There are about 6 times 10^19 atoms in a grain of sand. A cubic foot of sand can hold about 4 times 10^9 grains of sand of average size. Using those figures, how many atoms would you expect to find in a cubic foot of sand? Write the answer in scientific notation

Answer 1

To find the total number of atoms in a cubic foot of sand, we can multiply the number of atoms in a grain of sand by the number of grains of sand in a cubic foot.

1. The number of atoms in a grain of sand is approximately \( 6 \times 10^{19} \).
2. The number of grains of sand in a cubic foot is approximately \( 4 \times 10^{9} \).

Now, we can perform the multiplication:

\[ \text{Total atoms} = (\text{atoms per grain}) \times (\text{grains per cubic foot}) \] \[ \text{Total atoms} = (6 \times 10^{19}) \times (4 \times 10^{9}) \]

We can multiply the coefficients and add the exponents:

\[ = 6 \times 4 \times 10^{19+9} \] \[ = 24 \times 10^{28} \]

Now, we need to express \( 24 \times 10^{28} \) in proper scientific notation. In scientific notation, the coefficient should be between 1 and 10. Therefore, we can rewrite \( 24 \) as \( 2.4 \times 10^{1} \):

\[ 24 \times 10^{28} = 2.4 \times 10^{1} \times 10^{28} \] \[ = 2.4 \times 10^{1 + 28} \] \[ = 2.4 \times 10^{29} \]

Thus, the expected number of atoms in a cubic foot of sand is:

\[ \boxed{2.4 \times 10^{29}} \]

Answer 2

A gallon of water weighs about 8.3 lb. An Olympic-sized swimming pool holds 6.6 times 10^5 gallons. What is the approximate weight of the water in an Olympic-sized swimming pool? Write the answer in scientific notation.

Answer 3

To find the approximate weight of the water in an Olympic-sized swimming pool, we can multiply the weight of one gallon of water by the total number of gallons in the pool.

1. The weight of one gallon of water is approximately \( 8.3 \) lb.
2. An Olympic-sized swimming pool holds approximately \( 6.6 \times 10^{5} \) gallons.

Now, we perform the multiplication:

\[ \text{Total weight} = (\text{weight per gallon}) \times (\text{gallons in pool}) \] \[ \text{Total weight} = (8.3) \times (6.6 \times 10^{5}) \]

We can first calculate the coefficient:

\[ 8.3 \times 6.6 = 54.78 \]

Next, we multiply this result by \( 10^{5} \):

\[ \text{Total weight} = 54.78 \times 10^{5} \]

Now we need to express \( 54.78 \) in proper scientific notation. We can rewrite \( 54.78 \) as \( 5.478 \times 10^{1} \):

\[ 54.78 \times 10^{5} = 5.478 \times 10^{1} \times 10^{5} \] \[ = 5.478 \times 10^{1 + 5} \] \[ = 5.478 \times 10^{6} \]

Now, rounding \( 5.478 \) to two significant figures to match the context of the problem, we obtain approximately \( 5.5 \).

Thus, the approximate weight of the water in an Olympic-sized swimming pool is:

\[ \boxed{5.5 \times 10^{6}} \text{ lb} \]

Answer 4

the choices are
54.78 times 10^5
5.478 times 10^6
5.478 times 10^5
54.78 times 10^6

Answer 5

Let’s clarify the calculation step by step and compare it to the provided options:

1. Weight of one gallon of water: \( 8.3 \) lb
2. Gallons in an Olympic-sized swimming pool: \( 6.6 \times 10^{5} \)

Now we calculate the total weight:

\[ \text{Total weight} = 8.3 \times (6.6 \times 10^{5}) \]

First, calculate \( 8.3 \times 6.6 \):

\[ 8.3 \times 6.6 = 54.78 \]

Now multiply by \( 10^{5} \):

\[ \text{Total weight} = 54.78 \times 10^{5} \]

This gives us the result \( 54.78 \times 10^{5} \).

Next, let’s compare this to the options given:

1. \( 54.78 \times 10^{5} \) → This is the result we calculated.
2. \( 5.478 \times 10^{6} \) → This is equivalent to \( 54.78 \times 10^{5} \) when you convert it (dividing \( 54.78 \) by \( 10^{1} \)).
3. \( 5.478 \times 10^{5} \) → This is not equivalent to our calculation.
4. \( 54.78 \times 10^{6} \) → This would be incorrect based on our calculation.

Considering the calculated result and the options provided, the correct choice from the list is:

\[ \boxed{54.78 \times 10^{5}} \]