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Brums [2.3K]
3 years ago
13

What differentiates an athletic trainer from a clinical exercise physiologist?

Engineering
2 answers:
Leokris [45]3 years ago
5 0

<u>An athletic trainer must have clinical experience.</u>

kozerog [31]3 years ago
3 0

Answer:

It's number 3

Explanation:

You might be interested in
A circular ceramic plate that can be modeled as a blackbody is being heated by an electrical heater. The plate is 30 cm in diame
MakcuM [25]

Answer:

Heater power = 425 watts

Explanation:

Detailed explanation and calculation is shown in the image below

6 0
3 years ago
5. A typical paper clip weighs 0.59 g and consists of BCC iron. Calculate (a) the number of
marta [7]

Answer:

(a) 3.185*10^{21} cells

(b) 6.37*10^{21} atoms

Explanation:

(a)

Volume, V of unit cell

V=(2.866*10^{-8})^{3}=2.354*10^{-23}

Number of unit cells, N

N=\frac {W_{mat}}{V\rho_{mat}} Where W_{mat} is weight of material and \rho_{mat} is density of material

N=\frac{0.59}{7.87*(2.354*10^{-23}}=3.185*10^{21} cells

(b)

Number of atoms in paper clip

This is a product of number of unit cells and number of atoms per cell

Since iron has 2 atoms per cell

Number of atoms of iron=3.185*10^{21} cells*2 atoms/cell=6.37*10^{21} atoms

8 0
3 years ago
What are the BENEFITS and RISKS of using automobiles?
Molodets [167]

Answer:

u could get hurt or it could not sence in but it easy to work with and u can just relax till u get were ur going.

Explanation:

3 0
3 years ago
For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t
saveliy_v [14]

Complete Question

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.

Answer:

The elongation is =21.29mm

Explanation:

In order to gain a good understanding of this solution let define some terms

True Stress

       A true stress can be defined as the quotient obtained when instantaneous applied load is divided by instantaneous cross-sectional area of a material it can be denoted as \sigma_T.

True Strain

     A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as \epsilon_T.

The mathematical relation between stress to strain on the plastic region of deformation is

              \sigma _T =K\epsilon^n_T

Where K is a constant

          n is known as the strain hardening exponent

           This constant K can be obtained as follows

                        K = \frac{\sigma_T}{(\epsilon_T)^n}

No substituting  345MPa \ for  \ \sigma_T, \ 0.02 \ for \ \epsilon_T , \ and  \ 0.22 \ for  \ n from the question we have

                     K = \frac{345}{(0.02)^{0.22}}

                          = 815.82MPa

Making \epsilon_T the subject from the equation above

              \epsilon_T = (\frac{\sigma_T}{K} )^{\frac{1}{n} }

Substituting \ 411MPa \ for \ \sigma_T \ 815.82MPa \ for \ K  \ and  \  0.22 \ for \ n

       \epsilon_T = (\frac{411MPa}{815.82MPa} )^{\frac{1}{0.22} }

            =0.0443

       

From the definition we mentioned instantaneous length and this can be  obtained mathematically as follows

           l_i = l_o e^{\epsilon_T}

Where

       l_i is the instantaneous length

      l_o is the original length

Substituting  \ 470mm \ for \ l_o \ and \ 0.0443 \ for  \ \epsilon_T

             l_i = 470 * e^{0.0443}

                =491.28mm

We can also obtain the elongated length mathematically as follows

            Elongated \ Length =l_i - l_o

Substituting \ 470mm \ for l_o and \ 491.28 \ for \ l_i

          Elongated \ Length = 491.28 - 470

                                       =21.29mm

4 0
4 years ago
Why is the back-work ratio much higher in the brayton cycle than in the rankine cycle?
zloy xaker [14]

The back-work ratio much higher in the Brayton cycle than in the Rankine cycle because a gas cycle is the Brayton cycle, while a steam cycle is the Rankine cycle. Particularly, the creation of water droplets will be a constraint on the steam turbine's efficiency. Since gas has a bigger specific volume than steam, the compressor will have to work harder while using gas.

<h3>What are modern Brayton engines?</h3>

Even originally Brayton exclusively produced piston engines, modern Brayton engines are virtually invariably of the turbine variety. Brayton engines are also gas turbines.

<h3>What is the ranking cycle?</h3>

A gas cycle is the Brayton cycle, while the Ranking cycle is a steam cycle. The production of water droplets will especially decrease the steam turbine's performance. Gas-powered compressors will have to do more work since gas's specific volume is greater than steam's.

Th

To know more about Rankine cycle, visit: brainly.com/question/13040242

#SPJ4

4 0
1 year ago
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