Physical fitness is related to our ability to carry out daily tasks without being too tired or sore. This statement is TRUE.
<h3>Further explanation
</h3>
Good physical fitness means that you can perform daily activities, occupations, and sports. This condition will be achieved through good nutrition, regular exercise, and have enough rest.
The benefit of physical activity and exercise can be immediate as well as long-term. Most importantly, regular activity can improve your quality of life.There are four types of physical fitness:
- Cardiovascular exercise: exercises that increase the work of the heart and lungs, such as walking, jogging, step aerobics, swimming, and biking. Cardio activity improves your heart/lung function and muscle mass.
- Muscular strength exercise: this exercise is to build overall strength and muscle mass.
- Joint flexibility exercise: this exercise to help the ability of the muscle group can be stretched or joint can be moved, for example: bending, lifting and driving. Stretching can reduce the risk of injury.
- Muscular endurance exercise: this exercise to find out how many repetitions of exercises a person can perform, such as push-ups and sit-ups.
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Keywords: physical fitness, the definition of physical fitness, physical activity
Answer:
The number of protons can be found by looking at the atomic number
Explanation:
Its at the very top of the little element box
The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.
The given parameters;
- <em>initial temperature of metals, = </em>
<em /> - <em>initial temperature of water, = </em>
<em> </em> - <em>specific heat capacity of copper, </em>
<em> = 0.385 J/g.K</em> - <em>specific heat capacity of aluminum, </em>
= 0.9 J/g.K - <em>both metals have equal mass = m</em>
The quantity of heat transferred by each metal is calculated as follows;
Q = mcΔt
<em>For</em><em> copper metal</em><em>, the quantity of heat transferred is calculated as</em>;

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>copper metal</em>;

<em>For </em><em>aluminum metal</em><em>, the quantity of heat transferred is calculated as</em>;

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>aluminum metal </em><em>;</em>

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.
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