# Please show all the set up and work for these numerical problems. Please make sure the final answer shows the correct number of significant figures.

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### 1 Answer

These are numerical problems testing your knowledge of unit conversions.

**1.** We can do it by using the fact that 1 acre contains 43560 ft^2.

Hence, area = `240 xx 10^6 acres xx 43560 ft^2/(acre) = 1.045 xx10^13 ft^2`

The same can also be solved by using the given information, 1 acre = 4046.86 m^2 and 1 m^2 = 10.76 ft^2

Since 240 million acres has only 2 significant figures, our answer will be rounded off to **1.0 x 10^13 ft^2**

**2. **1 lbs = 0.4536 kg

so the mass of the baby is 14 lbs = 14 x 0.4536 kg = 6.35 kg

Using the recommended dosage of 15 mg/kg, the baby will need 15 x 6.35 mg of drug = 95.25 mg

Using the given information that 80 mg drug is present in 0.8 ml

Using unitary method, 95.25 mg will be contained in 0.8 x 95.25/80 ml = **0.95 ml**. (two significant figures)

**3**. Here the numerator and denominator are solved separately and then divided. Remember, addition and subtraction does not change the units, while for multiplication, the units are also multiplied.

Hence, numerator : 2.54 g + 0.0028 g = 2.5428 g (we are not rounding off for significant figures, otherwise, this would be written as 2.54 gm, to account for least number of decimal places)

and denominator: 0.0105 cm^2 x 0.06 cm = 0.00063 cm^3 ( will be rounded off to 0.0006 to account for least number of significant figures in the LHS).

And the answer is: 2.54 gm/0.0006 cm^3 = 4233.33 gm/cm^3

and if we correct for significant figures, the answer would be **4000 gm/cm^3** (only 1 significant figure).

**4**. `(7.87 ml) / (16.1 ml-8.44 ml) = 7.87/7.66 = 1.03`

(two significant figures only since that is the minimum number of significant figures in numerator and denominator)

**5**. 100 m in 9.58 sec

Speed = distance/time = 100 /9.58 m/sec = 10.44 m/sec (if we use the rule of significant figures, the answer will be 10 m/sec, since answer will have only 1 significant figure)

1 mile contains 1609.344 m and 1 hour contains 3600 sec

`10.44 m/sec = 10.44 m/sec xx [(1 mi)/(1609.344 m)]/[(1 hr)/(3600 sec)] = 23.35 (mi)/(hr)`

(if we correct it for significant figures and round it off to 1 significant figure, the answer will be **20 mi/hr**).

Hope this helps.