(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Suppose a spring has a natural length of 20 cm. If a 25-N force is required to keep it stretched to a length of 30 cm, how much work is required to stretch it from 20 cm to 25 cm? (b) Find the area of the region enclosed by one loop of the curve r=2sin(5θ).

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve is .

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

Where

, - Initial and final position, respectively, measured in meters.

- Force as a function of position, measured in newtons.

Given that and the fact that when , the spring constant (), measured in newtons per meter, is:

Now, the work function is obtained:

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be . The area of the region enclosed by one loop of the curve is given by the following integral:

By using trigonometrical identities, the integral is further simplified:

The area of the region enclosed by one loop of the curve is .

the answer for apex is the lowest

hi,

what occurs during a solar eclipse?

there are small tides across earth.

the moon is between earth and the sun.

p.s. im not sure if the first answer is correct because during a solar ecliphse there are spring tides and i dont know if they are the same.