. We have a slightly different value for G because on the moon, the gravitational forces, not as much as it is on Earth. And so we get a different thrust force. We don&#x27;t eat quite as much because there&#x27;s not as much pulling the Rocket back. So instead of 10 plus 9.1, we&#x27;re gonna do 10 plus 1.62 this time and we&#x27;ll get a value of about 2.3 since we&#x27;re only really we only really have one significant digit weaken. Round that to about two Newtons for the thrust force required on the men

Robert Zaballa · Answer

in this problem. We want to know what the amount of time was in accelerating to the takeoff speed. Given that that speed was 320 kilometers per hour and the mean acceleration was 2.9 meters per second squared. So I&#x27;ve written that is the final speed. But we also need to know what the initial speed Waas. But we can already make one observation that will tell us that the plane had to be at rest initially. So the initial speed was simply zero. So we simply right v I is equal to zero kilometers per hour and after that we can use the equation that the mean acceleration is equal. Teoh the difference in velocities. I refer to these A speeds but really we can also just in one dimension in this case, call these velocities for this problem. Ah, this mean acceleration is the difference in velocities over the time interval that the velocity changes. So we&#x27;re gonna find this time interval and to do so, we can invert this equation. So if we multiply both sides of the mean acceleration times the time interval, then it will cancel on the right side and we can also divide by the mean acceleration on both sides. It&#x27;ll cancel on the left side and we&#x27;re left with the equation for the time interval. That&#x27;s the equation will use. The only thing we need to do now is convert the final velocity two meters per second. Since the acceleration is in meters per second squared. So we already have kilometers and we can convert that two meters first. Then we&#x27;ll convert hours two seconds. So let&#x27;s start with 320 kilometers per hour. And if I write one kilometer down here, this is equivalent to 1000 meters. If I write one hour appear, this is equivalent to 60 minutes and if I write one minute up here, this is equivalent to 60 seconds. So this is equal to 320. Well, we&#x27;ll continue with that shortly. I&#x27;m just gonna illustrate how this method will cancel the units properly. So kilometer will cancel here. Um, meters will stay, ours will cancel, minutes will cancel and seconds will stay. So now we have 320 times 1000 with the meter unit remaining divided by 60 times 60 with the second unit remaining. So now we&#x27;re in meters per second and when you carry out this calculation you obtained 320 over 3.6 meters per second, which is equal to you, 88 0.89 meters per second. So now that we have VF and we know that b I zero, we can go ahead and get Delta T. So I&#x27;ll just draw a line here to separate. Delta T is equal to 88 0.89 minus zero, divided by 2.9 and this is equal to 30.65 seconds and that&#x27;s the amount of time needed to accelerate.

Problem

A stone is thrown straight up from the edge of a …

Problem 15 Easy Difficulty

A catapult can launch a plane from the deck of an aircraft carrier from 0 to $260 \mathrm{km} / \mathrm{h}$ in $2.0 \mathrm{s}$. How many $\mathrm{g}$ 's is the average acceleration for such a launch? $\left(1 \mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}$.) \right.

Answer

$3.7 \mathrm{g}^{\prime}$

Related Courses

Basic Technical Mathematics with Calculus

Related Topics

Discussion

Top Calculus 2 / BC Educators

Catherine R.

Kayleah T.

Caleb E.

Samuel H.

Calculus 2 / BC Courses

Integration Review - Intro

Definite vs Indefinite

Recommended Videos

Watch More Solved Questions in Chapter 26

Video Transcript

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

Catherine R.

Kayleah T.

Caleb E.

Samuel H.

Integration Review - Intro

Definite vs Indefinite

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

$3.7 \mathrm{g}^{\prime}$

Basic Technical Mathematics with Calculus

Catherine R.

Kayleah T.

Caleb E.

Samuel H.

Integration Review - Intro

Definite vs Indefinite

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

Catherine R.

Kayleah T.

Caleb E.

Samuel H.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus