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trasher [3.6K]
3 years ago
6

What are 5 tactics that can reduce the likelihood of injury?

Engineering
1 answer:
Nat2105 [25]3 years ago
7 0
1Talk to your doctor

Don't start any exercise program without first checking with your primary care provider. Your doctor can determine whether you're healthy enough to exercise, and what, if any, modifications you'll need to make to your program. "Exercise programs should be customized to the individual whenever possible to account for any limitations and ongoing medical conditions," Dr. Berkson advises.

2 Choose your workout carefully

High-impact exercise programs aren't ideal for women with conditions like arthritis or osteoporosis. Non-impact exercises, including swimming or using an elliptical exercise machine, will give you aerobic conditioning without stressing your joints.

3 Learn the proper technique

Don't start any new exercise without first learning the correct form. On the right is an example of proper squat technique. To learn the right form, work with a trainer at home or in the gym, or consult a physical therapist to help you tailor a workout to your health conditions and physical capabilities.

4 Get the right gear

Buy a pair of sturdy, comfortable sneakers that provide good arch support and have a cushioned heel to absorb shock. Wear loose, comfortable clothing that gives you room to move and breathe.

5 Start gradually

Don't jump into a new exercise program. "The greatest risk of injury comes with changing an exercise program or adding a new exercise," Dr. Berkson says. Start slowly. If you're cycling, for example, set the bike's controls on the lowest speed and tension, and pedal for just a few minutes your first few times. Gradually increase the speed and intensity only when you feel ready.
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An open-open organ pipe is 73.5 cm long. An open-closed pipe has a fundamental frequency equal to the third harmonic of the open
Anon25 [30]

Answer:

The length of the  open closed pipe is 12.25 cm

Explanation:

In open - open organ pipe, third harmonic has Antinode to Node, Node to Node, Node to Node and Node to Antinode.

The length of the open-open organ pipe is equal to the sum of wavelength in Antinode to Node, Node to Node, Node to Node and Node to Antinode.

L = A→N + N→N + N→N + N→A

L =\frac{\lambda }{4} + \frac{\lambda }{2} + \frac{\lambda }{2} + \frac{\lambda }{4} \\\\L = \frac{ 3\lambda }{2}\\\\\lambda = \frac{2L}{3} \\\\F = \frac{V}{\lambda} = \frac{3V}{2L}

In open-closed pipe, Fundamental frequency has Antinode to Node.

Thus, length of the open-closed organ pipe is equal to the wavelength in Antinode to Node.

L = A→N

L_o = \frac{\lambda}{4} \\\\\lambda =4L_o\\\\F_o = \frac{V}{4L_o}

From the information in the question, fundamental frequency of open-closed pipe is to the third harmonic of the open-open pipe.

F₀ = F

\frac{V}{4L_o} =\frac{3V}{2L} \\\\\frac{1}{4L_o} = \frac{3}{2*73.5 \ cm}\\\\\frac{1}{L_o} = \frac{3*4}{2*73.5 \ cm}\\\\\frac{1}{L_o}  = \frac{12}{147 \ cm}\\\\L_o =  \frac{147 \ cm}{12} = 12.25 \ cm

Therefore, the length of the  open closed pipe is 12.25 cm

4 0
3 years ago
Read 2 more answers
A closed rigid tank contains water initially at 10,000 kPa and 520ºC and is cooled to a final temperature of 270° C. Determine t
jasenka [17]

Answer:

final pressure is 6847.41 kPa

Explanation:

given data:

P_1 =  10,000 kPa

T_1 =520\ degree\ celcius = 793 K

T_2  = 270 degree celcius = 543 K

as we can see all temperature are more than 100 degree, it mean this condition is refered to superheated stream

for ischoric process we know that

\frac{P_1}{T_1} =\frac{P_2}{T_2}

\frac{10*10^6}{793} = \frac{P_2}{543}

P_2 = 6.84741*10^6 Pa

final pressure is 6847.41 kPa

5 0
3 years ago
A 20 mm diameter rod made of ductile material with a yield strength of 350 MN/m2 is subjected to a torque of 100 N.m, and a bend
vodka [1.7K]

Answer:

a) 42.422 KN

b) 44.356 KN

Explanation:

Given data :

Diameter = 20 mm

yield strength = 350 MN/m^2

Torque ( T )  = 100 N.m

Bending moment = 150 N.m

<u>Determine the value of the applied axial tensile force when yielding of rod occurs </u>

first we will calculate the shear stress and normal stress

shear stress ( г ) = Tr / J = [( 100 * 10^3)  * 10 ]  /  \pi /32 * ( 20)^4  

                                       = 63.662 MPa

Normal stress(  Гb + Гa )  = MY/ I  +  P/A

= [( 150 * 10^3)  * 10 ]  /  \pi /32 * ( 20)^4   + 4P / \pi  * 20^2

= 190.9859 + 4P / \pi  * 20^2  MPa

<u>a) Using MSS theory </u>

value of axial force = 42.422 KN

solution attached below

<u>b) Using MDE  theory </u>

value of axial force = 44.356 KN

solution attached below

5 0
3 years ago
In an experiment, one selected two samples of copper-silver alloy. One sample has 40 wt% of silver and 60wt% of copper and the o
mote1985 [20]
I belive it’s 3 sorry if it’s wrong
6 0
3 years ago
Is a water proof material used around tubs and
Leviafan [203]
I think it’s hard board
4 0
3 years ago
Read 2 more answers
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