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2 years ago

The emt must assume that any unwitnessed water-related incident is accompanied by:________

Physics

1 answer:

natta225 [31]2 years ago

3 0

The EMT must assume that any unwitnessed water-related incident is accompanied by potential spinal damage.

<h3>What is spinal damage?</h3>

Nerves or the spinal cord in any way damaged at the end of the spinal canal.

A rapid strike or cut to the spine can cause a traumatic spinal cord damage.

Below the damage site, a spinal cord injury frequently results in a lifelong loss of strength, feeling, and function.

A lot of people with spinal cord injuries may lead productive, independent lives with the help of rehabilitation and assistive technology.
Symptom-reducing medications and spinal stabilisation surgery are used as treatments.

Herniated discs are among the common injuries and diseases of the spine. Stenosis of the lower back and Scoliosis are others.

After taking part in a rehabilitation programme, over 80% of people with incomplete spinal cord injury (SCI) can walk again.

Learn more about spinal cord here:

brainly.com/question/23916836

#SPJ4

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You attach a meter stick to an oak tree, such that the top of the meter stick is 2.27 meters above the ground. later, an acorn f

Alexandra [31]

The acorn was at a height of <u>4.15 m</u> from the ground before it drops.

The acorn takes a time t to fall through a distance h₁, which is the length of the scale. When the acorn reaches the top of the scale, its velocity is u.

Calculate the speed of the acorn at the top of the scale, using the equation of motion,

$s=ut+ \frac{1}{2} at^2$

Since the acorn falls freely under gravity, its acceleration is equal to the acceleration due to gravity g.

Substitute 2.27 m for s (=h₁), 0.301 s for t and 9.8 m/s² for a (=g).

$s=ut+ \frac{1}{2} at^2\\ (2.27 m)=u(0.301s)+\frac{1}{2}(9.8m/s^2)(0.301s)^2\\ u=\frac{1.8261m}{0.301s} =6.067m/s$

If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.

Use the equation of motion,

$v^2=u^2+2as$

Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.

$v^2=u^2+2as\\ (6.067m/s)^2=(0m/s)^2+2(9.8m/s^2)h_2\\ h_2=\frac{(6.067m/s)^2}{2(9.8m/s^2)} =1.878 m$

The height h above the ground at which the acorn was is given by,

$h=h_1+h_2=(2.27 m)+(1.878 m)=4.148 m$

The acorn was at a height <u>4.15m</u> from the ground before dropping down.

3 0

3 years ago

Which of the following is the MOST common cause of death in middle adulthood?

Anestetic [448]

Suicide. I hope this helps!!

6 0

3 years ago

Read 2 more answers

Sylvia and Jadon now want to work a problem. Imagine a puck of mass 0.5 kg moving in a circular simulation. Suppose that the ten

leva [86]

Answer:

Tangential speed, v = 2.64 m/s

Explanation:

Given that,

Mass of the puck, m = 0.5 kg

Tension acting in the string, T = 3.5 N

Radius of the circular path, r = 1 m

To find,

The tangential speed of the puck.

Solution,

The centripetal force acting in the string is balanced by the tangential speed of the puck. The expression for the centripetal force is given by :

$F=\dfrac{mv^2}{r}$

$v=\sqrt{\dfrac{Fr}{m}}$

$v=\sqrt{\dfrac{3.5\ N\times 1\ m}{0.5\ kg}}$

v = 2.64 m/s

Therefore, the tangential speed of the puck is 2.64 m/s.

3 0

3 years ago

Describe how water molecules do not travel with the wave and the transfer of energy in water waves.

abruzzese [7]

The water molecules only oscillate up and down and not across the wave as water waves are transverse waves. Thus, the oscillations of the particles are perpendicular to the direction of wave propagation.

6 0

3 years ago

Add these two velocity vectors to find the magnitude of their resultant vector

a_sh-v [17]

<u>Answer</u>

4.6 m/s to the right.

<u>Explanation</u>

Velocity is a vector quantity. It has both magnitude and direction.

Taking the 5.8 m/s moving to the right as positive, then the 1.2 m/s moving to the left will be negative.

∴ 5.8 + -1.2 = 5.8 - 1.2

= 4.6 m/s.

The resultant velocity vector is 4.6 m/s to the right.

7 0

3 years ago