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2 years ago

What’s the importance of fluid balance while exercising

Physics

1 answer:

attashe74 [19]2 years ago

3 0

Answer:

The maintenance of fluid balance during exercise. Sodium also promotes retention of ingested fluids and leads to an increased plasma volume response during rehydration. The primary goal of supplementation should be considered, fluid vs carbohydrate provision, and the beverage composition altered accordingly.

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What is electron capture

USPshnik [31]

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3 years ago

Peak wavelength of light coming from a star with a temp of 4300k

PSYCHO15rus [73]

from Wein's law

$\frac{k}{ \lambda_{max} } = T \\ \lambda_{max} = \frac{k}{T} \\ \lambda_{max} = \frac{3.898 \times {10}^{ - 3} }{4300} \\ = 9.065 \times {10}^{ - 7}$

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2 years ago

If it requires 2.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be re

valina [46]

Answer:

16 J

Explanation:

It is given that,

Work done, W = 2 J

A spring is stretched by 2.0 cm from its equilibrium length

We need to find how much more work will be required to stretch it an additional 4.0 cm.

Let k is the spring constant of the spring. When W = 2J, and x = 2 cm, then energy required to stretch the spring is :

$U=\dfrac{1}{2}kx^2\\\\k=\dfrac{2U}{x^2}\\\\k=\dfrac{2(2)}{(0.02)^2}\\\\k=10000\ N/m$

The energy required to stretch the spring from 2 cm to additional 4 cm i.e. 2+4= 6 cm.

$W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\=\dfrac{1}{2}\times 10000\times ((0.06)^2-(0.02)^2)\\\\W=16\ J$

So, the required work done is 16 J.

7 0

2 years ago

A sphere of mass of 1.55 kg is accelerated upwards by a string to which the sphere is attached. Its speed increases from 2.81 m/

Mama L [17]

Answer:

The tension in the string is 16.24 N

Explanation:

Given;

mass of the sphere, m = 1.55 kg

initial velocity of the sphere, u = 2.81 m/s

final velocity of the sphere, v = 4.60 m/s

duration of change in the velocity, Δt = 2.64 s

The tension of the string is calculated as follows;

$T = ma + mg\\\\T = m(a + g)\\\\where;\\\\a \ is \ upward \ acceleration \ of \ the \ sphere\\\\g \ is \ acceleration \ due \ to \ gravity =9.8 \ m/s^2\\\\a = \frac{\Delta V}{\Delta t} = \frac{v- u}{ t} = \frac{4.6 - 2.81 }{2.64} = 0.678 \ m/s^2$

T = 1.55(0.678 + 9.8)

T = 1.55(10.478)

T = 16.24 N

Therefore, the tension in the string is 16.24 N

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2 years ago

6. What do the key results indicate?

slega [8]

the key result indicate that how you will achieve the your goal

7 0

3 years ago

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