The question here is how long does it take for a falling
person to reach the 90% of this terminal velocity. The computation is:
The terminal velocity vt fulfills v'=0. Therefore vt=g/c,
and so c=g/vt = 10/(100*1000/3600) = 36,000/100,000... /s. Incorporating the
differential equation shows that the time needed to reach velocity v is
t= ln [g / (g-c*v)] / c.
With v=.9 vt =.9 g/c,
t = ln [10] /c = 6.4 sec.
I am pretty sure the answer is D, but I'm not 100% sure
Answer: The height of the building is 50.75 feet.
Step-by-step explanation:
The ratio between the height of the object and the casted shadow must be equal for all the objects, as the angle at which the source if light impacts them is the same.
For the person, we know that it is 5.8ft tall, and the shadow is 3.2ft long.
The ratio will be: 5.8ft/3.2ft = 1.8125
Now, if H is the height of the building, and the shadow that the building casts is 28ft, we must have:
H/28ft = 1.8125
Now we can solve this for H.
H = 1.8125*28ft = 50.75 ft
Then the height of the building is 50.75 feet.
Answer: These sides are labeled in relation to an angle. The opposite side is across from a given angle. The adjacent side is the non-hypotenuse side that is next to a given angle.
Step-by-step explanation:
Yes it is it's been that way for years
