Assuming that the angle is the same for both ropes, then D. is the answer. You have to consider also if the ropes are close together or far apart and if the force to move the object is in line with the ropes or perpendicular to them.
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Answer:
Option A is correct.
(The faster object encounters more resistance)
Explanation:
Option A is correct. (The faster object encounters more resistance)
Air resistance depends on various factors:
- Speed of the object
- Cross-sectional area of the object
- Shape of the object
Formula:
As the speed of the object increases the amount of Air resistance/drag increases on the object, as the above formula shows direct relation between Air resistance/drag and velocity i.e F ∝ v^2.
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
