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laila [671]
2 years ago
7

How could you determine the injuries or ill person in a secondary assessment?

Physics
2 answers:
Natasha_Volkova [10]2 years ago
7 0

Signs look, listen, feel and smell for any signs of injury such as swelling, deformity, bleeding, discolouration or any unusual smells. When checking them you should always compare the injured side of the body with the uninjured side. Are they able to perform normal functions such as standing or moving their limbs

Secondary assessments are used in order to determine the injury, how the injury occurred, how severe the injury is, and to eliminate further injury.

Glad to help!

lana66690 [7]2 years ago
6 0

Answer:

just in internet tho

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An infinite line of charge with linear density λ1 = 8.2 μC/m is positioned along the axis of a thick insulating shell of inner r
bixtya [17]

1) Linear charge density of the shell:  -2.6\mu C/m

2)  x-component of the electric field at r = 8.7 cm: 1.16\cdot 10^6 N/C outward

3)  y-component of the electric field at r =8.7 cm: 0

4)  x-component of the electric field at r = 1.15 cm: 1.28\cdot 10^7 N/C outward

5) y-component of the electric field at r = 1.15 cm: 0

Explanation:

1)

The linear charge density of the cylindrical insulating shell can be found  by using

\lambda_2 = \rho A

where

\rho = -567\mu C/m^3 is charge volumetric density

A is the area of the cylindrical shell, which can be written as

A=\pi(b^2-a^2)

where

b=4.7 cm=0.047 m is the outer radius

a=2.7 cm=0.027 m is the inner radius

Therefore, we have :

\lambda_2=\rho \pi (b^2-a^2)=(-567)\pi(0.047^2-0.027^2)=-2.6\mu C/m

 

2)

Here we want to find the x-component of the electric field at a point at a distance of 8.7 cm from the central axis.

The electric field outside the shell is the superposition of the fields produced by the line of charge and the field produced by the shell:

E=E_1+E_2

where:

E_1=\frac{\lambda_1}{2\pi r \epsilon_0}

where

\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m is the linear charge density of the wire

r = 8.7 cm = 0.087 m is the distance from the axis

And this field points radially outward, since the charge is positive .

And

E_2=\frac{\lambda_2}{2\pi r \epsilon_0}

where

\lambda_2=-2.6\mu C/m = -2.6\cdot 10^{-6} C/m

And this field points radially inward, because the charge is negative.

Therefore, the net field is

E=\frac{\lambda_1}{2\pi \epsilon_0 r}+\frac{\lambda_2}{2\pi \epsilon_0r}=\frac{1}{2\pi \epsilon_0 r}(\lambda_1 - \lambda_2)=\frac{1}{2\pi (8.85\cdot 10^{-12})(0.087)}(8.2\cdot 10^{-6}-2.6\cdot 10^{-6})=1.16\cdot 10^6 N/C

in the outward direction.

3)

To find the net electric field along the y-direction, we have to sum the y-component of the electric field of the wire and of the shell.

However, we notice that since the wire is infinite, for the element of electric field dE_y produced by a certain amount of charge dq along the wire there exist always another piece of charge dq on the opposite side of the wire that produce an element of electric field -dE_y, equal and opposite to dE_y.

Therefore, this means that the net field produced by the wire along the y-direction is zero at any point.

We can apply the same argument to the cylindrical shell (which is also infinite), and therefore we find that also the field generated by the cylindrical shell has no component along the y-direction. Therefore,

E_y=0

4)

Here we want to find the x-component of the electric field at a point at

r = 1.15 cm

from the central axis.

We notice that in this case, the cylindrical shell does not contribute to the electric field at r = 1.15 cm, because the inner radius of the shell is at 2.7 cm from the axis.

Therefore, the electric field at r = 1.15 cm is only given by the electric field produced by the infinite wire:

E=\frac{\lambda_1}{2\pi \epsilon_0 r}

where:

\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m is the linear charge density of the wire

r = 1.15 cm = 0.0115 m is the distance from the axis

This field points radially outward, since the charge is positive . Therefore,

E=\frac{8.2\cdot 10^{-6}}{2\pi (8.85\cdot 10^{-12})(0.0115)}=1.28\cdot 10^7 N/C

5)

For this last part we can use the same argument used in part 4): since the wire is infinite, for the element of electric field dE_y produced by a certain amount of charge dq along the wire there exist always another piece of charge dq on the opposite side of the wire that produce an element of electric field -dE_y, equal and opposite to dE_y.

Therefore, the y-component of the electric field is zero.

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

4 0
3 years ago
A loudspeaker produces a sound wave of frequency 50hz.
viktelen [127]
Hey mate
Here is your answer
Option A)
Explanation:
The larger the amplitude of the waves, the louder the sound. Pitch (frequency) – shown by the spacing of the waves displayed. The closer together the waves are, the higher the pitch of the sound.
Pls mark as brainliest
3 0
2 years ago
A 0.10 newton spring toy with a spring constant of 160 newtons per meter is compressed 0.05 meter before it is launched. When re
forsale [732]

Answer:

(1) V = 0.2 J (2) 0.05J

Explanation:

Solution

Given that:

K = 160 N/m

x = 0.05 m

Now,

(1) we solve for the  initial potential energy stored

Thus,

V = 1/2 kx² = 0.5 * 160 * (0.05)²

Therefore V = 0.2 J

(2)Now, we solve for how much of the internal energy is produced as the toy springs up to its maximum height.

By using the energy conversion, we have the following

ΔV = mgh

=(0.1/9.8) * 9.8 * 1.5 = 0.15J

The internal energy = 0.2 -0.15

=0.05J

8 0
3 years ago
Two coils of wire are placed close together. Initially, a current of 3.04 A exists in one of the coils, but there is no current
Alchen [17]

Answer:

M=0.0247H

Explanation:

Given data

V_{voltage}=4.29V\\I_{current}=3.04A\\t_{time}=1.75*10^{-2}s

To find

Mutual inductance of the two-coil system

Solution

The mutual inductance given as:

M= (-VΔt)/ΔI

Substitute the given values

So

M=-\frac{4.29V*1.75*10^{-2}s}{(0-3.04A)}\\ M=0.0247H

4 0
3 years ago
in a longitudinal wave, the motion of the disturbance is in what direction relative to the wave motion?
VARVARA [1.3K]

Answer:

Longitudinal waves have the same direction of vibration as their direction of travel. This means that the movement of the medium is in the same direction as the motion of the wave.

4 0
2 years ago
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