Wind resistance is jointly proportional to velocity and surface area of the object. If an object is traveling 45 mph and has a surface area of 49 square feet it will experience a wind resistance of 496.125 Newtons. How fast must an object be moving to experience a wind resistance of 164.25 Newtons if it has a surface area of 10 square feet

Respuesta :

Answer:

an object will be moving 73 mph to experience a wind resistance of 164.25 Newtons if it has a surface area of 10 square feet

Explanation:

given information:

object speed, v₁ = 45 mph

surface area, s₁ = 49 ft²

wind resistance, R₁ = 496.125 N

if

R₂ = 164.25 N

s₂ = 10 ft²

v₂ =  ?

first we need to find the k from the first case. we know that

R₁ = k v₁ s₁

k = R/(v s)

  = 496.125/(45 x 49)

  = 0.225

now we can find the v₂

R₂ = k v₂ s₂

v₂ = R₂/ k s₂

   = 164.25/(0.225 x 10)

   = 73 mph

Answer:

Explanation:

Given:

Velocity, v1 = 45 mph

Converting from mph to m/s,

45 mi/h × 1609.3 m/1 mi × 1 h/3600 s

= 20.12 m/s

surface area, A1 = 49 square feet

Converting ft^2 to m^2,

49 ft^2 × (0.305 m)^2/(1 ft)^2

= 4.56 m^2

Wind resistance, R1 = 496.125 Newtons.

Wind resistance, R2 = 164.25 Newtons

Surface area, A2 = 10 square feet

Converting ft^2 to m^2,

10 ft^2 × (0.305 m)^2/(1 ft)^2

= 0.93 m^2

Wind resistance, R is jointly proportional to velocity, v and surface area of the object, A

R = k (v × A)

R1/(v1 × A1) = R2/(v2 × A2)

v2 = R2 × (v1 × A1)/(R1 × A2)

= 164.25 × (20.12 × 4.56)/(496.125 × 0.93)

= 32.66 m/s

Converting from m/s to mph,

32.66 m/s × 1 mi/1609.3 m × 3600 s/1 h

= 73.06 mph