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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
Question
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
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Physics
4 years
2021-08-12T22:47:17+00:00
2021-08-12T22:47:17+00:00 2 Answers
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Answers ( )
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
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