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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

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

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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