Answer:
2.464 cm above the water surface
Explanation:
Recall that for the cube to float, means that the volume of water displaced weights the same as the weight of the block.
We calculate the weight of the block multiplying its density (0.78 gr/cm^3) times its volume (11.2^3 cm^3):
weight of the block = 0.78 * 11.2^3 gr
Now the displaced water will have a volume equal to the base of the cube (11.2 cm^2) times the part of the cube (x) that is under water. Recall as well that the density of water is 1 gr/cm^3.
So the weight of the volume of water displaced is:
weight of water = 1 * 11.2^2 * x
we make both weight expressions equal each other for the floating requirement:
0.78 * 11.2^3 = 11.2^2 * x
then x = 0.78 * 11.2 cm = 8.736 cm
This "x" is the portion of the cube under water. Then to estimate what is left of the cube above water, we subtract it from the cube's height (11.2 cm) as follows:
11.2 cm - 8.736 cm = 2.464 cm
Lines of longitude and latitude . Surprisingly, almost all sunspots are found in two bands on the Sun, just north and south of the equator. At the start of the cycle, the sunspots appear at latitudes of about 30 degrees, both north and south<span> of the equator.</span>
<span>C. Reduce the friction between its moving parts.</span>
Answer:
The magnitude of the electric field is 129375 N/C toward south.
Explanation:
Given that,
Electric force
(A). We need to calculate the magnitude of the electric field
Using formula of electric field
Where, q = charge of proton
E = electric field
(B). The direction of the electric field is toward the direction of the force.
So, The direction of the electric field is toward south
Hence, The magnitude of the electric field is 129375 N/C toward south.