Answer:
0.000136kg/m3
Explanation:
13.6 / 1000 = 0.0136kg/cm3
0.0136 / 100 = 0.000136kg/m3
Given :
Initial velocity, u = 12.5 m/s.
Height of camera, h = 64.3 m.
Acceleration due to gravity, g = 9.8 m/s².
To Find :
How long does it take the camera to reach the ground.
Solution :
By equation of motion :
Putting all given values, we get :
t = 2.56 and t = −5.116.
Since, time cannot be negative.
t = 2.56 s.
Therefore, time taken is 2.56 s.
Hence, this is the required solution.
The average speed is determined by the following formula:
average speed = [sum of (speed * time for which that speed was traveled)] / total time
average speed = [(83 * 26 + 41 * 52 + 60 * 45 + 0 * 15) / 60] / [(26 + 52 + 45 + 15) / 60]
*note: The division by 60 is to convert minutes to hours. We see that the 60 cancels from the top and bottom of the division
average speed = 50.65 km/hr
The total distance traveled is equivalent to the numerator of the fraction we used in the first part. This is:
Distance = (83 * 26 + 41 * 52 + 60 * 45 + 0 * 15) / 60
Distance = 116.5 kilometers
"Celestial" = anything to do with the sky
("Cielo" ..... Spanish for "sky"
"Ceiling" ... that thing up over your head
"Caelum" .. Latin for "heaven")
"Terrestrial" = anything to do with the Earth
("Terra" ... Latin for "Earth")
<span>Well, since it's in the shape of a wheel and the person walks around the edge of it, they must have a centripetal acceleration. Since a=v^2/r you can solve for "v" using 2.20 as your "a" and 59.5 as your "r" (r=half of the diameter).
</span> a=v^2/r
v=(a*r)^(1/2)=((2.20)*(59.5))^(1/2)=<span>
<span>11.44 m/s.
</span></span><span> After you get "v," plugged that into T=2 pi r/ v. This will give you the 1rev per sec.
</span> T=2 pi r/ v= T=(2)*(pi)*(59.5)/(11.44)= <span>
<span>32.68 rev/s
</span></span> Use dimensional analysis to get rev per min (1rev / # sec) times (60 sec/min).
(32.68 rev/s)(60 s/min)=<span>
<span>1960.74 rev/min
</span></span>
