Answer with Explanation:
We are given that
Weight of an ore sample=17.5 N
Tension in the cord=11.2 N
We have to find the total volume and the density of the sample.
We know that
Tension, T=![W-F_b](https://tex.z-dn.net/?f=W-F_b)
=buoyancy force
T=Tension force
W=Weight
By using the formula
![11.2=17.5-F_b](https://tex.z-dn.net/?f=11.2%3D17.5-F_b)
N
![F_b=V_{object}\times \rho_{water}\cdot g](https://tex.z-dn.net/?f=F_b%3DV_%7Bobject%7D%5Ctimes%20%5Crho_%7Bwater%7D%5Ccdot%20g)
Where
=Volume of object
=Density of water
=Acceleration due to gravity
Substitute the values then we get
![6.3=9.8\times 1000\times V_{object}](https://tex.z-dn.net/?f=6.3%3D9.8%5Ctimes%201000%5Ctimes%20V_%7Bobject%7D)
![V_{object}=\frac{6.3}{9.8\times 1000}=6.43\times 10^{-4} m^3](https://tex.z-dn.net/?f=V_%7Bobject%7D%3D%5Cfrac%7B6.3%7D%7B9.8%5Ctimes%201000%7D%3D6.43%5Ctimes%2010%5E%7B-4%7D%20m%5E3)
Volume of sample=![6.43\times 10^{-4} m^3](https://tex.z-dn.net/?f=6.43%5Ctimes%2010%5E%7B-4%7D%20m%5E3)
Density of sample,![\rho_{object}=\frac{Mass}{volume_{object}}](https://tex.z-dn.net/?f=%5Crho_%7Bobject%7D%3D%5Cfrac%7BMass%7D%7Bvolume_%7Bobject%7D%7D)
Where mass of ore sample=1.79 kg
Substitute the values then, we get
![\rho_{object}=\frac{1.79}{6.43\times 10^{-4}}=2.78\times 10^3 kg/m^3](https://tex.z-dn.net/?f=%5Crho_%7Bobject%7D%3D%5Cfrac%7B1.79%7D%7B6.43%5Ctimes%2010%5E%7B-4%7D%7D%3D2.78%5Ctimes%2010%5E3%20kg%2Fm%5E3)
Density of the sample=![2.78\times 10^{3} kgm^{-3}](https://tex.z-dn.net/?f=2.78%5Ctimes%2010%5E%7B3%7D%20kgm%5E%7B-3%7D)
It involves electrons.
The cathode is the electrode where electron deficient ions move to.
While the anode is electrode where electron excess ions move to.
So the relationship between Cathode and Anode involves electrons.
C.
While the answer is that it does, it transmits light VERY poorly. Most of the light bounces off it and the rest is refracted into the ocean. This is why you can't see much that is far away from you in the ocean unlike if you're just on land.