13) The frequency of a vibrating string varies inversely as its length. A string 3 feet long vibrates 175 cycles per second. Find the frequency of a 5 foot string. 14) The force of the wind blowing on a vertical surface varies jointly as the area of the surface and the square of the velocity. If a wind blowing at 50 mph exerts a force of 75 pounds on a surface of 500 ft2, how much force will a wind of 75 mph place on a surface of 10 ft2?

13) f2 = 105 cycles/sec

14) F2 = 3.375 lbs

Explanation:

Part 13)

- The frequency (f) of a vibrating string varies inversely as its length (L):

f = k / L

Where, k = proportionality constant

- Set up proportionality constant to equations as follows:

k = f1*L1 = f2*L2

Where, f1 = 175 cycles/s, L1 = 3 ft

f2 = ?, L2 = 5 ft

- Using the relationship we determine f2:

f2 = f1*L1 / L2

f2 = 175*3/5

f2 = 105 cycles/sec

Part 14)

- The force (F) of the wind blowing on a vertical surface varies jointly as the area of the surface (A) and the square of the velocity (v^2):

F = k*A*v^2

F / A*v^2 = k

Where, k = proportionality constant

- Set up proportionality constant to equations as follows:

F1 / A1*v1^2 = F2 / A2*v2^2

Where, F1 = 75 lbs, A1 = 500 ft^2, v = 50 mph

F2 = ?, A2 = 10 ft^2, v = 75 mph

- Using the relationship we determine F2:

F2 = F1 * (A2*v2^2 / A1*v1^2)

F2 = 75 * (10*75^2 / 500*50^2)

F2 = 3.375 lbs