Question

1. find the density of the box in #1 if it has a mass of 20 g.

Explanation:

To solve this, we need the volume of the box from problem #1. Assuming from a typical problem #1, if the box (a rectangular prism) has dimensions, say, length \( l \), width \( w \), height \( h \), volume \( V = l \times w \times h \). Let's assume in #1, the box had dimensions (for example, if it was a 2cm x 3cm x 4cm box, but since we don't have #1, let's note the general formula: density \(
ho = \frac{\text{mass}}{\text{volume}} \).

Since the problem refers to #1, we need the volume from that. Let's suppose in #1, the box's volume was calculated as \( V \) (e.g., if #1 had a box with length 2 cm, width 2 cm, height 2 cm, volume \( V = 2 \times 2 \times 2 = 8 \, \text{cm}^3 \), but we need the actual #1 data. Wait, maybe a common #1: if the box is a cube with side 2 cm, volume \( V = 2^3 = 8 \, \text{cm}^3 \). Then density \(
ho = \frac{20 \, \text{g}}{8 \, \text{cm}^3} = 2.5 \, \text{g/cm}^3 \). But since we don't have #1, we need the volume from that problem.

Assuming the volume from #1 is \( V \), then density \(
ho = \frac{20}{V} \).

If, for example, in #1 the box had dimensions 1cm x 2cm x 5cm, volume \( V = 1 \times 2 \times 5 = 10 \, \text{cm}^3 \), then density \(
ho = \frac{20}{10} = 2 \, \text{g/cm}^3 \).

But since we need the #1 data, let's state the formula:

Step 1: Recall density formula

Density \(
ho = \frac{\text{mass}}{\text{volume}} \), where mass \( m = 20 \, \text{g} \), and volume \( V \) is from problem #1.

Step 2: Get volume from #1

Suppose in #1, the box (rectangular prism) had length \( l \), width \( w \), height \( h \), so \( V = l \times w \times h \). Let's say in #1, the box was 2cm x 2cm x 3cm, so \( V = 2 \times 2 \times 3 = 12 \, \text{cm}^3 \). Then \(
ho = \frac{20}{12} \approx 1.67 \, \text{g/cm}^3 \).

But since we don't have #1, we need the volume from that. Once we have \( V \), divide 20g by \( V \) to get density.

For example, if #1's box had volume \( 8 \, \text{cm}^3 \), then \(
ho = \frac{20}{8} = 2.5 \, \text{g/cm}^3 \).

Since the problem is incomplete without #1's volume, but using the formula:

If we assume #1's box volume is \( V \), then density is \( \frac{20}{V} \).

(Note: To solve, obtain the volume from problem #1 and apply \(
ho = \frac{20}{V} \).)

Answer:

To solve this, we need the volume of the box from problem #1. Assuming from a typical problem #1, if the box (a rectangular prism) has dimensions, say, length \( l \), width \( w \), height \( h \), volume \( V = l \times w \times h \). Let's assume in #1, the box had dimensions (for example, if it was a 2cm x 3cm x 4cm box, but since we don't have #1, let's note the general formula: density \(
ho = \frac{\text{mass}}{\text{volume}} \).

Since the problem refers to #1, we need the volume from that. Let's suppose in #1, the box's volume was calculated as \( V \) (e.g., if #1 had a box with length 2 cm, width 2 cm, height 2 cm, volume \( V = 2 \times 2 \times 2 = 8 \, \text{cm}^3 \), but we need the actual #1 data. Wait, maybe a common #1: if the box is a cube with side 2 cm, volume \( V = 2^3 = 8 \, \text{cm}^3 \). Then density \(
ho = \frac{20 \, \text{g}}{8 \, \text{cm}^3} = 2.5 \, \text{g/cm}^3 \). But since we don't have #1, we need the volume from that problem.

Assuming the volume from #1 is \( V \), then density \(
ho = \frac{20}{V} \).

If, for example, in #1 the box had dimensions 1cm x 2cm x 5cm, volume \( V = 1 \times 2 \times 5 = 10 \, \text{cm}^3 \), then density \(
ho = \frac{20}{10} = 2 \, \text{g/cm}^3 \).

But since we need the #1 data, let's state the formula:

Step 1: Recall density formula

Density \(
ho = \frac{\text{mass}}{\text{volume}} \), where mass \( m = 20 \, \text{g} \), and volume \( V \) is from problem #1.

Step 2: Get volume from #1

Suppose in #1, the box (rectangular prism) had length \( l \), width \( w \), height \( h \), so \( V = l \times w \times h \). Let's say in #1, the box was 2cm x 2cm x 3cm, so \( V = 2 \times 2 \times 3 = 12 \, \text{cm}^3 \). Then \(
ho = \frac{20}{12} \approx 1.67 \, \text{g/cm}^3 \).

But since we don't have #1, we need the volume from that. Once we have \( V \), divide 20g by \( V \) to get density.

For example, if #1's box had volume \( 8 \, \text{cm}^3 \), then \(
ho = \frac{20}{8} = 2.5 \, \text{g/cm}^3 \).

Since the problem is incomplete without #1's volume, but using the formula:

If we assume #1's box volume is \( V \), then density is \( \frac{20}{V} \).

(Note: To solve, obtain the volume from problem #1 and apply \(
ho = \frac{20}{V} \).)