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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? 450 pounds 500 pounds 45 pounds 5 pounds

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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? 450 pounds 500 pounds 45 pounds 5 pounds

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? 450 pounds 500 pounds 45 pounds 5 pounds

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HonorMaster · Tutor for 5 years

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### 500 pounds

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## Step1: Define the inverse relationship<br />### The pressure $P$ and distance $d$ have an inverse relationship, defined as $P \cdot d = k$, where $k$ is the constant.<br /><br />## Step2: Calculate the constant $k$<br />### Given $P = 150$ pounds and $d = 10$ inches, we substitute these values into the equation: $k = P \cdot d = 150 \cdot 10 = 1500$. <br /><br />## Step3: Solve for new pressure at 3 inches<br />### Using the constant $k = 1500$ and new distance $d = 3$ inches, we solve for $P$: <br />\[ P = \frac{k}{d} = \frac{1500}{3} = 500 \text{ pounds} \]
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