make sure to show all work to receive credit. you may need a separate sheet of paper.
1. find the volume of a box measuring 2 cm by 7 cm by 3 cm.
2. an object placed in a graduated cylinder raises the volume from 12.2 ml to 14.5 ml. find the volume of the object.
3. find the volume of a cube measuring 5 cm on each side.
4. find the density of the box in #1 if it has a mass of 20 g.
5. find the density of the object in #2 if it has a mass of 4 g.
6. find the density of the cube in #3 if it has a mass of 100 g.
7. find the mass of an object that has a density of 1.5 g/cm³ and has a volume of 8 cm³.
8. find the volume of an object with a density of 3.1 g/ml and a mass of 12 g.
9. find the mass of a cube that has a density of 2.7 g/ml and measures 3 cm on each side.
10. challenge: find the density of a soda can with a radius of 3.25 cm, a height of 12.2 cm, and a mass of 40 g.

Question

Accepted Answer

### 1.
# Explanation:
## Step1: Recall the formula for the volume of a rectangular box (which is a rectangular prism), $ V = l \times w \times h $, where $ l $ is length, $ w $ is width, and $ h $ is height.
Here, $ l = 2 \, \text{cm} $, $ w = 7 \, \text{cm} $, $ h = 3 \, \text{cm} $.
## Step2: Substitute the values into the formula.
$ V = 2 \times 7 \times 3 $
$ V = 42 \, \text{cm}^3 $
# Answer:
$ 42 \, \text{cm}^3 $

### 2.
# Explanation:
## Step1: The volume of the object is the difference between the final volume and the initial volume in the graduated cylinder. The formula is $ V_{\text{object}} = V_{\text{final}} - V_{\text{initial}} $.
Here, $ V_{\text{initial}} = 12.2 \, \text{mL} $, $ V_{\text{final}} = 14.5 \, \text{mL} $.
## Step2: Substitute the values.
$ V_{\text{object}} = 14.5 - 12.2 $
$ V_{\text{object}} = 2.3 \, \text{mL} $
# Answer:
$ 2.3 \, \text{mL} $

### 3.
# Explanation:
## Step1: Recall the formula for the volume of a cube, $ V = s^3 $, where $ s $ is the length of a side.
Here, $ s = 5 \, \text{cm} $.
## Step2: Substitute the value into the formula.
$ V = 5^3 $
$ V = 125 \, \text{cm}^3 $
# Answer:
$ 125 \, \text{cm}^3 $

### 4.
# Explanation:
## Step1: Recall the formula for density, $
ho = \frac{m}{V} $, where $
ho $ is density, $ m $ is mass, and $ V $ is volume. From problem 1, the volume of the box $ V = 42 \, \text{cm}^3 $, and $ m = 20 \, \text{g} $.
## Step2: Substitute the values into the density formula.
$
ho = \frac{20}{42} \approx 0.48 \, \text{g/cm}^3 $
# Answer:
$ \approx 0.48 \, \text{g/cm}^3 $

### 5.
# Explanation:
## Step1: Recall the density formula $
ho = \frac{m}{V} $. From problem 2, the volume of the object $ V = 2.3 \, \text{mL} $, and $ m = 4 \, \text{g} $.
## Step2: Substitute the values.
$
ho = \frac{4}{2.3} \approx 1.74 \, \text{g/mL} $
# Answer:
$ \approx 1.74 \, \text{g/mL} $

### 6.
# Explanation:
## Step1: Recall the density formula $
ho = \frac{m}{V} $. From problem 3, the volume of the cube $ V = 125 \, \text{cm}^3 $, and $ m = 100 \, \text{g} $.
## Step2: Substitute the values.
$
ho = \frac{100}{125} = 0.8 \, \text{g/cm}^3 $
# Answer:
$ 0.8 \, \text{g/cm}^3 $

### 7.
# Explanation:
## Step1: Recall the density formula $
ho = \frac{m}{V} $, and rearrange it to solve for mass: $ m =
ho \times V $. Here, $
ho = 1.5 \, \text{g/cm}^3 $, $ V = 8 \, \text{cm}^3 $.
## Step2: Substitute the values.
$ m = 1.5 \times 8 $
$ m = 12 \, \text{g} $
# Answer:
$ 12 \, \text{g} $

### 8.
# Explanation:
## Step1: Recall the density formula $
ho = \frac{m}{V} $, and rearrange it to solve for volume: $ V = \frac{m}{
ho} $. Here, $
ho = 3.1 \, \text{g/mL} $, $ m = 12 \, \text{g} $.
## Step2: Substitute the values.
$ V = \frac{12}{3.1} \approx 3.87 \, \text{mL} $
# Answer:
$ \approx 3.87 \, \text{mL} $

### 9.
# Explanation:
## Step1: First, find the volume of the cube. The formula for the volume of a cube is $ V = s^3 $, where $ s = 3 \, \text{cm} $. So $ V = 3^3 = 27 \, \text{cm}^3 $ (and since $ 1 \, \text{cm}^3 = 1 \, \text{mL} $, $ V = 27 \, \text{mL} $). Then, use the density formula $
ho = \frac{m}{V} $ and rearrange to $ m =
ho \times V $. Here, $
ho = 2.7 \, \text{g/mL} $, $ V = 27 \, \text{mL} $.
## Step2: Substitute the values.
$ m = 2.7 \times 27 $
$ m = 72.9 \, \text{g} $
# Answer:
$ 72.9 \, \text{g} $

### 10.
# Explanation:
## Step1: A soda can is a cylinder, so first find the volume of the cylinder using the formula $ V = \pi r^2 h $, where $ r = 3.25 \, \text{cm} $, $ h = 12.2 \, \text{cm} $.
## Step2: Calculate the volume. $ V = \pi \times (3.25)^2 \times 12.2 $. First, $ (3.25)^2 = 10.5625 $. Then, $ 10.5625 \times 12.2 = 128.8625 $. Then, $ V = \pi \times 128.8625 \approx 3.1416 \times 128.8625 \approx 404.7 \, \text{cm}^3 $ (or $ 404.7 \, \text{mL} $).
## Step3: Use the density formula $
ho = \frac{m}{V} $, where $ m = 40 \, \text{g} $, $ V \approx 404.7 \, \text{mL} $.
## Step4: Substitute the values. $
ho = \frac{40}{404.7} \approx 0.10 \, \text{g/mL} $
# Answer:
$ \approx 0.10 \, \text{g/mL} $

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QUESTION IMAGE

Question

Explanation:

1.

Step1: Recall the formula for the volume of a rectangular box (which is a rectangular prism), \( V = l \times w \times h \), where \( l \) is length, \( w \) is width, and \( h \) is height.

Step2: Substitute the values into the formula.

Step1: The volume of the object is the difference between the final volume and the initial volume in the graduated cylinder. The formula is \( V_{\text{object}} = V_{\text{final}} - V_{\text{initial}} \).

Step2: Substitute the values.

Step1: Recall the formula for the volume of a cube, \( V = s^3 \), where \( s \) is the length of a side.

Step2: Substitute the value into the formula.

Answer:

2.