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Question 122 IptsPart 1: For M1 185g and M2 =125g what is the magnitude of the [acceleration) in m/s??...

Question

Question 122 IptsPart 1: For M1 185g and M2 =125g what is the magnitude of the [acceleration) in m/s??

Question 12 2 Ipts Part 1: For M1 185g and M2 =125g what is the magnitude of the [acceleration) in m/s??



Answers

(II) A sports car moving at constant speed travels 110 $\mathrm{m}$ in
5.0 $\mathrm{s}$ . If it then brakes and comes to a stop in 4.0 $\mathrm{s}$ , what is
the magnitude of its acceleration in $\mathrm{m} / \mathrm{s}^{2},$ and in $g^{\prime} \mathrm{s}$
$\left(g=9.80 \mathrm{m} / \mathrm{s}^{2}\right) ?$

So a certain force f gives an acceleration ace of one to the first block of 12 meters per second squared. It gives a second block an acceleration of 3.3 meters per second squared and and we know that the forces the same on either of them. So the net force is the same on block One and blocked too. And so we can say that F is equaling EMS of one ace of one equaling m sub too. Ace. Up to now, we have two unknowns and we can say that the accelerations for the system basically, we can say that the accelerations for EMS up to plus up someone and m sub two minus m someone can be solved by substituting where I'm someone is equaling half over a sub one. And of course I'm sad too is given by F over Ace up to. So now we can say for part a to seek the acceleration of object of mass m sub two minus m someone we can say that the force, the acceleration a for party would be equaling that force f divided by m sub two minus m. Someone substituting in. This is going to give us the force over F over a sub two minus f over Ace of one. And so this is going to give us a sub one a sub two divided by ace of one minus ace up to. And so this is giving us 12.0 meters loosely. You lose the units so 12.0 meters per second squared multiplied by 3.30 meters per second squared, divided by 12.0 minus 3.30 And we find that the acceleration a is going to be four 0.55 meters per second squared. Now for part B, it's gonna be the exact same. So the exact same. However, here we have the acceleration equaling the force over. I'm sub two. Plus I'm someone. So now this is going to give us a sub one ace up two divided by ace of one plus a sub too. So this would be 12.0, 3.30 divided by 12.0 plus 3.30 and the acceleration here is found to be 2.59 meters per second squared. This would be our final answer for part B. That is the end of the solution. Thank you for watching

Hello. So for this question From new test the questioning because you plus 80. Right? So if you isolate um a you know, grandma lets you by the Chilean time. Right? So, in this question, Yes. And and they're changing time for both journeys are the site. So I'm going to actually changing time. Well, that's why I'm just going to reveal and see you over. Okay. So there's going to be for the first one. So this is 111 Then the other one. The second one will be V two. And with you too over a two, I deserve the changing times. They are the same. Right? So, I can equate these two equations. They have everyone when this you won over everyone cause 32 My miss You too. Over 82. In the question, they said the body started from rest so it just goes to zero. He's also close to zero. They let everyone over a one it goes to over 82 started final velocity for the second motion is twice That are the first item. Is that V 2? Because to not twice 2.5. Do you want? All right. So if I read your question as V one over a one was 2.5, you know, I'm goodbye to this. Everyone can cancel this, everyone. Yeah, there's one over a one plus 2.5. Oh, my baby, too. Right. They do the first situation as straight as 1/3 equals 2.5 A two defining a too right For another 8 to which ends with three times 2.5. And there will be 7.5. Uh That is the 2nd square. This is the acceleration of the second. Thank you.

In this problem, we have centripetal motion. Now centripetal motion, as we may recall, is when we execute a rotation and the forests is perpendicular to the velocity. We also remember that in net force is equal to mass times acceleration and in the case of centripetal motion, this is also equal to mass times velocity squared all over the radius of the turn. Now masses cancel out. We and we end up simply with acceleration equals two velocity squared all over the radios and we're told the velocity we know that it is 10 m per second squared over 25 m, which is the radius of the train 10 square 100 divided by 25. That's going to give us four meters per second square.

So when the motorcycle travels a circular part and here is the motorcycle, then the centripetal force and the acceleration points two words. The center and the velocity points tangent to the bat as long the direction off motion.


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