Answered

Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Join our Q&A platform to get precise answers from experts in diverse fields and enhance your understanding. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

Answer to number 8 please

Answer To Number 8 Please class=

Sagot :

First segment: The airplane is descending slowly, with time and distance at a 3:1 ratio. It is probably approaching the airport

Second segment: The airplane is maintaining its altitude. It is probably waiting for the other planes to clear away

Third segment:The airplane descending rapidly, with time and distance at a 1:2 ratio. It is probably landing

Sorry, my bad.
To find acceleration, we use v2-v1 /t
40mph - 20mph / 10 min
20mph/10min
We can convert this to 20/6 miles per 10 minutes, and cancel out the 10 min to 20/6 miles, which is 10/3 miles

AL2006

Acceleration = (change in speed) / (time for the change)

(Note:  That's all the Physics there is to this problem.
The rest is all arithmetic.)
=======================================

change in speed = (40 - 20) miles/hour = -20 miles/hour
time for the change = 10 minutes

Acceleration =  (-20 miles/hour) / (10 minutes) =

                          -2 miles/hour per minute .

That's a perfectly good and technically correct expression for acceleration.
But obviously the units might make some people dizzy.  So let's try to
clean it up a little.

Notice that 10 minutes is 1/6 of an hour.
So we could write the acceleration as

Acceleration = (-20 miles/hour) / (1/6 hour) = -120 miles/hour per hour =

                         -120 miles/hour² .

You could convert this into any units you like.  It's really not a physics problem
any more, it's just an exercise in converting units.