Of 1.25 m/s², what is its orbital speed? 7. A race car goes...

of 1.25 m/s², what is its orbital speed? 7. A race car goes around a level, circular track with a radius of 1.00 km at a speed 120 km/h. What is the centripetal acceleration of the car? 8. The moon revolves around the Earth in 27.3 days in a nearly circular orbit with a radius of 3.80 x 10? km. Assume that the Moon's motion is uniform. With what acceleration is it falling toward Earth? 9. A car enters a circular curve with a radius of curvature of 0.400 km at a constant speed of 83.0 km/h. If the friction

Transcript

0:00 All right.

00:01 So for question seven, we're told that we have a car going around a level circular track.

00:07 The radius of the track is one kilometer in standard si units of meters, which multiply that by a thousand.

00:15 It's a thousand meters.

00:17 The speed of the car around the track is 120 kilometers per hour.

00:23 In si units meters per second, it's going to be a thousand, two hundred, well, a hundred, two hundred, well, times a thousand meters times 60 minutes in an hour, 60 seconds in a minute.

00:53 Oh, so actually we're dividing that.

00:58 So 100 ,000 divided by 60 divided by 60, and that's about 33 and a third, so 33 .33.

01:06 So 33 .33 meters per second.

01:08 We're asked to calculate the centripetal acceleration.

01:11 This is given by the velocity squared over the radius of the circular path.

01:17 So using our standard units so that everything cancels out accordingly, we have 33 .33 meters per second squared divided by a thousand meters.

01:34 This comes out to 1 .11.

01:37 One of the meters cancels out so we're just with meters on the top divided by second squared as expected for the units of acceleration.

01:45 Now for question eight, we have the moon revolves around the earth in a period of revolution of 27 .3 days, we're assuming a nearly circular orbit, and the radius is equal to 3 .80 times 10 to the 5th kilometers, which is 3 .80 times 10 to the 8th meters...

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