Explanation:
Clear rendering reads;
"Make a table, like the one in the example, with the results obtained in the tests you carried out in previous activities. Then answer: What do these results indicate regarding your physical condition and how do they relate to your health? From your answers, consider the challenge of improving your physical condition by maintaining or improving your exercise routine or permanent practice of physical activity".
So the incomplete text above it seems you've been instructed to perform an experiment and then state your result/analysis.
Complete Question
The complete question is shown on the uploaded image
Answer:
The tension on the shank is 
Explanation:
From the question we are told that
The strain on the strain on the head is 
The contact area is
Looking at the first diagram
At 600 MPa of stress
The strain is 
At 450 MPa of stress
The strain is 
To find the stress at
we use the interpolation method

Substituting values



Generally the force on each head is mathematically represented as

Substituting values


Now the tension on the bolt shank is as a result of the force on the 6 head which is mathematically evaluated as



Answer:
Velocity of the electron at the centre of the ring, 
Explanation:
<u>Given:</u>
- Linear charge density of the ring=

- Radius of the ring R=0.2 m
- Distance of point from the centre of the ring=x=0.2 m
Total charge of the ring

Potential due the ring at a distance x from the centre of the rings is given by

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

Let
be the change in potential Energy given by

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

So the electron will be moving with 
Hi! Decomposers (mainly soil bacterium, fungus, or invertebrate)<span /> are categorized as consumers due to the fact that they consume dead organic matter such as plants and animals. They differ from producers (green plants and some bacteria) because they do not produce their own food using photosynthesis or chemosythensis.
Hope this helped!
Answer:
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Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.