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Arada [10]
3 years ago
13

When exercising in the heat, which of the following hydration strategies is best for temperature regulation during an event (e.g

., 10K race)
Physics
1 answer:
Olegator [25]3 years ago
3 0

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.

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A body moving with an acceleration 2 m/s?then what is the change in velocity in 4sec.​
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Answer:

As Per Provided Information

Moving body has 2m/s² acceleration

Time taken by body is 4 second

We are asked to find the 'change in velocity' ( ∆V) by the body.

<u>Formula Used here</u>

\boxed{\bf{\Delta \: V \:  =  acceleration \:  \times time \:}}

<u>Substituting </u><u>the </u><u>given </u><u>value</u>

<u>\sf\longrightarrow\Delta\:V \:  = 2 \times 4 \\  \\  \\ \sf\longrightarrow\Delta\:V \:  =8m {s}^{ - 1}</u>

<u>Therefore</u><u>,</u>

  • <u>Change </u><u>in </u><u>velocity </u><u>is </u><u>8</u><u> </u><u>m/</u><u>s</u>
7 0
2 years ago
A note of frequency 200Hz has a velocity of 400m/s. what is the wavelength of the note​
Xelga [282]

Answer:

\huge\boxed{\sf \lambda = 2 m}

Explanation:

<h3><u>Given data:</u></h3>

Frequency = f = 200 Hz

Velocity = v = 400 m/s

<h3><u>Required:</u></h3>

Wavelength = λ = ?

<h3><u>Formula:</u></h3>

v = fλ

<h3><u>Solution:</u></h3>

Put the givens in the formula

400 = (200)λ

Divide 200 to both sides

400/200 = λ

2 m = λ

λ = 2 m

\rule[225]{225}{2}

8 0
1 year ago
Suppose we repeat the experiment from the video, but this time we use a rocket three times as massive as the one in the video, a
shusha [124]

Answer:

2/3

Explanation:

In the case shown above, the result 2/3 is directly related to the fact that the speed of the rocket is proportional to the ratio between the mass of the fluid and the mass of the rocket.

In the case shown in the question above, the momentum will happen due to the influence of the fluid that is in the rocket, which is proportional to the mass and speed of the same rocket. If we consider the constant speed, this will result in an increase in the momentum of the fluid. Based on this and considering that rocket and fluid has momentum in opposite directions we can make the following calculation:

Rocket speed = rocket momentum / rocket mass.

As we saw in the question above, the mass of the rocket is three times greater than that of the rocket in the video. For this reason, we can conclude that the calculation should be done with the rocket in its initial state and another calculation with its final state:

Initial state: Speed ​​= rocket momentum / rocket mass.

Final state: Speed ​​= 2 rocket momentum / 3 rocket mass. -------------> 2/3

8 0
3 years ago
PLEASE HELP ASAP!!!!!!!!!!
Marianna [84]
Hey,


I think the answer's are 1,3


Hope this helpss

~Girlygir101~
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