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
<h2>
Speed with which it return to its initial level is 100 m/s</h2>
Explanation:
We have equation of motion v² = u² + 2as
Initial velocity, u = 100 m/s
Acceleration, a = -9.81 m/s²
Final velocity, v = ?
Displacement, s = 0 m
Substituting
v² = u² + 2as
v² = 100² + 2 x -9.81 x 0
v² = 100²
v = ±100 m/s
+100 m/s is initial velocity and -100 m/s is final velocity.
Speed with which it return to its initial level is 100 m/s
It’s not Cereal and Milk, carbonated drinks, and smoky air
The answer would be C. (:
