Explanation:
It is given that,
The volume of a right circular cylindrical, ![V=108\ cm^3](https://tex.z-dn.net/?f=V%3D108%5C%20cm%5E3)
We know that the volume of the cylinder is given by :
![V=\pi r^2 h](https://tex.z-dn.net/?f=V%3D%5Cpi%20r%5E2%20h)
............(1)
The upper area is given by :
![A=32r^2+2\pi rh](https://tex.z-dn.net/?f=A%3D32r%5E2%2B2%5Cpi%20rh)
![A=32r^2+2\pi r\times \dfrac{108}{\pi r^2}](https://tex.z-dn.net/?f=A%3D32r%5E2%2B2%5Cpi%20r%5Ctimes%20%5Cdfrac%7B108%7D%7B%5Cpi%20r%5E2%7D)
![A=32r^2+\dfrac{216}{r}](https://tex.z-dn.net/?f=A%3D32r%5E2%2B%5Cdfrac%7B216%7D%7Br%7D)
For maximum area, differentiate above equation wrt r such that, we get :
![\dfrac{dA}{dr}=64r-\dfrac{216}{r^2}](https://tex.z-dn.net/?f=%5Cdfrac%7BdA%7D%7Bdr%7D%3D64r-%5Cdfrac%7B216%7D%7Br%5E2%7D)
![64r-\dfrac{216}{r^2}=0](https://tex.z-dn.net/?f=64r-%5Cdfrac%7B216%7D%7Br%5E2%7D%3D0)
![r^3=\dfrac{216}{64}](https://tex.z-dn.net/?f=r%5E3%3D%5Cdfrac%7B216%7D%7B64%7D)
r = 1.83 m
Dividing equation (1) with r such that,
![\dfrac{h}{r}=\dfrac{108}{\pi r}](https://tex.z-dn.net/?f=%5Cdfrac%7Bh%7D%7Br%7D%3D%5Cdfrac%7B108%7D%7B%5Cpi%20r%7D)
![\dfrac{h}{r}=\dfrac{108}{\pi 1.83}](https://tex.z-dn.net/?f=%5Cdfrac%7Bh%7D%7Br%7D%3D%5Cdfrac%7B108%7D%7B%5Cpi%201.83%7D)
![\dfrac{h}{r}=59 \pi](https://tex.z-dn.net/?f=%5Cdfrac%7Bh%7D%7Br%7D%3D59%20%5Cpi)
Hence, this is the required solution.
Rainbows are caused by the dispersion of light, which itself consists of a combination of refraction and reflection of light around little droplets of water.
Choice C
Potential energy = mass x gravity x height
P.E = 4 x 9.8 x 3
P.E = 117.6 J
Answer:
the magnitude of acceleration will be 1.50m/s^2
Explanation:
To calculate your acceleration, you can use your formula that states that the net force on an object is equal to the mass of the object multiplied by the acceleration of the object. Fnet=ma
if you draw out this situation and label the forces you will have your vector towards the right with a magnitude of 20.0N and then your friction vector will be pointing to the left (in other words, in the negative direction) (opposing the direction of movement) with a magnitude of 5.00N, with the 10.0 kg box in the middle.
The net force will be calculated using F1+F2=Fnet where your F1=20.0N and F2= -5.00N (since it is towards the negative direction).
you will find that Fnet=15.0N
With that, plug in the values you know to calculate the acceleration of the block:
Fnet=ma
(15.0N)=(10.0kg)a from her you can divide both sides by 10 to isolate a:
1.50=a (and now make sure to label the units of your answer)
a=1.50m/s^2 (which is the typical unit for acceleration)
106.68 centimetres are in 3.50 feet