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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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onsider two projectiles that can be shot upward by spring guns. Object A is made of solid aluminum and has a mass of 50 grams. O
BARSIC [14]

Answer:

B

Explanation:

From Newton's law of motion, we have:

V^2 = U^2 + 2gH

Where V and U are final and initial velocity respectively.

H is the height.

For the object to have a sustain a maximum height it means the final velocity of the object is zero.

By computing the height of the object sustain by A, we have:

0^2 = 2^2 -2×10×H

0= 4 -20H

4 = 20H;

H= 0.2m

For object B we have;

0^2 = 1^2 -2×10×H

0 = 1 -20H

H = 1/20= 0.05m

From computing the height sustain by both objects, we see object B is projected at a shorter height into atmosphere than A.

Hence object B will return to the ground first.

8 0
2 years ago
Which of the following are true statements? A. Like charges repel B. Unlike charges repel C. Unlike charges attract or D. Charge
Brums [2.3K]

Answer: The correct answers are (A) and (C).

Explanation:

The expression from electrostatic force is as follows;

F=\frac{kq_{1}q_{1}}{r^{2} }

Here, F is the electrostatic force, k is constant, r is the distance between the charges and q_{1},q_{1} are the charges.

The electrostatic force follows inverse square law. It is inversely proportional to the square of the distance between the charges. It is directly proportional to the product of the charges.

Like charges repel each other. There is a force of electrostatic repulsion between the like charges. Unlike charges attract each other. There is a force of electrostatic attraction between unlike charges.

The charges are induced on the neutral object when it is placed nearby the charged object without actually touching it.

Therefore, the true statements from the given options are as follows;

Like charges repel.

Unlike charges attract.

7 0
3 years ago
The picture is attached.
VLD [36.1K]

Answer:

C

Explanation:

5 0
3 years ago
Read 2 more answers
Select the correct answer.
11111nata11111 [884]

Answer:

Implicit memory

Explanation:

Implicit memory stores skill-related data by repeating an activity that always follows the same pattern. It includes all motor, sensory, and intellectual skills, as well as every form of conditioning. The capacity thus acquired does not depend on consciousness. We are able to perform sometimes complex tasks with our thinking turned to something completely different. This explains why Adam types so fast without even remembering the exact location of the letters on the keyboard.

8 0
3 years ago
A train started from rest and moved with constant acceleration. At one time it was traveling 27 m/s, and 150 m farther on it was
AlekseyPX

Explanation:

(a) Given:

Δx = 150 m

v₀ = 27 m/s

v = 54 m/s

Find: a

v² = v₀² + 2aΔx

(54 m/s)² = (27 m/s)² + 2a (150 m)

a = 7.29 m/s²

(b) Given:

Δx = 150 m

v₀ = 0 m/s

a = 7.29 m/s²

Find: t

Δx = v₀ t + ½ at²

150 m = (0 m/s) t + ½ (7.29 m/s²) t²

t = 6.42 s

(c) Given:

v₀ = 0 m/s

v = 27 m/s

a = 7.29 m/s²

Find: t

v = at + v₀

27 m/s = (7.29 m/s²) t + 0 m/s

t = 3.70 s

(d) Given:

v₀ = 0 m/s

v = 27 m/s

a = 7.29 m/s²

Find: Δx

v² = v₀² + 2aΔx

(27 m/s)² = (0 m/s)² + 2 (7.29 m/s²) Δx

Δx = 50 m

7 0
2 years ago
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