Answer:
When air rises in the atmosphere it gets cooler and is under less pressure. When air cools, it's not able to hold all of the water vapor it once was. Air also can't hold as much water when air pressure drops. The vapor becomes small water droplets or ice crystals and a cloud is formed.
Explanation:
hope this helps.
Answer:
Explanation:
E = σ/ε = (F/A) / (ΔL/L)
E = (mg/(πd²/4) / (ΔL/L)
E = (4mg/(πd²) / (ΔL/L)
E = 4Lmg/(πd²ΔL)
E = 4(30.0)(90)(9.8)/(π(0.01²)0.25)
E = 1.35 x 10⁹ Pa or 1.35 GPa
Answer:
V = V_0 - (lamda)/(2pi(epsilon_0))*ln(R/r)
Explanation:
Attached is the full solution
now you justhave to solve the last numbers because...Iam sooo Lazy right now
Answer:
Your question was incomplete so here is the complete question and answer.
Q. When exercising in the heat, which of the following hydration strategies is best for temperature regulation during an event (e.g., 10K race)
a) plain water
b) 5-7 percent glucose solution
c) Glucose polymer solution of 6-8 percent
d) There appears to be no difference among these different forms of hydration techniques relative to temperature regulation.
Ans. d) There appears to be no difference among these different forms of hydration techniques relative to temperature regulation.
Explanation:
Temperature Regulation is an important phenomenon for the person exposed to extreme hot conditions or weather. Exercising in hot conditions increase the body temperature. Greater and intense exercise, greater the production of heat. Then the heat dissipation takes place in the form of excessive sweating which results in dehydration. That was just the brief overview of temperature regulation. Above mentioned techniques are equally good hydration techniques so there is no difference at all. You can have a plain water or glucose solutions of above mentioned percentages.
