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ollegr [7]
3 years ago
11

What is oxygenated blood? De-oxygenated blood?

Physics
2 answers:
WITCHER [35]3 years ago
6 0
Oxygenated blood that has oxygen in them while de-oxygenated blood has carbon dioxide. in which the oxygenated blood carries the oxygen throughout the body since that cells need oxygen to function. called "gas exchange." once the cells got their required oxygen. the carbon dioxide needs somewhere to go, thus having deoxygenated blood. and that carbon dioxide needs to get out of the body

Butoxors [25]3 years ago
6 0
Blood with rich amount of oxygen is known as "Oxygenated Blood" whereas blood with less amount of oxygen are referred to as "De-oxygenated Blood".

Hope this helps!
You might be interested in
A disadvantage of some geophysical techniques based on magnetism is that their use in urban areas may be subject to distortion r
julsineya [31]

Answer:

Resistivity Method

Explanation:

A disadvantage of some geophysical techniques based on magnetism is that their use in urban areas may be subject to distortion resulting from power lines and metals in the immediate vicinity. One method that can be used in such areas relies on soil moisture and the ground's resistance to electricity and is known as Resistivity.

It is usually carried out by injecting a small electrical current through the earth and measured a subtle  sub-surface change in resistance over a given area.

3 0
3 years ago
An aircraft flying at a steady velocity of 70 m/s eastwards at a height of 800 m drops a package of supplies.
iren2701 [21]

The motion of the package can be described as the motion of a projectile,

given that it has an horizontal velocity and it is acted on by gravity.

  • a) \vec{v} = 70·i
  • b) The package will reach ground in approximately <u>12.77 seconds</u>.
  • c) The speed of the package as it lands is approximately <u>145.51 m/s</u>.
  • d) The path of the package based on a stationary frame of reference is <u>parabolic</u>
  • e) The path of the package as seen from the plane is <u>directly vertical</u> downwards

Reasons:

Velocity of the aircraft = 70 m/s

Direction of flight of the aircraft = Eastward

Height from which the aircraft drops the package, h = 800 m

a) The initial velocity of the package, \vec{v} = 70·i

b) The time it will take the package to reach the ground, <em>t</em>, is given by the formula;

\displaystyle h = \mathbf{\frac{1}{2} \cdot g \cdot t^2}

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

Therefore;

\displaystyle t = \mathbf{\sqrt{ \frac{2 \cdot h}{g} }}

Which gives;

\displaystyle t = \sqrt{ \frac{2 \times 800}{9.81} } \approx \mathbf{12.77}

The time it will take the package to reach the ground, t ≈ <u>12.77 seconds</u>

c) The vertical velocity just before the package reaches the ground, v_y, is given as follows;

v_y^2 = 2·g·h

Therefore;

v_y = √(2·g·h)

Which gives;

v_y = √(2 × 9.81 × 800) ≈ 125.28

v_y ≈ 125.28 m/s

Which gives; \vec{v} = 70·i - 125.28·j

Therefore, |v| = √(70² + (-125.28)²) ≈ 143.51

The speed of the package as it lands, |v| ≈ <u>143.51 m/s</u>

d) The motion of the package that includes both horizontal and vertical motion is a projectile motion.

Therefore;

The path of the package is the path of a projectile, which is a <u>parabolic shape</u>.

e) As seen by someone on the aeroplane, the horizontal velocity will be

zero, therefore, the package will appear as accelerating <u>directly vertical</u>

downwards.

Learn more about projectile motion here:

brainly.com/question/1130127

6 0
2 years ago
A model of an atom is shown above. The blue dots represent electrons, and the red dot represents the nucleus. This model accurat
Anvisha [2.4K]
I think the model being described is the Rutherford atom model. This is where he posit that atoms have a nucleus which contains positively charged protons and surrounded by electrons. 


Thank you for your question. Please don't hesitate to ask in Brainly your queries. 
4 0
3 years ago
Read 2 more answers
What average mechanical power (in W) must a 64.5 kg mountain climber generate to climb to the summit of a hill of height 285 m i
kakasveta [241]

Answer:61.275 W

Explanation:

Given

Mass of climber m=64.5 kg

Height to climb h=285 m

Time taken t=49 min \approx 2940 s

Now Using  Work-energy Theorem work done is given by

W=m g h

W=64.5\times 9.8 \times 285

W=180148.5 J

W=180.148 kJ

Now  Average Mechanical Power is given by

P=\frac{W}{\Delta t}

P=\frac{180.148}{2940}=0.0612 kW

P=61.275 W

7 0
3 years ago
Please answer and explain the equations of how you got the answer
Dima020 [189]

Answer:

a) H=277113.45m

b) t = 142.09 seconds  

Solution

In this question we have given

initial speed, v= 1.7x10^3m/s

angle of projection with ground, α= 55 degree

a) Neglecting air resistance, horizontal component of velocity is given as

V(h)=vcos\alpha............(1)

put value of v and α in equation(1)

V(h)=cos 55\times1.7x10^3m/s

V(h) = 975.08 m/s  

Now, horizontal range is given as

H = V(h) \times t

H = 975.08 \times t

= 975.08t m .... (1)  


b) Time of flight

Vertical component of velocity is given as

V(v)=vsin\alpha............(1)

V(v)=vsin55\times1.7x10^3m/s

V(v) = 1392.5585 m/s  

We know that time to reach at maximum height is given by formula

t=\frac{V(v)}{g}

t= (1392.5585 m/s)/9.8m/s^2

time to reach at maximum height,t = 142.09seconds  

Total time in air 2t = 284.19 seconds.

put value of t in equation (1)

Therefore horizontal range is given as

H=975.08m/s\times284.19s\\H=277113.45m

3 0
3 years ago
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