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4vir4ik [10]
1 year ago
5

Prompt: Explain how your physical, emotional, and mental state affects your performance in your exercise. Use specific examples

from your experience
Physics
1 answer:
Lady_Fox [76]1 year ago
3 0

Being physically, emotionally, and mentally healthy can keep you healthy and help you avoid significant health issues during exercises.

<h3>How to illustrate the information?</h3>

Exercise boosts mental health by lowering anxiety, depression, and depressive symptoms, as well as by raising self-esteem and increasing cognitive abilities. Additionally, it has been discovered that exercise helps with symptoms including poor self-esteem and social withdrawal.

Exercise on a regular basis can significantly improve symptoms of sadness, anxiety, and ADHD. Additionally, it lowers stress, enhances memory, promotes sound sleep, and uplifts your mood in general. It should be noted that it's important to be in the right state of mind during exercise to prevent injuries.

Therefore, being physically, emotionally, and mentally healthy can keep you healthy and help you avoid significant health issues during exercises.

Learn more about exercises on:

brainly.com/question/1365564

#SPJ1

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4 0
3 years ago
Air at 30°C and 2MPa flows at steady state in a horizontal pipeline with a velocity of 25 m/s. It passes through a throttle valv
erastovalidia [21]

Answer:

V_2=159.9\ m/s

T_2=290.6K

Explanation:

At initial condition

P=2 MPa

T=30°C

V=25 m/s

At final condition

P=0.3 MPa

Now from first law for open system

h_1+\dfrac{V_1^2}{2000}=h_2+\dfrac{V_2^2}{2000}

We know that for air

h= 1.010 x T  KJ/kg

1.01\times 303+\dfrac{25^2}{2000}=1.01\times T+\dfrac{V_2^2}{2000}                               -----1

Now from mass balance

\rho_1A_1V_1=\rho_2A_2V_2

We also know that

\rho=\dfrac{P}{RT}

\dfrac{P_1}{RT_1}V_1=\dfrac{P_2}{RT_2}V_2

\dfrac{2}{R\times 303}\times 25=\dfrac{0.3}{RT_2}V_2

T_2=1.81V_2                  ----------2                                                                                                

Now from equation 1 and 2

V_2^2+3673.749V-612916.49=0

So we can say that

V_2=159.9\ m/s

This is the outlet velocity.

Now by putting the values in equation 2

T_2=1.81\times 159.9  

T_2=290.6K

This is the outlet temperature.

4 0
4 years ago
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