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.
Answer:
Transition has to cross between solid and liquid in gray zone.No indoor organized public events and social gatherings are allowed, except with members of the same household.
Given:
Inductance, L = 150 mH
Capacitance, C = 5.00 mF
= 240 V
frequency, f = 50Hz
= 100 mA
Solution:
To calculate the parameters of the given circuit series RLC circuit:
angular frequency,
= 
a). Inductive reactance,
is given by:

b). The capacitive reactance,
is given by:

c). Impedance, Z = 

d). Resistance, R is given by:



e). Phase angle between current and the generator voltage is given by:




The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C
R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

Substitute numerical values:

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.
As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).
Learn more about Gaussian sphere here:
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