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1 year ago

Caffeine concentration is 1.99 mg/oz how many cans would be leathal if 10g was leathal and there where 12oz in a can

1 answer:

5 0

The number of cans that would be considered lethal if 10g was lethal and there where 12oz in a can is 419 cans.

<h3>How to convert mass?</h3>

According to this question, caffeine concentration is 1.99 mg/oz.

1.99 milligrams can be converted to grams as follows:

1.99milligrams ÷ 1000 = 0.00199grams

This means that 0.00199grams per oz is the caffeine concentration.

If there were 12 oz in a can, then, 0.00199grams × 12 = 0.02388 grams in 1 can.

This means that if 10grams is considered lethal, 10grams ÷ 0.02388 grams = 419 cans would be lethal for consumption.

Therefore, the number of cans that would be considered lethal if 10g was lethal and there where 12oz in a can is 419 cans.

Learn more about conversion factor at: brainly.com/question/14479308

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Average speed is the total distance divided by the

seraphim [82]

<h2>Answer:</h2>

Time

<h2>Explanation:</h2>

The average speed of an object that is moving is defined as the distance traveled divided by the time of travel. You can measure the distance with a ruler and the time with a stopwatch. This can be expressed as the following formula:

$v=\frac{\Delta x}{\Delta t}$

For instance, if an object travels a distance $\Delta x=100m$ in 4 seconds, the the average speed is:

$v=\frac{100m}{4s} \\ \\ \therefore \boxed{v=25m/s}$

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3 years ago

An object is placed 18 cm in front of spherical mirror.if the image is formed at 4cm to the right of the mirror, calculate it's

ivolga24 [154]

1) Focal length

We can find the focal length of the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i}$ (1)
where
f is the focal length
$d_o$ is the distance of the object from the mirror
$d_i$ is the distance of the image from the mirror

In this case, $d_o = 18 cm$ , while $d_i=-4 cm$ (the distance of the image should be taken as negative, because the image is to the right (behind) of the mirror, so it is virtual). If we use these data inside (1), we find the focal length of the mirror:
$\frac{1}{f}= \frac{1}{18 cm}- \frac{1}{4 cm}=- \frac{7}{36 cm}$
from which we find
$f=- \frac{36}{7} cm=-5.1 cm$

2) The mirror is convex: in fact, for the sign convention, a concave mirror has positive focal length while a convex mirror has negative focal length. In this case, the focal length is negative, so the mirror is convex.

3) The image is virtual, because it is behind the mirror and in fact we have taken its distance from the mirror as negative.

4) The radius of curvature of a mirror is twice its focal length, so for the mirror in our problem the radius of curvature is:
$r=2f=2 \cdot 5.1 cm=10.2 cm$

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3 years ago

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Answer:

$\frac{V_{e}}{V_{h}}=0.428*10^{2}$

Explanation:

From conservation of energy states that

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3 years ago

Explain the relationship between energy and the potential to do work

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C. Should be the answer it worked for me

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3 years ago

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