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3 years ago

When you cut your knee what part of the blood helps the blood stop flowing?

Chemistry

2 answers:

raketka [301]3 years ago

7 0

Platelets

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IgorC [24]3 years ago

4 0

Answer:

the red blood cells

Explanation:

white blood cells fight off infection

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Give the result of the following expression with the correct number of significant figures

Marianna [84]

84 m/s x 31.221 s = 2600 m
-rounded from 2622.564 m because 84 m/s has only two significant figures

4 0

3 years ago

If the volume of a cylinder is reduced from 8.0 liters to 4.0 liters, the pressure of the gas in the cylinder will change from 7

Elza [17]

Answer:

P₂ = 140 KPa

Explanation:

Given data:

Initial volume = 8.0 L

Final volume = 4.0 L

Initial pressure = 70 KPa

Final pressure = ?

Solution:

According to Boyle's law

P₁V₁ = P₂V₂

P₂ = P₁V₁ / V₂

P₂ = 70 KPa ×8.0 L/4.0 L

P₂ = 560 KPa .L / 4.0 L

P₂ = 140 KPa

6 0

3 years ago

Which statement describes a chemical property that can be used to distinguish between compound A and compound B?

AURORKA [14]

Answer:

Explanation:

(4) A does not burn in air, and B does burn in air is your answer. Flammability is a chemical property while the rest of these are physical properties.

Still stuck? Get 1-on-1 help from an expert tutor now.

7 0

2 years ago

What is a matter a wave travels through called?

tresset_1 [31]

Answer:

Medium

Explanation:

It is called a medium

5 0

3 years ago

A 10 mm Brinell hardness indenter is used for some hardness testing measurements of a steel alloy. a) Compute the HB of this mat

Lady_Fox [76]

Explanation:

(a) The given data is as follows.

Load applied (P) = 1000 kg

Indentation produced (d) = 2.50 mm

BHI diameter (D) = 10 mm

Expression for Brinell Hardness is as follows.

HB = $\frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}$

Now, putting the given values into the above formula as follows.

HB = $\frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}$ = $\frac{2 \times 1000 kg}{3.14 \times 10 mm [D - \sqrt{((10 mm)^{2} - (2.50)^{2})}]}$

= $\frac{2000}{9.98}$

= 200

Therefore, the Brinell HArdness is 200.

(b) The given data is as follows.

Brinell Hardness = 300

Load (P) = 500 kg

BHI diameter (D) = 10 mm

Indentation produced (d) = ?

d = $\sqrt{(D^{2} - [D - \frac{2P}{HB} \pi D]^{2})}$

= $\sqrt{(10 mm)^{2} - [10 mm - \frac{2 \times 500 kg}{300 \times 3.14 \times 10 mm}]^{2}}$

= 4.46 mm

Hence, the diameter of an indentation to yield a hardness of 300 HB when a 500-kg load is used is 4.46 mm.

5 0

2 years ago