Answer:
1. If another need to continue his work
2. For documentation purposes for research validity
3. It enchance the visibility of quality assurance
Common saying "IF YOU HAVE NOT DOCUMENTED IT, YOU HAVE NOT DONE IT"
1. In the first image, you can see two cacti and a javelina in the background.
2. In picture one, neither of the plants are being disturbed. In picture two, however, the smaller plant is indeed being disturbed by the javelina.
3. The javelina is not eating the cactus on the right because it is stronger in color (?).
4. The cactus the javelina was eating has disappeared.
5. In picture 3, the idle cactus blossomed.
To determine the density, we use the following formula:
![density= \frac{mass}{volume}](https://tex.z-dn.net/?f=density%3D%20%5Cfrac%7Bmass%7D%7Bvolume%7D%20)
mass= 2.813 grams
volume= ?
we need to first calculate the volume before we can solve for the density. since the question states that the sample has a spherical shape, we can use the volume formula of spheres to find it.
![Sphere_{volume}= \frac{4}{3} \pi r^{3}](https://tex.z-dn.net/?f=Sphere_%7Bvolume%7D%3D%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20r%5E%7B3%7D%20)
radius (r)=
![\frac{diameter}{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bdiameter%7D%7B2%7D%20)
=
![\frac{8.0 mm}{2} = 4.0 mm](https://tex.z-dn.net/?f=%20%5Cfrac%7B8.0%20mm%7D%7B2%7D%20%3D%204.0%20mm)
let's plug in the values
V=
![\frac{4}{3} \pi 4^{3} = 268 mm^3](https://tex.z-dn.net/?f=%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%204%5E%7B3%7D%20%3D%20268%20mm%5E3)
we need to change the
![mm^3](https://tex.z-dn.net/?f=mm%5E3)
to
![cm^3](https://tex.z-dn.net/?f=cm%5E3)
using the following conversion
1 cm= 10 mm
![268 mm^3 (\frac{1 cm}{10 mm} )^3 = 0.268 cm^3](https://tex.z-dn.net/?f=268%20mm%5E3%20%28%5Cfrac%7B1%20cm%7D%7B10%20mm%7D%20%29%5E3%20%3D%200.268%20cm%5E3)
now we can find the density.
density= 2.813 grams/ 0.268 cm=
10.5 g/cm3
Answer:
Here the force is 10N and the mass is 5 kg. Dividing both sides by 5kg, we get a = 2 m/s^2.
The letter T represents groundwater