Answer:
The best option is for the following option m = 15 [g] and V = 5 [cm³]
Explanation:
We have that the density of a body is defined as the ratio of mass to volume.
where:
Ro = density = 3 [g/cm³]
Now we must determine the densities with each of the given values.
<u>For m = 7 [g] and V = 2.3 [cm³]</u>
![Ro=7/2.3\\Ro=3.04 [g/cm^{3} ]](https://tex.z-dn.net/?f=Ro%3D7%2F2.3%5C%5CRo%3D3.04%20%5Bg%2Fcm%5E%7B3%7D%20%5D)
<u>For m = 10 [g] and V = 7 [cm³]</u>
<u />![Ro=10/7\\Ro=1.42[g/cm^{3} ]\\](https://tex.z-dn.net/?f=Ro%3D10%2F7%5C%5CRo%3D1.42%5Bg%2Fcm%5E%7B3%7D%20%5D%5C%5C)
<u />
<u>For m = 15 [g] and V = 5 [cm³]</u>
<u />![Ro=15/5\\Ro=3[g/cm^{3} ]\\](https://tex.z-dn.net/?f=Ro%3D15%2F5%5C%5CRo%3D3%5Bg%2Fcm%5E%7B3%7D%20%5D%5C%5C)
<u />
<u>For m = 21 [g] and V = 8 [cm³]</u>
<u />![Ro=21/8\\Ro=2.625[g/cm^{3} ]\\](https://tex.z-dn.net/?f=Ro%3D21%2F8%5C%5CRo%3D2.625%5Bg%2Fcm%5E%7B3%7D%20%5D%5C%5C)
<u />
We can rearrange the mirror equation before plugging our values in.
1/p = 1/f - 1/q.
1/p = 1/10cm - 1/40cm
1/p = 4/40cm - 1/40cm = 3/40cm
40cm=3p <-- cross multiplication
13.33cm = p
Now that we have the value of p, we can plug it into the magnification equation.
M=-16/13.33=1.2
1.2=h'/8cm
9.6=h'
So the height of the image produced by the mirror is 9.6cm.
Answer:
In general solids are easier to transport than liquids, but the above metal example is a valid one and the only other one that comes to mind is that of concrete. It is mixed as a liquid and transported as such, but then sprayed or laid down to dry and form a solid surface or filler.
Explanation:
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