Answer:
Tension in the cable is T = 16653.32 N
Explanation:
Give data:
Cross section Area A = 1.3 m^2
Drag coefficient CD = 1.2
Velocity V = 4.3 m/s
Angle made by cable with horizontal =30 degree
Density
Drag force FD is given as
Drag force = 14422.2 N acting opposite to the motion
As cable made angle of 30 degree with horizontal thus horizontal component is take into action to calculate drag force
TCos30 = F_D
T = 16653.32 N
a) The kinetic energy (KE) of an object is expressed as the product of half of the mass (m) of the object and the square of its velocity (v²):
It is given:
v = 8.5 m/s
m = 91 kg
So:
b) We can calculate height by using the formula for potential energy (PE):
PE = m*g*h
In this case, h is eight, and PE is the same as KE:
PE = KE = 3,287.4 J
m = 91 kg
g = 9.81 m/s² - gravitational acceleration
h = ? - height
Now, let's replace those:
3,287.4= 91 * 9.81 * h
⇒ h = 3,287.4/(91*9.81) = 3,287.4/892.7 = 3.7 m