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3 years ago

PLEASE HELP ME EXPLAIN THESE QUESTIONS IT'S FOR ANATOMY AND PHYS

Physics

1 answer:

NeX [460]3 years ago

4 0

Answer:

I tried

Explanation:

You have to check a 12 year olds respiration rate by Siting them down and trying to relax. It's best to take the respiratory rate while sitting up in a chair or in bed. Measure their breathing rate by counting the number of times their chest or abdomen rises over the course of one minute. Then Record this number. Now you have to answer the first few questions based on that.

Heart rate, blood pressure, respiratory rate and temperature are the big four vital signs.

8. Secondary assessments are used in order to determine the injury, how the injury occurred, how severe the injury is, and to eliminate further injury and that is why it is important.

9. It should only be performed when a person shows no signs of life or when they are unconscious, unresponsive, not breathing or not breathing normally.

In order to perform CPR, you need to check the scene and the person. Make sure the scene is safe, then tap the person on the shoulder and shout "Are you OK?" to ensure that the person needs help. Then pen the airway, Check for breathing, Push hard, push fast, deliver rescue breaths, continue CPR steps.

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3. If an undetected air bubble was trapped inside the gas collection tube, how would this affect your calculated percent yield?

soldi70 [24.7K]

When an air bubble gets trapped inside the gas collection tube, then the gas that is actually collected becomes impure. This will definitely give an inaccurate result and the calculated percentage yield will be wrong. I hope that this is the answer that you were looking for and it has come to your help.

5 0

3 years ago

A Galilean telescope adjusted for a relaxed eye is 36.2 cm long. If the objective lens has a focal length of 39.5 cm , what is t

GuDViN [60]

Answer:

The magnification is $m = 12$

Explanation:

From the question we are told that

The object distance is $u = 36.2 \ cm$

The focal length is $v = 39.5 \ cm$

From the lens equation we have that

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

=> $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$

substituting values

$\frac{1}{v} = \frac{1}{39.5} - \frac{1}{36.2}$

$\frac{1}{v} = -0.0023$

=> $v = \frac{1}{0.0023}$

=> $v =-433.3 \ cm$

The magnification is mathematically represented as

$m =- \frac{v}{u}$

substituting values

$m =- \frac{-433.3}{36.2}$

$m = 12$

6 0

3 years ago

Scientific laws and theories did not begin as questions and hypothesis? True or false

Nesterboy [21]

False. That's exactly how scientific theories begin

8 0

3 years ago

Air at 20ºC with a convection heat transfer coefficient of 20 W/m2·K blows over a pond. The surface temperature of the pond is 2

dimulka [17.4K]

Answer:

The heat flux between the surface of the pond and the surrounding air is<em> 60 W/</em> $m^{2}$ <em> </em>

Explanation:

Heat flux is the rate at which heat energy moves across a surface, it is the heat transferred per unit area of the surface. This can be calculated using the expression in equation 1;

q = Q/A ...............................1

since we are working with the convectional heat transfer coefficient equation 1 become;

q = h ( $T_{sf} -T_{sd}$ ) ........................2

where q is the heat flux;

Q is the heat energy that will be transferred;

h is the convectional heat coefficient = 20 W/ $m^{2}$ .K;

$T_{sf}$ is the surface temperature = $23^{o}$ C 23°C + 273.15 = 296.15 K;

$T_{sd}$ is the surrounding temperature = $20^{o}$ C = 20°C + 273.15 = 293.15 K;

The values are substituted into equation 2;

q = 20 W/ $m^{2}$ .K ( 296.15 K - 293.15 K)

q = 20 W/ $m^{2}$ .K ( 3 K)

q = 60 W/ $m^{2}$

Therefore the heat flux between the surface of the pond and the surrounding air is 60 W/ $m^{2}$

3 0

4 years ago

How can a better understanding of friction help predict geological events, like earthquakes?

ANTONII [103]

Answer:

Friction plays a key role in how ruptures unzip faults in the earth's crust," says Vito Rubino, research scientist at Caltech's Division of Engineering and Applied Science (EAS). "Assumptions about dynamic friction affect a wide range of earthquake science predictions, including how fast ruptures will occur, the nature of ground shaking, and residual stress levels on faults. Yet the precise nature of dynamic friction remains one of the biggest unknowns in earthquake science."

hope it helps

6 0

3 years ago