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Eddi Din [679]
3 years ago
15

What should a student do if the hydrochloric acid splashes in their eyes? .

Physics
2 answers:
Brilliant_brown [7]3 years ago
7 0

Answer:

immediately wash eyes with water

gayaneshka [121]3 years ago
3 0

First aid suggestions include:

Hold your face under running water for 15 to 20 minutes and allow the water stream to flood into your eyes. Use your fingers to hold your eyelids apart (make sure there is no trace of the chemical on your fingers).

If you wear contact lenses, remove them as soon as possible.

Seek immediate medical advice. Medical staff will need to know what chemical was involved, particularly whether it was acid or alkaline, liquid or powder.

Do not judge the seriousness of your eye injury on the degree of pain. For example, alkali chemicals don’t usually cause significant symptoms, but can seriously damage the eye.

Powder or particulate (granular matter, like wet concrete) chemicals can be particularly damaging since they are more difficult to flush out.

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About 2% of our solar nebula consisted of elements besides hydrogen and helium. However, the very first generation of star syste
Mariulka [41]

Answer:

Answer

Explanation:

The very first generation of star systems in the universe probably consisted only of hydrogen and helium. This means that,

There were no comets or asteroids in these first-generation star systems. As the comets and asteroids were the only soruce of elements other than Hydrogen and Heliums ( As starts mainly consists of helium and hydrogen).

3 0
3 years ago
2. What would be the acceleration of the clown at 5 s? (A) 1.6 m/s2 (B) 8.0 m/s2 (C) 2.0 m/s2 (D) 3.4 m/s2 3. After 12 seconds,
Neko [114]
Is there an image that goes with this question?
7 0
2 years ago
Two forces and are applied to an object whose mass is 11.8 kg. The larger force is . When both forces point due east, the object
S_A_V [24]

Can you please fill in whatever goes in the blanks ?

Without them, the question makes no sense and has no answer.

Two forces (___) and (___) are applied to an object whose mass is 11.8 kg. The larger force is (___) . When both forces point due east, the object's acceleration has a magnitude of 0.408 m/s2. However, when (___) points due east and (___) points due west, the acceleration is 0.227 m/s2, due east. Find (a) the magnitude of (___) and (b) the magnitude of (___) .

6 0
3 years ago
The variable that the experimenter decides to change to see if there is or is not an effect is the Independent Variable.
EastWind [94]
The correct answer is:  [A]:  "TRUE" .
_________________________________________________
6 0
3 years ago
a 63 kg object needs to be lifted 7 in a matter of 5 Seconds approximately how much horsepowers is required to achieve this task
kupik [55]
We need to lift an object from a point A to another point B. So, we have these variables:

For the object:
m = 63Kg

Height:
h = 7in

Time:
t = 5s

Horsepower (hp<span>) is a </span>unit of measurement<span> of </span>power. So, t<span>he term was adopted in the late 18th century by </span>Scottish<span> engineer </span>James Watt<span> to compare the output of </span>steam engines<span> with the power of </span>draft horses<span>. So:

</span>P =  \frac{W}{t}

So, we need to calculate the work W. A force<span> is said to do work if, when acting, there is a </span>displacement<span> of the point of application in the direction of the force. In a mathematical language:
</span>
W=Fd

Given that the displacement is vertically, this force will be:

F=mg=(63kg)(9.8m/s^{2})=617.4N

We need to convert the unit in into m, so:

1in = 0.0254m
 
Then:
7in = 0.1778m

Therefore, as the displacement d = h:

W = (617.4)(0.1778)=109.77(J)

Finally:

P= \frac{109.77}{5}=21.95W

The result is expressed in Watts, so we need to convert it into hp, so:

1hp = 746W

P= 21.95W(\frac{1hp}{746W})=0.03hp





7 0
3 years ago
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