Refer to the diagram shown below.
The mass of the car is 1170 kg, therefore its weight is
W = (1170 kg)*(9.8 m/s²) = 11466 N
The component of the weight acting down the incline is
F = W sin(25°) = (11466 N)*sin(25°) = 4845.7 N
The normal reaction from the inclined plane is
N = W cos(25°) = (11466 N) cos(25°) = 1039.2 N
T = tension in the cable, acting at 31° above the surface of the ramp.
The Free Body Diagram on the right shows all the forces (friction is ignored)
and they FDB is sufficient for determining the value of T which establishes equilibrium.
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Answer:
None of the above forces on air drag on him is equal to his weight
Explanation:
In the velocity-time graph,the gradient of the curve where it is flatten shows the parachutist reaches the terminal velocity when it reaches terminal velocity which means the parachutist reaches constant velocity or speed,indicating that the acceleration of free fall(g) is zero.And according to the resultant force formula weight - air drag= mass*acceleration. so when accelerate is zero,resultant force is zero. And hence the equation will be like this: weight= air drag
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