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natta225 [31]
3 years ago
9

What is the purpose of cpr??

Physics
2 answers:
Ket [755]3 years ago
8 0
Its main purpose is to restore partial flow of oxygenated blood to the brain and heart in a short amount of time.
Marta_Voda [28]3 years ago
7 0

Answer:

To restore a heart beat

Explanation:

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A 65-kg bungee jumper, who is attached to one end of an 85-m long bungee cord that has its other end tied to a bridge, jumps off
irinina [24]

Answer:

The impulse delivered to the bungee jumper is 1.32 kN.s

Explanation:

The situation can be shown graphically as shown in the figure.

Impulse delivered to the bungee jumper = Area under the curve.

The curve represents a triangle and the area of traiangle = (1/2)base×height

The base of the triangle from the graph = 1.2 seconds.

The height of the triangle from the graph = 2.2 kN

Thus,

<u>Impulse = (1/2)×(1.2 seconds)×(2.2 kN) = 1.32 kN.s</u>

3 0
3 years ago
A fisherman has caught a very large, 5.0kg fish from a dock that is 2.0m above the water. He is using lightweightfishing line th
agasfer [191]

Answer:

t = 2 s

Explanation:

As we know that fish is pulled upwards with uniform maximum acceleration

then we will have

T - mg = ma

here we know that maximum possible acceleration of so that string will not break is given as

T = 54 N

now we have

54 - (5 \times 9.8) = 5 a

a = 1 m/s^2

now for such acceleration we can use kinematics

d = \frac{1}{2}at^2

2 = \frac{1}{2}(1) t^2

t = 2 s

7 0
3 years ago
You are part of a design team assigned the task of making an electronic oscillator that will be the timing mechanism of a micro-
Snowcat [4.5K]

Solution :

We assume that there is a ring having a charge +Q and radius r. Electric field due to the ring at a point P on the axis is given by :

E_P=\int dE \cos

E_P=\int \frac{KdQ}{(\sqrt{r^2+x^2})^2}\times \frac{x}{\sqrt{r^2+x^2}}

\vec{E_P}=\frac{Kx}{r^2+x^2} \int dQ

\vec{E_P}=\frac{KxQ}{(r^2+x^2)^{3/2}} \hat{i}

If we put an electron on point P, then force on point e is :

\vec{F}=-|e|\vec{E_P}

F= \frac{-eKQx}{(r^2+x^2)^{3/2}}= \frac{-eKQx}{r^3[1+\frac{x^2}{r^2}]^{3/2}}

If r >> x , then    $\frac{x^2}{r^2} \approx 0$

Then,  $\frac{-eKQ}{r^3}x$

$ma =\frac{-eKQ}{r^3}x$

$a =\frac{-eKQ}{mr^3}x$

Compare, a = -ω²x

We get,

$\omega^2 = \frac{eKQ}{R^3m}$

$\omega = \sqrt{\frac{eKQ}{r^3m}}$

$2 \pi f = \sqrt{\frac{eKQ}{r^3m}}$

$f = \frac{1}{2 \pi}\sqrt{\frac{eKQ}{mr^3}}$

6 0
3 years ago
Which of The following is the best example of water changing from a liquid to gas
expeople1 [14]

Answer:

There are no examples but this should be evaporation

Explanation:

4 0
3 years ago
Guys please helpp!!!!1
Setler79 [48]

Answer:

Position A/Position E

K = E, U = 0

Position B/Position D

K = (1-x)\cdot E, U = x\cdot E, for 0 < x < 1

Position C

K = 0, U = E

Explanation:

Let suppose that ball-Earth system represents a conservative system. By Principle of Energy Conservation, total energy (E) is the sum of gravitational potential energy (U) and translational kinetic energy (K), all measured in joules. In addition, gravitational potential energy is directly proportional to height (h) and translational kinetic energy is directly proportional to the square of velocity.

Besides, gravitational potential energy is increased at the expense of translational kinetric energy. Then, relative amounts at each position are described below:

Position A/Position E

K = E, U = 0

Position B/Position D

K = (1-x)\cdot E, U = x\cdot E, for 0 < x < 1

Position C

K = 0, U = E

3 0
2 years ago
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