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frosja888 [35]
3 years ago
5

An AED should only be used if a person is too tired to continue with chest compressions.

Physics
1 answer:
malfutka [58]3 years ago
6 0

The given statement "An AED should only be used if a person is too tired to continue with chest compressions" is false.

Answer: Option B

<u>Explanation:</u>

Most people think that they are very tired after applying the compress for 2-3 minutes. When the compressor is tired, it tends to compress more slowly. For this reason, rescuers are advised to do compressions in every 2 min to prevent fatigue and optimise the compressions quality. This should only be done if the person does not show signs of life or is dead, does not respond and breathes normally.

For example, consider you come across the victim in a narrow place and you have two helpers: Rescuer 2 arrives with the AED (automated external defibrillator) and puts it on the opposite side from Rescuer 1, who does chest compressions. Rescuer 2 turns on the AED and fixes the electrodes to the victim's chest, connecting wires to the AED if necessary. Rescuer 1 can continue CPR (Cardiopulmonary resuscitation) by placing electrodes until the victim's heart rate has been analysed.

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"a 10 kg rock is pushed off the edge of a bridge 50 meters above the ground. what was the kinetic energy of the rock at the midw
valentinak56 [21]
Let's call h the initial height of the rock (h=50 m, the height of the bridge).

Initially, the rock has only gravitational potential energy, which is given by
U_i=mgh
where m=10 kg is the mass of the rock while g is the gravitational acceleration. So, the total mechanical energy of the rock at this point is
E_i = U_i = mgh=(10 kg)(9.81 m/s^2)(50 m)=4905 J

At midway point of its fall, its height is \frac{h}{2}, so its potential energy is
U_f = mg \frac{h}{2} = (10 kg)(9.81 m/s^2)(25 m)=2452.5 J
But now the rock is also moving by speed v, so it also has kinetic energy:
K_f =  \frac{1}{2}mv^2
So the total energy at the midway point of the fall is
E_f = U_f + K_f (1)

The mechanical energy must be conserved, so E_i = E_f, so we can rewrite (1) and solve it to find the kinetic energy of the rock at midway point of its fall:
E_i = U_f + K_f
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8 0
3 years ago
A boat traveled upstream 90 miles at an average speed of (v-3) miles per hour and then traveled the same distance downstream at
MAVERICK [17]

Answer:

A) 2.5

Explanation:

Extracting vital information from the question;

speed upstream = (v-3)mile/hr, distance traveled = 90 mile, speed downstream = (v+3), time for downstream in hours = t while time for upstream = t + 0.5 hr since the upstream journey is half hour longer.

speed = distance / time = 90 / (t + 0.5)

( v - 3) = 90 / ( t + 0.5)

cross multiply

(v-3) (t + 0.5) = 90 equation (1) for upstream motion

( v+3) = 90 / t

cross multiply

t(v+3) = 90 equation (2) for downstream motion

make v subject of the formula in equation 2

vt + 3t = 90

vt = 90 - 3t

divide both side by t

vt/t = (90 - 3t) / t

v = (90 - 3t) / t

 substitute for t in equation 1

(( 90 - 3t) / t) - 3) (t + 0.5) = 90

solve through finding l.c.m ( lowest common multiple)

(90 - 3t - 3t)/ t (t + 0.5) = 90

(( 90 - 6t ) / t )(t + 0.5) = 90

open the brackets and cross multiply

90t + 45 - 6t² - 3t = 90 t

rearrange  and collect the like terms

- 6t² - 3t + 45 = 90t - 90t

- 6t² - 3t + 45 = 0 multiply both side by -1

6t² + 3t - 45 = 0

divide both side by 3

2t² + t - 15 =  0

factorize the expression by multiplying - 15 by 2t² = - 30t²

find factors of - 30t² that adds to t  = 6t × (-5t)

replace t with (+6t - 5t) in the equation

2t²+ 6t - 5t - 15 = 0

factorize

2t ( t + 3) - 5 ( t + 3) =0

(2t -5)(t + 3) = 0  

2t - 5  = 0 or t + 3 = 0

2t = 5 or t = -3

divide through 2

2t / 2 = 5/ 2 = 2.5 or t = -3 since time cannot be negative

them t = 2..5 seconds

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