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Question ID: 974926

The pressure applied to a leverage bar varies inversely as the distance from the object. If 150 pounds is required for a distance of 10 inches from the object, how much pressure is needed for a distance of 3 inches?


Sagot :

To determine the pressure needed when the distance from the object is 3 inches, given that the pressure varies inversely as the distance, we can use the relationship of inverse proportionality.

1. Understand Inverse Proportionality:
When two quantities are inversely proportional, the product of the two quantities remains constant. In this case:
[tex]\[ \text{pressure} \times \text{distance} = \text{constant} \][/tex]

2. Given Data:
We are told that 150 pounds of pressure is required when the distance is 10 inches.
Let's denote the initial pressure as \( P_1 \) and the initial distance as \( D_1 \):
[tex]\[ P_1 = 150 \text{ pounds}, \quad D_1 = 10 \text{ inches} \][/tex]

3. Set Up the Relationship:
From the proportionality statement, we set up the relationship:
[tex]\[ P_1 \times D_1 = \text{constant} \][/tex]
Plugging in the given values:
[tex]\[ 150 \times 10 = \text{constant} \][/tex]
[tex]\[ \text{constant} = 1500 \][/tex]

4. Find the Required Pressure:
Let's denote the required pressure as \( P_2 \) and the new distance as \( D_2 \):
[tex]\[ D_2 = 3 \text{ inches} \][/tex]
Using the inverse proportionality, we establish the equation:
[tex]\[ P_2 \times D_2 = \text{constant} \][/tex]
We already know the constant is 1500:
[tex]\[ P_2 \times 3 = 1500 \][/tex]

5. Solve for \( P_2 \):
[tex]\[ P_2 = \frac{1500}{3} \][/tex]

6. Calculate the Result:
[tex]\[ P_2 = 500 \text{ pounds} \][/tex]

Therefore, the pressure needed for a distance of 3 inches from the object is [tex]\( 500 \)[/tex] pounds.
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