#$&*
course Phy 201
Im not sure if my last submission of this assignment was titled correctly, so heres the correct one.
Question 1: 235km/3.25h = 72.3km/h as the speedQuestion 2: 25km/h to 25km/60minutes which is 0.416km/minute then divide 15km by 0.416km/min to get 36minutes
Question 3: 110km/h * 1h/60minutes = 110km/60minutes * 1minute/60s = 110km/3600seconds = .031km/s. then multiply by 2seconds to get .062km traveled
Question 4: 35mi/h * 1.609344km/mi = 56.33km/h. then 56.33km/h * 1000m/km * 1h/60minutes * 1minute/60seconds = 15.65m/s. then 15.65m/s * 3.3feet/m = 51.64ft/s.
Question 5: ‘ds = 7.6cm ‘dt = 3.1s. 7.6cm/3.1s = 2.45cm/s
Question 6: ‘dt = 6.5s ‘ds = 5.1cm. 5.1cm/6.5s = 0.78cm/s Yes the speed should be able to be calculated at the specific time intervals.
Question 9: 2miles/12.5minutes = 0.16mile/min *1.6km/mile * 1min/60s = 4.2x10^-3 km/s * 1000m/km = 4.2m/s
Question 11: 8.5km/2 = 4.25km * 1h/95km = 0.044h * 60minutes/h = 2.64 minutes
Question 16: 95km/h * 1000m/km = 95000m/h * 1h/60minutes *1min/ 60s = 26.4m/s divided by 6.2s = 4.3m/s^2
Question 22: 85m / 23m/s = 3.7s then 23m/s divided by 3.7s = 6.2 m/s^2
Question 27: 85km/h * 1000m/km = 85000m/h divided by 3600 = 23.6m/s divided by 0.8m = 29.5s so then 23.6m/s divided by 29.5s = 0.8m/s^2 and then you take 9.8m/s^2 and divide it by 0.8 to get 12.25, then move the decimal place to the left two places to get .1225g
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You need to say more about what you are thinking:
85km/h * 1000m/km = 85000m/h divided by 3600 = 23.6m/s could be a little more rigorous with units but this is OK
divided by 0.8m = 29.5s it's unclear how you connect this calculation to the definitions; however moving 0.8 m at over 20 m/s the time interval is much less than 1 second. You aren't identifying your quantities. Are you treating 23.6 m/s as an average velocity? Is this really the average velocity? Have you identified initial and final velocities, etc. and established that this is an average velocity?
so then 23.6m/s divided by 29.5s = 0.8m/s^2 Why would you do this? If you identify 23.6 m/s as change in velocity and 29.5 s as the corresponding time interval, this would give you the acceleration, and you would want to identify this as an acceleration. I think this is what you intend, but you need to state it.
and then you take 9.8m/s^2 and divide it by 0.8 to get 12.25, then move the decimal place to the left two places to get .1225g 9.8 is 12.25 times as great as 0.8, but that doesn't mean that 0.8 is .1225 times as great as 9.8
This really doesn't look bad, but you are submitting a string of calculations (most correct) without explaining yourself. I've inserted notes into a copy of your last solution, which should clarify what I mean. I can't refer back and forth to copies of the problems, but if you flesh out your calculations with a bit of explanation I can easily follow your thinking without having to do that, and will then be glad to give you feedback on your work.
In any case you should be submitting the queries as assigned. Having worked through the problems that won't take you long.