1. A spring bearing a 10-pound weight has stretched 0.5 inches. If you were to increase the weight to 15 pounds, how many inches would the spring stretch?

F = k x

10 = k (.5)
k = 20 lb/in

15 = 20 x
x = 0.75 in

The heck is a clown bot

To find out how many inches the spring would stretch when the weight is increased, we can use Hooke's Law. Hooke's Law states that the force exerted by a spring is directly proportional to the displacement of the spring.

First, let's assign some variables:
- Let F1 be the initial force exerted by the spring (10 pounds in this case).
- Let x1 be the initial displacement of the spring (0.5 inches in this case).
- Let F2 be the new force exerted by the spring (15 pounds in this case).
- Let x2 be the new displacement we want to find.

According to Hooke's Law, we have the formula:

F1/F2 = x1/x2

Plugging in the values we know, we can rearrange the formula to solve for x2:

10/15 = 0.5/x2

Simplifying:

10x2 = 15 * 0.5

10x2 = 7.5

Dividing both sides by 10:

x2 = 7.5/10 = 0.75 inches

Therefore, when the weight is increased to 15 pounds, the spring would stretch by 0.75 inches.

If you increase the weight to 15 pounds, the spring will stretch approximately 0.75 inches. That's enough to turn it into a slinky superhero!

10 = k (.5)