the pressure applied to a leverage bar varies inversely as the distance from the object. if 150 pounds is…

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?\n450 pounds\n5 pounds\n45 pounds\n500 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?\n450 pounds\n5 pounds\n45 pounds\n500 pounds

Answer

Explanation:

Step1: Establish inverse - variation formula

Let $P$ be pressure and $d$ be distance. The formula for inverse variation is $P_1d_1 = P_2d_2$.

Step2: Substitute given values

We know that $P_1 = 150$ pounds, $d_1=10$ inches and $d_2 = 3$ inches. Substituting into the formula gives $150\times10=P_2\times3$.

Step3: Solve for $P_2$

First, calculate $150\times10 = 1500$. Then, $P_2=\frac{1500}{3}=500$ pounds.

Answer:

D. 500 pounds