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University of Michigan - Ann Arbor
Hope College
University of Winnipeg
01:40
Keshav Singh
(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?
07:43
Kathleen Tatem
(II) A person has a reasonable chance of surviving an automobile crash if the deceleration is no more than 30 $g$'s. Calculate the force on a 65-kg person accelerating at this rate.What distance is traveled if brought to rest at this rate from 95 km/h?
0:00
Muhammed Shafi
(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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welcome to our first example video on what to do with centripetal acceleration in this video will consider a classic problem. Physics that goes something like this. You have a record player that's spinning at in omega of approximately 40 rpm. Say it's rotating this direction. Notice. If you try the right hand rule here, you'll find that your right your thumb for me, right finger. Your right hand is pointing out of the screen towards you, so we'll recall this positive 40 rpm. So let's say that we have a bug sitting on this record player that's walking back and forth on. We want to find out. What is the acceleration of this bug, This centripetal acceleration of this bug at different radio. We'll say that it starts out somewhere near the middle and that it has a radius of 10 centimeters, and then at some point later, it's going to have a radius of we'll say 18 centimeters now, in order to calculate the centripetal acceleration. Remember, we had to equations. We have V squared over R, and we have Omega squared multiplied by our and neither one of these will work. Since we have omega, we might as well Use it but notice that our omega is in rotations per minute. Our radius is in centimeters and we probably want an acceleration that's in terms of meters per second squared. So we're going to need to do some conversions here in order to get this right. So first of all, let's go ahead and convert our omega. We have 40 rpm and we want to multiply that by We have 60 seconds in one minute and then in one revolution or one rotation, there are two pi radiance. So this will get us units of radiance per second. Yeah. Now we want to convert from 10 centimeters, too 1 m into meters. So remember that 10 centimeters a centimeter is 10 to the negative 2 m, which means it's going to be equal to we need to move the decimal over two times 0.1 meters or for 18 centimeters. That will be equal to 0.18 meters. Now, in order to calculate a C, then we're gonna multiply it by omega squared Times are so in the first case that will be Omega Square. It is 40 times two pi divided by 60 radiance per second. Okay, square all of that and multiply by our, which is 0.1 m. Notice that the units will come out be meters per second squared and then for the second. Well, it centripetal acceleration is equal to omega squared times are which is 40 times two pi divided by 60 all times 0.18 meters. And so I encourage you to plug these in and calculate what they might be you could buy. On the other hand, also have calculated what your visa are you would save equals R times omega. Yeah, and then you could plug it in, uh, to A C is equal to v squared over r if you prefer That way, though, it's, ah, little more button pushing on your calculator.
Newton's Laws of Motion
Applying Newton's Laws
Work
Kinetic Energy
Potential Energy
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