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3 years ago

A mixture that appears to contain only one substance is a(n)

Physics

2 answers:

bekas [8.4K]3 years ago

8 0

Answer:

mixtures that appear to be the same throughout are homogenous mixtures

Goryan [66]3 years ago

4 0

Mixtures that seem to contain only one substance are homogeneous mixtures. You can figure this out because an element or compound aren’t mixtures. And the prefix “homo” means same while the prefix “hetero” means different. So homogeneous is a mixture that looks the same. Hope this helps, have a good day :)

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How is light refracted inside a crystal ball ?

Bingel [31]

The light refects ofthe sides

7 0

3 years ago

Find the difference between two masses measured as 123.6 grams and 115.972 grams. Express the answer to the correct number of si

xxTIMURxx [149]

Answer:

The difference is 7.6 grams.

Explanation:

In mathematics the difference of two numbers is express as the subtraction between them:

$a-b$

So to find out the difference between the two measured masses, a will be represented by 123.6 grams since is the bigger number, and b by 115.972 grams.

Therefore, it is get:

$123.6grams-115.972grams = 7.6grams$

<u>Hence, the difference is 7.6 grams. </u>

The result of 7.628 will be expressed as 7.6 to have the correct number of significant figures.

Notice how that can be express in units of kilograms too since there is 1000 gram in 1 kilogram:

$7.6grams . \frac{1Kg}{1000grams}$ ⇒ $7.6x10^{-3}Kg$

8 0

3 years ago

Blood in a carotid artery carrying blood to the head is moving at 0.15 m/s when it reaches a section where plaque has narrowed t

sp2606 [1]

Answer:

26.9 Pa

Explanation:

We can answer this question by using the continuity equation, which states that the volume flow rate of a fluid in a pipe must be constant; mathematically:

$A_1 v_1 = A_2 v_2$ (1)

where

$A_1$ is the cross-sectional area of the 1st section of the pipe

$A_2$ is the cross-sectional area of the 2nd section of the pipe

$v_1$ is the velocity of the 1st section of the pipe

$v_2$ is the velocity of the 2nd section of the pipe

In this problem we have:

$v_1=0.15 m/s$ is the velocity of blood in the 1st section

The diameter of the 2nd section is 74% of that of the 1st section, so

$d_2=0.74d_1$

The cross-sectional area is proportional to the square of the diameter, so:

$A_2=(0.74)^2 A_1=0.548 A_1$

And solving eq.(1) for v2, we find the final velocity:

$v_2=\frac{A_1 v_1}{A_2}=\frac{A_1 (0.15)}{0.548 A_1}=0.274 m/s$

Now we can use Bernoulli's equation to find the pressure drop:

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$

where

$\rho=1025 kg/m^3$ is the blood density

$p_1,p_2$ are the initial and final pressure

So the pressure drop is:

$p_1 - p_2 = \frac{1}{2}\rho (v_2^2-v_1^2)=\frac{1}{2}(1025)(0.274^2-0.15^2)=26.9 Pa$

8 0

3 years ago

Binding of a signaling molecule to which type of receptor leads directly to a change in the distribution of substances on opposi

Mrac [35]

When two sides of a membrane are in contact with each other, the distribution of ions will alter as a result of the binding of a signal molecule to a ligand-gated ion channel.

<h3>What is a ligand-gated ion channel?</h3>

Ligand-gated ion channels (LGICs) are membrane proteins that are structurally integral and feature a pore that permits the controlled passage of particular ions across the plasma membrane. The electrochemical gradient for the permeant ions drives the passive ion flux.

When a chemical ligand, such as a neurotransmitter, attaches to the protein, ligand-gated ion channels open. Changes in membrane potential cause voltage channels to open and close. When a receptor physically deforms, as in the case of pressure and touch receptors, mechanically-gated channels open.

Learn more about ligand-gated ion channel here:

brainly.com/question/15215628

#SPJ4

3 0

2 years ago

The diagram shows a position-time graph What is the displacement of the object

algol13

The object is moving, so at different times, it has different displacement. I'm guessing that you probably want to know the displacement at the end of the time on the graph ... 5 seconds.

Displacement is the distance and the direction FROM (the position at the beginning) TO (the position at the end).

At the beginning ... time=0 ... the position is 1 meter.

At the end ... time=5 ... the position is zero.

The distance FROM the beginning TO the end is (zero - 1m) . That's <em>-1m </em>.

5 0

3 years ago