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.
Answer:
μ=0.151
Explanation:
Given that
m= 3.5 Kg
d= 0.96 m
F= 22 N
v= 1.36 m/s
Lets take coefficient of kinetic friction = μ
Friction force Fr=μ m g
Lets take acceleration of block is a m/s²
F- Fr = m a
22 - μ x 3.5 x 10 = 3.5 a ( take g =10 m/s²)
a= 6.28 - 35μ m/s²
The final speed of the block is v
v= 1.36 m/s
We know that
v²= u²+ 2 a d
u= 0 m/s given that
1.36² = 2 x a x 0.96
a= 0.963 m/s²
a= 6.28 - 35μ m/s²
6.28 - 35μ = 0.963
μ=0.151
<span>An object is located 51 millimeters from a diverging lens the object has a height og 13 millimeters and the image height is 3.5 millimeters how far in front of the lens is image located?</span>
Answer:
Vprom = 0.00347[km/min]
Explanation:
We can calculate each of the average speeds and then perform the overall average between the two speeds.
V1 = 6/54
V1 = 0.111[km/min]
V2 = 1/16
V2 = 0.0625[km/min]
![V_{prom} = \frac{V_{1} + V_{2}}{2} \\V_{prom} = \frac{0.1111 + 0.0625}{2}\\V_{prom} = 0.00347 [km/min]](https://tex.z-dn.net/?f=V_%7Bprom%7D%20%3D%20%5Cfrac%7BV_%7B1%7D%20%2B%20V_%7B2%7D%7D%7B2%7D%20%20%5C%5CV_%7Bprom%7D%20%3D%20%5Cfrac%7B0.1111%20%2B%200.0625%7D%7B2%7D%5C%5CV_%7Bprom%7D%20%3D%200.00347%20%5Bkm%2Fmin%5D)
