The surface area of a cube is:
S = 6l², making l the subject:
l = √(S/6)
While the volume is:
V = l³
If we substitute l to get:
V = (√(S/6))³
√(S/6) = ³√V
S/6 = (³√V)²
S = 6(³√V)² ⇒ 6V^(2/3)
Usually in the deep sea and underwater caves where there is no light
Answer:
where are the statements?
Explanation:
The amount of heat needed to raise the temperature of a substance by

is given by

where
m is the mass of the substance
Cs is its specific heat capacity

is the increase in temperature
For oxygen, the specific heat capacity is approximately

The variation of temperature for the sample in our problem is

while the mass is m=150 g, so the amount of heat needed is
I'm going to assume this is over a horizontal distance. You know from Newton's Laws that F=ma --> a = F/m. You also know from your equations of linear motion that v^2=v0^2+2ad. Combining these two equations gives you v^2=v0^2+2(F/m)d. We can plug in the given values to get v^2=0^2+2(20/3)0.25. Solving for v we get v=1.82 m/s!