Answer:
a. Power requirement, P = 30.63 W
b. Force, F = 15.68N
Explanation:
Given.
Velocity = v = 150km/hr
v = 150km/hr / 3600hr/1000km (m/s)
v = 150 / 3.6 m/s
v = 41.67m/s
Density = d = 1.225kg/m³ (at standard temperature pressure)
a. Power required to tow the banner.
Power required to tow the banner is given as FV.
Where F = the drag force and V ,= velocity
The drag force is calculated as:
F = ½dACa
Where d = bℓ = 0.06
A = Area of banner (length = 25m, height = 0.8m)
A = 25 * 0.8
A = 20m²
Cd = Coefficient of drag = density = 1.225
So, F = ½* 0.06 * 20 * 1.225
F = 0.735N
Power = FV = 0.735 * 41.67
Power = 30.62745W
Power = 30.63W -- Approximated
b. Force required to tow the banner on a rigid surface
The drag force is calculated as:
F = ½dACa
Where d = bℓ = 0.06
A = Area of banner (length = 25m, height = 0.8m)
A = 25 * 0.8
A = 20m²
Cd = Coefficient of drag = density = 1.28
So, F = ½* 1.28 * 20 * 1.225
F = 15.68N
A) The estimated power required to tow the banner is : 30.63 watts
B) The estimated drag force on the banner is : 0.768 N
Given data
Height of banner ( b ) = 0.8 m
Length ( ℓ ) of banner = 25.0 m
velocity = 150 km/hr = 41.67m/s
Density ( d ) = 1.225 kg/m³
Given that the drag coefficient = 0.06m
Power = F * V ------- ( 1 )
F = drag force, V = velocity
F ( drag force ) = ½ *d * A* Cd
where : d = 0.06 , A = 20 m² , Cd = 1.225
therefore F = 1/2 * 0.06 * 20 * 1.225
= 0.735 N
back to equation ( 1 )
Power = 0.735 * 41.67
≈ 30.63 watts
Drag force ( F ) = ½ *d * A* Cd
= ½* 0.06 * 20 * 1.28
= 0.768 N
Note :
Cd for a rigid flat plate = 1.28
Hence we can conclude that The estimated power required to tow the banner is : 30.63 watts and The estimated drag force on the banner is : 0.768 N
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