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Tatiana [17]
2 years ago
12

A cylindrical tube 12.5 cm high and 1.5 cm in diameter is used to collect blood samples. how many cubic decimeters (dm3) of bloo

d can it hold (v of a cylinder = r2h)?
Physics
1 answer:
Reil [10]2 years ago
6 0

0.176 cubic decimeters (dm3) of blood it can hold (v of a cylinder = r2h).

<h3>What is a cylindrical tube?</h3>

Round and hollow Cylinders i.e., when the general change in cross-sectional region is significantly less than the overall change in tension along the cylinder. At the end of the day, the cylinder region variety should be slower than the spatial variety of the actual wave. Round and hollow cylinders are generally utilized as transport compartments to safeguard interior items. The holders are expected to keep from imploding in any event, when they would tumble to the ground and be affected. In certain cases, the terms might be utilized conversely, but there is one vital contrast among cylinder and line, especially in how the material is requested and resilience. Tubing is utilized in underlying applications, so the external breadth turns into the significant aspect.

Learn more about cylindrical tube, refer:

brainly.com/question/27220701

#SPJ4

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If the average velocity during the athlete's walk back
goblinko [34]

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

T=\frac{d}{1.50}

From the question we are told

If the average velocity during the athlete's walk back  to the starting line in Guided Example 2.5 is – 1.50 m/s,

Generally the equation Time spent  is mathematically given as

T=\frac{d}{v}

Therefore

T=\frac{d}{1.50}

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

T=\frac{d}{1.50}

For more information on this visit

brainly.com/question/22271063

4 0
3 years ago
A sealed tank containing seawater to a height of 12.8 m also contains air above the water at a gauge pressure of 2.90 atm . Wate
exis [7]

Answer:

The velocity of water at the bottom, v_{b} = 28.63 m/s

Given:

Height of water in the tank, h = 12.8 m

Gauge pressure of water, P_{gauge} = 2.90 atm

Solution:

Now,

Atmospheric pressue, P_{atm} = 1 atm = 1.01\tiems 10^{5} Pa

At the top, the absolute pressure, P_{t} = P_{gauge} + P_{atm} = 2.90 + 1 = 3.90 atm = 3.94\times 10^{5} Pa

Now, the pressure at the bottom will be equal to the atmopheric pressure, P_{b} = 1 atm = 1.01\times 10^{5} Pa

The velocity at the top, v_{top} = 0 m/s, l;et the bottom velocity, be v_{b}.

Now, by Bernoulli's eqn:

P_{t} + \frac{1}{2}\rho v_{t}^{2} + \rho g h_{t} = P_{b} + \frac{1}{2}\rho v_{b}^{2} + \rho g h_{b}

where

h_{t} -  h_{b} = 12.8 m

Density of sea water, \rho = 1030 kg/m^{3}

\sqrt{\frac{2(P_{t} - P_{b} + \rho g(h_{t} - h_{b}))}{\rho}} =  v_{b}

\sqrt{\frac{2(3.94\times 10^{5} - 1.01\times 10^{5} + 1030\times 9.8\times 12.8}{1030}} =  v_{b}

v_{b} = 28.63 m/s

5 0
3 years ago
Which illusion or test was most difficult for you to perceive correctly? Why do you think this particular illusion was so challe
Alecsey [184]

Answer:

test 5 seemed to be the hardest for me to perceive in account i only saw three f's when there was indeed 6 it was very difficult to find the f's even going very slowly.  

Explanation:

correct on edge

4 0
3 years ago
Ricardo and Jane are standing under a tree in the middle of a pasture. An argument ensues, and they walk away in different direc
tankabanditka [31]

Answer:

a) 24.33 m of distance.

b) 34.55° east of the south.

Explanation:

The question is incomplete. The whole exercise is the following:

<em>"Ricardo and Jane are standing under a tree in the middle of a pasture. An argument ensues, and they walk away in different directions. Ricardo walks 28.0 m in a direction 60.0° west of north. Jane walks 12.0 m in a direction 30.0° south of west. They then stop and turn to face each other. </em>

<em>(a) What is the distance between them? </em>

<em>(b) In what direction should Ricardo walk to go directly toward Jane?</em><em>"</em>

Now that we know what we need to do in this question, let's head for every part of the problem.

<u>a) Distance between Ricardo and Jane</u>

In this case, we need to analyze the given data:

Ricardo (which we will call R) is 28 m from the starting point at 60° west of north, and Jane (J) is 12 m at 30° south of west. So the distance between them, will be the point where they both stop and face each other. This point can be seen in the image attached (See picture).

Let's call the distance between them as "D", to get the distance of D, according to the picture will be:

D = J - R   (1)

However, as they are facing in different angles and directions, we cannot do the difference of their values distance just like that. In order to do that, we need to calculate the components in the "x" and "y" axis of each vector. In that way, we can get the components of x and y of the Distance D, and then, the whole distance between them will be:

D = √Dx² + Dy²     (2)

So, let's get the components of x and y of R and J.

For Ricardo (R):

Rx = R sin60° = 28 sin60° = -24.25 m

Ry = R cos60° = 28 cos60° = 14 m

The sign "-" it's because R it's on the second quadrant, therefore in x, we'll have to add the negative.

For Jane (J):

Jx = J cos30° = 12 cos30° = -10.39 m

Jy = J sin30° = 12 sin30° = -6 m

Again, the negative is added because J is on the third quadrant.

Now that we have the components, let's calculate vector D using expression (1):

Dx = -10.39 - (-24.25) = 13.86 m

Dy = -6 - 14 = -20 m

Now, using expression (2) we can finally know the distance between Jane And Ricardo:

D = √(-20)² + (13.86)²

<h2>D = 24.33 m</h2>

This is the distance between Jane and Ricardo.

b) Direction of Ricardo walking to Jane

In this case, we already have the components of x and y of the distance between them, so, to know the direction:

Tanα = Dy/Dx

α = tan⁻¹ (Dy/Dx)

Replacing the values we have:

α = tan⁻¹ (-20/13.86)

α = 55.45°

Which should south of east or:

β = 90 - 55.45

<h2>β = 34.55°</h2>

Ricardo should walk 34.55° east of south

Hope this helps

8 0
3 years ago
The area of the eye where there is no version​
fomenos
It is called the fovea centralis.
4 0
3 years ago
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