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3 years ago

Sleep is a component of which part of the health triangle

Physics

1 answer:

katrin [286]3 years ago

5 0

Answer:

physical health

Explanation:

Physical health has many components including: exercise, nutrition, sleep, alcohol & drugs, and weight management.

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Help can’t find the answer no where

katen-ka-za [31]

i'm stuck on that question also

5 0

3 years ago

. Determine if approximate cylindrical symmetry holds for the following situations. State why or why not. (a) A 300-cm long copp

MA_775_DIABLO [31]

Answer:

a) Yes

b) No

Explanation:

In the first case, part a, yes we can say for certainty that cylinderical symmetry holds. Why so? You may ask. This is because from the question, we are told that the length of the rod is 300 cm. And this said length is longer than the distance to the point from the center of the rod, which is 5 cm.

In the second half of the question, I beg to disagree that cylindrical symmetry holds. Again, you may ask why, this is because the length of the rod in this case, is having the same order of magnitude as the distance to the center of the rod. Thus, it is not symmetrical.

6 0

3 years ago

Which of the following best represents a chemical reaction?

Sonja [21]

Answer:

The answer to your question should be D.

Explanation:

reactants are on the laft side of arrow and products are on right side of arrow

7 0

3 years ago

In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates i

RSB [31]

a) The length of the arm of the centrifuge is 10.9 m

b) The angular acceleration is $2.7 rad/s^2$

Explanation:

In a uniform circular motion, the centripetal acceleration is given by

$a_c=\omega^2 r$

where:

$\omega$ is the angular speed of the circular motion

r is the radius of the circle

For the centrifuge in this problem, we have:

$\omega=1.7 rad/s$ is the angular speed

The centripetal acceleration is 3.2 times the acceleration due to gravity ( $g=9.8 m/s^2$ ), so:

$a_c=3.2 g = 3.2(9.8)=31.4 m/s^2$

Therefore, we can re-arrange the previous equation to find r, the radius of the circle (which corresponds to the length of the arm of the centrifuge):

$r=\frac{a_c}{\omega^2}=\frac{31.4}{1.7^2}=10.9 m$

In the second part of the exercise, the centrifuge speeds up from an initial angular speed of 0 to a final angular speed of 1.7 rad/s. The total acceleration experienced at the final moment is

$a=4.4 g$