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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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ear the end of a marathon race, the first two runners are separated by a distance of 45.0 m. The front runner has a velocity of
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Answer:

a) V_{2/1}=0.8m/s

b) The second runner will win

c) d = 10.54m

Explanation:

For part (a):

V_{2/1} = V_{2} - V_{1} = 0.8m/s

For part (b) we will calculate the amount of time that takes both runners to cross the finish line:

t_{1} = \frac{X_{1}}{V_{1}}=\frac{250}{3.45}=72.46s

t_{2} = \frac{X_{2}}{V_{2}}=\frac{250+45}{4.25}=69.41s

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For part (c), we know how much time it takes the second runner to win, so we just need the position of the first runner in that moment:

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The length of the wire is 36 m.

<u>Explanation:</u>

Given, Diameter of sphere = 6 cm

We know that, radius can be found by taking the half in the diameter value. So,

       \text { sphere radius, } R=\frac{D}{2}=\frac{6}{2}=3 \mathrm{cm}=3 \times 10^{-2} \mathrm{m}

Similarly,

      \text { wire radius, } r=\frac{0.2}{2}=0.1 \mathrm{mm}=1 \times 10^{-3} \mathrm{m}

We know the below formulas,

          \text {volume of sphere}=\frac{4}{3} \times \pi \times R^{3}

          \text {volume of wire}=\pi \times r^{2} \times l

When equating both the equations, we can find length of wire as below, where \pi=\frac{22}{7}

          \frac{4}{3} \times \pi \times R^{3}=\pi \times r^{2} \times l

         \frac{4}{3} \times \frac{22}{7} \times\left(3 \times 10^{-2}\right)^{3}=\frac{22}{7} \times\left(1 \times 10^{-3}\right)^{2} \times l

The \pi value gets cancelled as common on both sides, we get

           \frac{4}{3} \times 27 \times 10^{-6}=10^{-6} \times l

The 10^{-6} value gets cancelled as common on both sides, we get

           l=4 \times 9=36 m

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3 years ago
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