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PSYCHO15rus [73]
3 years ago
13

Which would be most reliable source for information about the toxity of an industrial chemical

Physics
2 answers:
amid [387]3 years ago
6 0
For example, a trade secret may<span> be a confidential device, pattern, </span>information<span>, or </span>chemical<span> make-up.</span>Chemical industry<span> trade secrets are generally formulas, process data, or a "specific </span>chemical<span> identity." The latter is the type of trade secret </span>information<span> referred to in the Hazard Communication Standard. The term includes</span>
inn [45]3 years ago
3 0

A scientific journal that was peer reviewed :)

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please help. Sound travels at 330 m/s. If a lightning bolt strikes the ground 5 m away from you, how long will it take for the s
Olenka [21]

Answer:

0.015 seconds

Explanation:

5/330=0.015

3 0
3 years ago
If you can answer my last post ill give you 75 points pls its very important and please make sure it correct!!!!!!!
Hoochie [10]

Answer:

I can't see the post :/

Explanation:

5 0
3 years ago
Read 2 more answers
What are non-contact forces please include example in your answer
Alina [70]

Answer:

Gravitational force.  Magnetic force.  Electrostatics.  Nuclear force.

Explanation:

Apple falling from a tree

raindrops falling from the sky

4 0
3 years ago
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An unknown immiscible liquid seeps into the bottom of an open oil tank. Some measurements indicate that the depth of the unknown
ira [324]

Answer:

The specific gravity of the unkown liquid is 15.

Explanation:

Gauge pressure, at the bottom of the tank in this case, can be calculated from

P_{bot}=\gamma_{oil}h_1 + \gamma_{unk}h_2,

where h_1 and h_2 are the height of the column of oil and the unkown liquid, respectively. Writing for \gamma_{unk}, we have

\gamma_{unk} = \frac{P_{bot} - \gamma_{oil}h_1}{h_2} = \frac{65\frac{kN}{m^2} - 8.5\frac{kN}{m^3}\times 5m}{1.5m} = \mathbf{15 \frac{kN}{m^3}}.

Relative to water, the unknow liquid specific weight is 15 times bigger, therefore this is its specific gravity as well.

3 0
3 years ago
If a person weighs 818 N on earth and 5320 N on the surface of a nearby planet, what is the acceleration due to gravity on that
alexandr402 [8]

Answer:

g'=63.74\ m/s^2

Explanation:

It is given that,

Weight of the person on Earth, W = 818 N

Weight of a person is given by the following formula as :

W=mg

g is the acceleration due to gravity on earth

m=\dfrac{W}{g}

m=\dfrac{818\ N}{9.8\ m/s^2}

m = 83.46 kg

The mass of an object is same everywhere. It does not depend on the location.

Let W' is the weight of the person on the surface of a nearby planet, W' = 5320 N

g' is the acceleration due to gravity on that planet. So,

g'=\dfrac{W'}{m}

g'=\dfrac{5320\ N}{83.46\ kg}                

g'=63.74\ m/s^2

So, the acceleration due to gravity on that planet is 63.74\ m/s^2. Hence, this is the required solution.                                                                    

6 0
3 years ago
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