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3 years ago

How do intensity, frequency, and time affect physical fitness?

1 answer:

5 0

Important bc it makes it more effective, the specific rate makes or breaks the fitness. Frequency is important to allow your body to rebuild and repair the damage from working out, it allows the body to adapt and time for rest/ healing. Intensity depends on how much your body breaks so the recover time and frequency must be adjusted. Time effects because of the distance between frequencies which plays a role.

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In general, what do you think are the benefits if you already achieve your body goals.

balandron [24]

Answer:

exercise dailyyy

Help you control your weight. ...

Reduce your risk of heart diseases. ...

Help your body manage blood sugar and insulin levels. ...

Help you quit smoking. ...

Improve your mental health and mood. ...

Help keep your thinking, learning, and judgment skills sharp as you age.

Explanation:

7 0

3 years ago

Question 15)

RUDIKE [14]

Answer:

k Nishant

Explanation:

i think option A

8 0

2 years ago

The concentration of sodium and potassium ions in the blood and body fluid is regulated by :

Liula [17]

Answer:

Kidney

Explanation:

One of the main function of the kidney is to maintain the homeostasis of sodium and potassium ions in the blood and body.

Aldosterone is a key steroid hormone that balances sodium and potassium ions in the blood and body fluid. Potassium and sodium ions generate electric impulse in the body which helps to perform different activities such as muscles flexing.

Kidney function for reabsorption and secretion, in which reabsorption of Na is done nd balances the sodium and potassium in the blood and body.

6 0

3 years ago

As air pressure decreases, what happens

Naya [18.7K]

Answer:

I believe it is B, not 100% sure though

Explanation:

5 0

2 years ago

The Heaviside function H is defined by H(t)={0 if t<0, 1 if t≥0 It is used in the study of electric circuits to represent the

Studentka2010 [4]

Answer:

$V(t)= 240V* H(t-5)$

Explanation:

The heaviside function is defined as:

$H(t) =1 \quad t\geq 0\\H(t) =0 \quad t$

so we see that the Heaviside function "switches on" when $t=0$ , and remains switched on when $t>0$

If we want our heaviside function to switch on when $t=5$ , we need the argument to the heaviside function to be 0 when $t=5$

Thus we define a function f:

$f(t) = H(t-5)$

The $-5$ term inside the heaviside function makes sure to displace the function 5 units to the right.

Now we just need to add a scale up factor of 240 V, because thats the voltage applied after the heaviside function switches on. ( $H(t-5) =1$ when $t\geq 5$ , so it becomes just a 1, which we can safely ignore.)

Therefore our final result is:

$V(t)= 240V* H(t-5)$

I have made a sketch for you, and added it as attachment.

5 0

3 years ago