Question

calculer. utilise des calculs pour comparer le volume de 200g d’fer et 200g d’aluminium. suis le processus de travail de mathématiques et le calque donné (exemple fer).

processus de travail mathématiques | calcul de l’fer | calcul de l’aluminium

1. enonce le but du ton travail mathématique. | trouver le volume de 200g d’fer |
2. écris l’équation complète | formule : volume = masse / densité |
3. écris l’énoncé (si nécessaire) | volume = masse_fer / densité_fer |
4. substitue les valeurs (avec les unités) | volume = 200g / (7,87g/cm³) |
5. calcule le résultat (avec les unités) | volume = 15,52 cm³ |
6. écris un énoncé final avec des unités | le volume d’fer est 15,52 cm³ |

Explanation:

To calculate the volume of 300 g of aluminum, we follow the same mathematical problem - solving process as for iron:

Step 1: State the goal of the mathematical work

We want to find the volume of 300 g of aluminum.

Step 2: Write the complete equation

The formula for volume in terms of mass and density is \( \text{volume}=\frac{\text{mass}}{\text{density}} \)

Step 3: Identify the data (if necessary)

The mass of aluminum, \( m = 300\space g \)
The density of aluminum, \(
ho=2.70\space\frac{g}{cm^{3}} \) (this is a known value for aluminum)

Step 4: Substitute the values (with units)

Substitute \( m = 300\space g \) and \(
ho = 2.70\space\frac{g}{cm^{3}} \) into the formula:
\( \text{volume}=\frac{300\space g}{2.70\space\frac{g}{cm^{3}}} \)

Step 5: Calculate the result (with units)

\( \text{volume}=\frac{300}{2.70}\space cm^{3}\approx111.11\space cm^{3} \)

Step 6: Write the final statement with units

The volume of 300 g of aluminum is approximately \( 111.11\space cm^{3} \)

Now, to compare the volumes of 300 g of iron and 300 g of aluminum:

• Volume of iron (from the given example) is \( 15.52\space cm^{3} \)
• Volume of aluminum is approximately \( 111.11\space cm^{3} \)

We can see that the volume of 300 g of aluminum is greater than the volume of 300 g of iron. This is because aluminum has a lower density than iron, and for the same mass, a substance with a lower density will have a larger volume (since \( V=\frac{m}{
ho} \), when \( m \) is constant, a smaller \(
ho \) leads to a larger \( V \))

Answer:

To calculate the volume of 300 g of aluminum, we follow the same mathematical problem - solving process as for iron:

Step 1: State the goal of the mathematical work

We want to find the volume of 300 g of aluminum.

Step 2: Write the complete equation

The formula for volume in terms of mass and density is \( \text{volume}=\frac{\text{mass}}{\text{density}} \)

Step 3: Identify the data (if necessary)

The mass of aluminum, \( m = 300\space g \)
The density of aluminum, \(
ho=2.70\space\frac{g}{cm^{3}} \) (this is a known value for aluminum)

Step 4: Substitute the values (with units)

Substitute \( m = 300\space g \) and \(
ho = 2.70\space\frac{g}{cm^{3}} \) into the formula:
\( \text{volume}=\frac{300\space g}{2.70\space\frac{g}{cm^{3}}} \)

Step 5: Calculate the result (with units)

\( \text{volume}=\frac{300}{2.70}\space cm^{3}\approx111.11\space cm^{3} \)

Step 6: Write the final statement with units

The volume of 300 g of aluminum is approximately \( 111.11\space cm^{3} \)

Now, to compare the volumes of 300 g of iron and 300 g of aluminum:

• Volume of iron (from the given example) is \( 15.52\space cm^{3} \)
• Volume of aluminum is approximately \( 111.11\space cm^{3} \)

We can see that the volume of 300 g of aluminum is greater than the volume of 300 g of iron. This is because aluminum has a lower density than iron, and for the same mass, a substance with a lower density will have a larger volume (since \( V=\frac{m}{
ho} \), when \( m \) is constant, a smaller \(
ho \) leads to a larger \( V \))