Discuss how personal choice can enhance your safety when exercising.

Physics

2 answers:

anzhelika [568]2 years ago

8 0

Answer and explanation;

In general, you can choose when, where, and how to exercise. Each of these choices can either positively or negatively affect your safety during physical activity.

To play it safe and reduce your risk of injury: Begin your exercise program slowly with low-intensity exercises. Wear appropriate shoes for your activity. Warm up before exercising, and cool down afterward. Pay attention to your surroundings when exercising outdoors.

Lady_Fox [76]2 years ago

5 0

<span>In general, you can choose when, where, and how to exercise. Each of these choices can either positively or negatively affect your safety during physical activity.</span>

You might be interested in

An object in a certain direction with an acceleration in the perpendicular direction

geniusboy [140]

Answer:

An object moving in certain direction with an acceleration in the perpendicular direction. The above condition is possible . Example of such situation in life would be when stone tied to a string whirling in a circular path

Hope this helps and pls mark as BRAINLIEST :)

3 0

3 years ago

Varg sees a spring that has a spring constant of 4 N/m that is stretched 5 m. He stretches the spring an additional 5 m. Conside

AysviL [449]

Answer:

Elastic potential energy, E = 200 J

Explanation:

It is given that,

Spring constant, K = 4 N/m

initial stretching in the spring, x = 5 m

Finally, it is stretched an additional 5 m i.e. x' = 5 m

Let E is the elastic energy in the spring after Varg stretches the spring. it is given by :

$E=\dfrac{1}{2}k(x+x')^2$

$E=\dfrac{1}{2}\times 4\times (10)^2$

E = 200 J

So, the elastic energy in the spring after Varg stretches the spring is 200 J. hence, this is the required solution.

8 0

3 years ago

Read 2 more answers

A force F=0.12N is aplied on spring and spring elongates by 3cm . specific constant of spring

PilotLPTM [1.2K]

The spring constant is 4 N/m

Explanation:

When a spring is stretched/compressed by the application of a force, the relationship between the magnitude of the force applied and the elongation of the spring is given by Hooke's law:

$F=kx$