Answer:
33 N
Explanation:
v = Velocity of fluid = 8+2 = 10 m/s
= Density of fluid = 1.2 kg/m³
C = Coefficient of drag = 1.1
A = Cross sectional area = 0.5 m²
Drag force is given by
The drag force on the athlete is 33 N
The amount of metal in a closed cylindrical can that is 10 cm high and 4 cm in diameter if the metal on the top and the bottom is 0.1 cm thick and the metal on the sides is 0.05 cm thick is 8.8 cm.
The formula for calculating the volume of a cylinder is given below.
V = πr^2 h
Get the differential of the volume as shown:
dV = V/ h dh + V / r dr
V/ h = πr^2
V/ h = 2 πr h
Now, the differential becomes
dV = πr^2dh + 2πrh dr
Given the following parameters i.e. diameter and height
dh = 0.1 + 0.1 = 0.2 cm
dr = 0.05 cm
h = 10 cm
d = 4 cm
r = 2cm
Substituting the values in the above equation, we get
dV = 3.14(2)^2(0.2) + 2(3.14)(2)(10)(0.05)
dV = 2.512 + 6.28
dV = 8.792 cm
dV = 8.8 cm
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Answer:
Given:
Object distance (u) = 25.0 cm
Image distance (v) = -50.0 cm
To Find:
Magnification (m)
Explanation:
Substituting values of Image distance(v) & Object distance (u) in the equation:
-(-50) = 50:
