Question 2 (1 point) According to Hooke's Law, the force needed to stretch a spring varies directly to the amount the spring is stretched 150 pounds of force
stretches a spring five inches, how much will the spring be stretched by a force of 180 pounds?
inches
Blank 1:

Explanation:

x = amount (in inches) the spring is stretched

y = force applied (in pounds)

y = kx for some constant k, since x and y vary directly

Plug in the given info that x = 5 and y = 150 pair up together

Solve for k

150 = k*5

150/5 = k

k = 30

The equation goes from y = kx to y = 30x. Whatever the stretching distance (x) is, you multiply by 30 to get the force needed (y) to pull the spring.

Plug in y = 180 to solve for x

y = 30x

180 = 30x

30x = 180

x = 180/30

x = 6

The spring will be stretched 6 inches if you apply a force of 180 pounds.

answer rather than having to add and subtract the measurements of the semi-circle, notice that the semicircle to the left of ad is actually identical in area to the blank semicircle-shaped space to the left of bc. so if we transferred the red semicircle into the blank space, we would end up with a square of side length 24 cm. the area of a square is s^2 = 24^2 = 676 cm^2.

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