What Is the Frictional Force Acting on a Skydiver at Constant Velocity?

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A skydiver descends at a constant velocity of 2 m/s, with a total mass of 70 kg. The gravitational force acting on the skydiver is calculated to be 686 N. Since the skydiver is at constant velocity, the net force is zero, indicating that the air's frictional force equals the gravitational force. The discussion emphasizes that the acceleration is effectively zero, allowing for the determination of the frictional force without needing distance or time. Understanding that the forces are balanced is key to solving the problem.
frigid
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A skydiver jumps from an airplane and descends at a constant velocity of 2m/s. The mass of the skydiver and his equipment is 70kg. Calculate the magnitude of the air's frictional force acting on the skydiver and his equipment.

F = 70kg x 9,8N/kg = 686N

I'm lost
 
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Can anyone please help me with this one real quick.

I know that Fnet is Fa - Ff = MA
His mass is 70kg and velocity 2m/s. But since i don't have the distance of his fall or the time i can't calculate the acceleration. So how do I figure out the magnitude of the air's frictional force which would be Ff?

:cry: :cry: :cry: :cry: :cry:
 
frigid said:
But since i don't have the distance of his fall or the time i can't calculate the acceleration.
Reread the problem statement carefully: the acceleration is given.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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