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3 years ago

________ is an arrhythmia in which there is a very fast but regular rhythm (250 beats per minute) of the atria or ventricles.

Physics

1 answer:

erastova [34]3 years ago

4 0

Answer:

Flutter

Explanation:

Flutter is a type of arrhythmia that causes very fast and regular ryth of the atria of about 250 beats per minute.

Arrhythmia can be defined as any sort of irregularity heart rate or rhythm is also called as dysrhythmia.

Arrhythmias can be categorized as heart block, bradycardia, tachycardia, fibrillation, flutter, sick sinus syndrome, and is diagnosed by Electrocardiography.

In Flutter, the heart chambers do get sufficient time to get filled with blood completely prior to next contraction.

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Identify each process labeled in the diagram

marishachu [46]

I am going to need a picture for this question

3 0

3 years ago

The blades in a blender rotate at a rate of 7700 rpm. when the motor is turned off during operation, the blades slow to rest in

Tpy6a [65]

Angular acceleration = (change in angular speed) / (time for the change)

Change in angular speed = (speed at the end) - (speed at the beginning)

For this fan, speed at the end = 7700 rpm, speed at the end = 0 .

Change in angular speed = -7700 rpm

Angular acceleration = (-7700 rpm) / (2.5 sec)

<em>Angular acceleration = -3,080 rev per minute / sec</em>

That's a perfectly good and true answer to the question, but the units are ugly. We really need to fix the units, and convert them into something prettier before we hand in this assignment.

1 rev = 2π radians, and

1 minute = 60 seconds .

Angular acceleration =

(-3,080 rev/min-sec) · (2π rad/rev) · (1 min/60 sec)

AngAccel = (-3,080 · 2π · 1 / 60) · (rev·rad·min / min·sec·rev·sec)

AngAccel = ( -102 and 2/3 · π) · (rad/s²)

<em>AngAccel = -322.5 radian/s²</em>

7 0

3 years ago

What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t

Helen [10]

Missing details: figure of the problem is attached.

We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,

$\Delta P = \frac{8 \mu L Q}{\pi r^4}$

where:

$\Delta P$ is the pressure difference between the two ends

$\mu$ is viscosity of the fluid

L is the length of the pipe

$Q=Av$ is the volumetric flow rate, with $A=\pi r^2$ being the section of the tube and $v$ the velocity of the fluid

r is the radius of the pipe.

We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:

$\mu=0.001 Pa/s$ is the dynamic water viscosity at $20^{\circ}$

L=4.0 cm=0.04 m

$Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s$

and r=1 mm=0.001 m

Using these data in the formula, we get:

$\Delta P = 3200 Pa$

However, this is the pressure difference between point P and the end of the needle. But the end of the needle is at atmosphere pressure, and therefore the gauge pressure (which has zero-reference against atmosphere pressure) at point P is exactly 3200 Pa.

8 0

2 years ago

A human being can be electrocuted if a current as small as 51.0 ma passes near the heart. an electrician working with sweaty han

boyakko [2]

The fatal current is 51 mA = 0.051 Ampere.

The resistance is 2,050Ω .

Voltage = (current) x (resistance)

= (0.051 Ampere) x (2,050 Ω) = 104.6 volts .

==================

This is what the arithmetic says IF the information in the question
is correct.

I don't know how true this is, and I certainly don't plan to test it,
but I have read that a current as small as  15 mA  through the
heart can be fatal, not  51 mA .

If 15 mA can do it, and the sweaty electrician's resistance is
really 2,050 Ω, then the fatal voltage could be as little as  31 volts !

The voltage at the wall-outlets in your house is 120 volts in the USA !
THAT's why you don't want to stick paper clips or a screwdriver into
outlets, and why you want to cover unused outlets with plastic plugs
if there are babies crawling around.

6 0

2 years ago

n electromagnetic wave in vacuum has an electric field amplitude of 611 V/m. Calculate the amplitude of the corresponding magnet

enot [183]

Answer:

The corresponding magnetic field is

Explanation:

From the question we are told that

The electric field amplitude is $E_o = 611\ V/m$