<span>Last choice on the list:
Object A has a net charge of 0 because the positive and negative
charges are balanced.
Object B has a net charge of –2 because there is an imbalance of
charged particles (2 more negative electrons than positive protons).</span>
Answer:
Explanation:
Asúmase que la patinadora experimenta una aceleración constante. La fuerza neta experimentada por la patinadora:
![F_{net} = (50\,kg)\cdot \left[\frac{\left(15\,\frac{m}{s}\right)^{2}-\left(0\,\frac{m}{s}\right)^{2} }{2\cdot (3000\,m)} \right]](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20%2850%5C%2Ckg%29%5Ccdot%20%5Cleft%5B%5Cfrac%7B%5Cleft%2815%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%5Cright%29%5E%7B2%7D-%5Cleft%280%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%5Cright%29%5E%7B2%7D%20%7D%7B2%5Ccdot%20%283000%5C%2Cm%29%7D%20%5Cright%5D)
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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
Answer:
B. Its density is lower than that of water
Explanation:
density = mass / volume
density of the liquid = 85 / 100 = 0.85 g/cm^3
now,
density of water is 1 g/cm^3 which is greater than the density of the given liquid ( 0.85 g/cm^3 )
