Answer:
The power for circular shaft is 7.315 hp and tubular shaft is 6.667 hp
Explanation:
<u>Polar moment of Inertia</u>
= 0.14374 in 4
<u>Maximum sustainable torque on the solid circular shaft</u>
=
= 3658.836 lb.in
= 
lb.ft
= 304.9 lb.ft
<u>Maximum sustainable torque on the tubular shaft</u>
= 
= 3334.8 lb.in
= 
lb.ft
= 277.9 lb.ft
<u>Maximum sustainable power in the solid circular shaft</u>
= 
= 4023.061 lb. ft/s
= 
hp
= 7.315 hp
<u>Maximum sustainable power in the tubular shaft</u>
= 
= 3666.804 lb.ft /s
= 
hp
= 6.667 hp
"Lightyear" is a <em>distance</em> ... specifically, the distance that light travels
through vacuum in one year of 365 days.
It's equivalent to 9.454 x 10^12 kilometers, or 5.875 x 10^12 miles .
Yes,and because not everyone can wink and often that someone can only wink with one eye only
Answer:
your in mr langfords class
Explanation:
bruh moment
Answer:
Explanation:
(a) It is given that Joseph jogs on a straight road of 300m in a time interval of 2 minutes and 30 seconds, which is equal to 150seconds. Therefore, when Joseph jogs from point A to point B, he covers a distance of 300m in time of 150seconds. Hence, his average speed is 300m/150s=2ms^−1. Since it is a straight road and he jogs in a single direction in this case, his displacement is equal to 300m. Since it is a straight road and he jogs in a single direction in this case, his displacement is equal to 300m.
Hence, his average velocity is 300m/150s=2ms^−1
(b) Then it is given that he turns back and points B and jogs on the same road but in the opposite direction for a time interval for 1 minute and covers a distance of 100m.If we consider the whole motion of Joseph, i.e. from point A to point C, then he covers a total distance of 300m+100m=400m. And he covers this total distance in a time interval of 2.5min+1min=3.5min=210s.
Therefore, his average speed for this journey is 400m210s=1.9ms−1.
For the same journey is displacement is equal to the distance between the points A and C,i.e. 300m−100m=200m.
Hence, his average velocity for this case is 200m/210s=0.95ms^−1
