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zysi [14]
3 years ago
5

If the caffeine concentration in a particular brand of soda is 2.97 mg/oz, 2.97 mg/oz, drinking how many cans of soda would be l

ethal? Assume that 10.0 g of caffeine is a lethal dose, and there are 12 oz in a can.
Physics
1 answer:
Ray Of Light [21]3 years ago
5 0

Explanation:

The given data is as follows.

     Concentration of caffeine = 2.97 mg/oz

     Number of oz in a can = 12 oz

Therefore, the concentration of caffeine in one can is calculated as follows.

                 = (12 \times 2.97) mg

                 = 35.64 mg

                 = 35.64 \times 10^{-3} g

Since, it is given that lethal dose is 10.0 g. Hence, number of cans are calculated as follows.

     No. of cans = \frac{\text{Lethal dose}}{\text{concentration in one can}}

                         = \frac{10 g}{35.64 \times 10^{-3} g}

                         = 280.58

                         = 281 (approx)

Thus, we can conclude that 281 cans of soda would be lethal.

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Gravitational energy or gravitational potential energy is the potential energy a huge item has corresponding to one more monstrous article because of gravity. It is the potential energy related with the gravitational field, which is delivered (changed over into active energy) when the articles fall towards one another. Gravitational potential energy increments when two articles are brought further apart. It is the potential energy related with the gravitational field, which is delivered (changed over into dynamic energy) when the items fall towards one another. Gravitational potential energy increments when two items are brought further separated.

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3 0
1 year ago
A 0.439 kg puck, initially at rest on a horizontal, frictionless surface, is struck by a 0.307 kg puck moving initially along th
lesya [120]

Answer:

u2 = 0.266 m/s

Explanation:

Let the left Puck mass at rest = m1 =0.307 Kg

mass of the right puck m2 = 0.439 kg

velocity of m1 before collision v1= 2.19 m/s

velocity of m2 before collision v2 = 0m/s

velocity of m1 after collision u1 =1.19 m/s

velocity of m2 after collision u2 = ? m/s

θ = 37°

<u>Solution:</u>

Before collision:

Momentum (y-axis ) before collision= 0 Kgm/s

Momentum (x-axis ) before collision= m1v1 + m2v2 = 0.307 Kg x 2.19 m/s + 0

= 0.672 Kgm/s

After collision:

Momentum (y-axis ) after collision= m1u1 sinθ  + m2u2 sinθ

= 0.307 x 1.19 m/s sin 37 °  + 0.439 x u2 sin 37°

= 0.22 + 0.26 u2

Momentum (x-axis ) after collision= m1u1 cosθ  + m2u2 cos θ

= 0.307 x 1.19 m/s cos 37 ° + 0.439 x u2 cos 37°

= 0.29 + 0.35 u2

According to law of conservation momentum

momentum before collision = momentum after collision.

0 + 0.672 Kgm/s =  0.22 Kgm/s  + 0.26 kg u2 + 0.29 Kgm/s + 0.35 kg u2

0.672 Kgm/s = 0.51 Kgm/s + 0.61 u2

u2 = 0.266 m/s

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The spacing between particles increases.

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3 years ago
Suppose you want to search for brown dwarfs using a space telescope. Will you design your telescope to detect light in the ultra
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Answer:

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