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2 years ago

Which technique is best for manual in-line stabilization of a person floating faceup on the surface?

Physics

2 answers:

zmey [24]2 years ago

7 0

Answer:

vise grip

Explanation:

Manual in-line stabilization (MILS) of the cervical spine is a type of airway management when dealing with patients in traumatic condition ..it is a means that is performed by grasping the mastoid process of the patient, so as to prevent the movement of the cervical column during intubation of the trachea

MLS provides a means of stability to the cervical column for a patient in trauma. During this technique, a patient is restricted from moving his or her cervical collar. The vise grip can be used for a patient with neck injury. The technique is used to roll a patient to face up to prevent further injuries.

Elodia [21]2 years ago

7 0

Answer: the technique which is best for manual in-line stabilization of a person floating faceup on the surface is vice grip.

Explanation:

Vice grip is a rescue technique used to prevent further injury to a guest who is suspected of having suffered a spinal injury as the typical posture is floating faceup on the surface.

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1 year ago

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2 years ago

The electric output of a power plant is 716 MW. Cooling water is the main way heat from the powerplant is rejected, and it flows

Stels [109]

Answer:

(a) 83475 MW

(b) 85.8 %

Explanation:

Output power = 716 MW = 716 x 10^6 W

Amount of water flows, V = 1.35 x 10^8 L = 1.35 x 10^8 x 10^-3 m^3

mass of water, m = Volume x density = 1.35 x 10^8 x 10^-3 x 1000

= 1.35 x 10^8 kg

Time, t = 1 hr = 3600 second

T1 = 25.4° C, T2 = 30.7° C

Specific heat of water, c = 4200 J/kg°C

(a) Total energy, Q = m x c x ΔT

Q = 1.35 x 10^8 x 4200 x (30.7 - 25.4) = 3 x 10^12 J

Power = Energy / time

Power input = $P = \frac{3 \times 10^{12}}{3600}=8.35 \times 10^{8}W$

Power input = 83475 MW

(b) The efficiency of the plant is defined as the ratio of output power to the input power.

$\eta =\frac{Power output}{Power input}$

$\eta =\frac{716}{83475}=0.858$

Thus, the efficiency is 85.8 %.

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2 years ago

Abby throws a ball straight up and times it. she sees that the ball goes by the top of a flagpole after 0.50 s and reaches the l

Arisa [49]

The time it takes for a ball to complete the trip is given to be 4.10 s. This implies that the trip going up of the ball is equal to 2.05 s. Using this time to determine the initial speed, we have,
Vf = Vo - gt
Vf is the final speed which is equal to 0 on the topmost location of the ball.
0 = Vo - (9.8 m/s²)(2.05 s)
Vo = 20.09 m/s

At 0.5 s,
Vf = 20.09 m/s - (9.8 m/s²)(0.5 s)
Vf = 15.19 m/s

Thus, the speed of the ball when the height is in level with the top of the flagpole is 15.19 m/s.

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3 years ago

The cannon on a battleship can fire a shell a maximum distance of 26.0 km.

Alla [95]

It is possible to demonstrate that the maximum distance occurs when the angle at which the projectile is fired is $\theta = 45^{\circ}$ .
In fact, the laws of motions on both x- and y- directions are
$S_x(t)= v_0 cos \theta t$
$S_y(t)= v_0 \sin \theta t - \frac{1}{2} gt^2$
From the second equation, we get the time t at which the projectile hits the ground, by requiring $S_y(t)=0$ , and we get:
$t= \frac{2 v_0 \sin \theta}{g}$
And inserting this value into Sx(t), we find
$S_x(t) = 2 \frac{v_0^2}{g} \sin \theta \cos \theta= \frac{v_0^2}{g} \sin (2\theta)$
And this value is maximum when $\theta=45^{\circ}$ , so this is the angle at which the projectile reaches its maximum distance.

So now we can take again the law of motion on the x-axis
$S_x(t)= \frac{v_0^2}{g} \sin (2\theta)$
And by using $S_x = 26 km=26000 m$ , we find the value of the initial velocity v0:
$v_0 = \sqrt{ \frac{S_x g}{\sin (2\theta)} } = \sqrt{ \frac{(26000m)(9.81m/s^2)}{\sin (2\cdot 45^{\circ})} } =505 m/s$

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2 years ago