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Sonja [21]
3 years ago
13

What does physiological mean

Physics
1 answer:
velikii [3]3 years ago
6 0

Answer: physiological  

ADJECTIVE

relating to the branch of biology that deals with the normal functions of living organisms and their parts.

"physiological research on the causes of violent behavior"

relating to the way in which a living organism or bodily part functions.

"slow down your body's physiological response to anger by breathing deeply"

Explanation:

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The basic barometer can be used as an altitude-measuring device in airplanes. The ground control reports a barometric reading of
kipiarov [429]

Answer:

Δh_air=714m

Explanation:

Given data

P_{1}=753mmHg\\P_{2}=690mmHg\\ p_{air}=1.2kg/m^{3}\\  g=9.8m/s^{2}

Solution

ΔP=P₁-P₂

=(ΔhHg)×pHg×g

=(Δh_air)× p_air ×g

Then

Δh_air=(pHg+ΔhHg)÷p_air

=\frac{13600*(753-690)*10^{-3} }{1.2}\\ =714m

Δh_air=714m

7 0
3 years ago
Where might water be found on the moon
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Water Could be Found in the frozen Ice on the moon. it would most likely be underground however
8 0
3 years ago
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A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o
Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field dE at the center of curvature of the arc is

dE = k\dfrac{dQ}{r^2}cos(\theta ) <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where r is the radius of curvature.

Now

dQ = \lambda rd\theta,

where \lambda is the charge per unit length, and it has the value

\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.

Thus, the electric field at the center of the curvature of the arc is:

E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta  }{r^2} cos(\theta)

E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.

Now, we find \theta_1 and \theta_2. To do this we ask ourselves what fraction is the arc length  3.0 of the circumference of the circle:

fraction = \dfrac{3.0m}{2\pi (2.3m)}  = 0.2076

and this is  

0.2076*2\pi =1.304 radians.

Therefore,

E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.

evaluating the integral, and putting in the numerical values  we get:

E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\

\boxed{ E = 31.329N/C.}

4 0
3 years ago
PLEASE HELP WILL GIVE 10 POINTS!!!!!
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Its readily available
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Hydrogen isn't very good fuel source due to its high flammability and can create a nasty mini hydrogen bomb. <span />
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What were two arguments or lines of evidence in support of the geocentric model?
Inessa05 [86]

Answer:

Two arguments or lines of evidence in support of the geocentric model are:

Geocentric model: This model describe that the earth is in the center of the universe and all the planets. Moon and the sun are revolved around the earth. This model explained the predominant description.

Heliocentric model: In this model the earth and the sun are consider moving and the sun are in the center of the solar system. And the sun is at the center position where all the planets are revolving around it, being dissimilar to the geocentric model.

7 0
3 years ago
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