Is it point a) you are missing?OscarF said:
No point a) was the only one I could do...greg_rack said:
What could you state for a body traveling at terminal velocity, in terms of forces and acceleration acting on it?OscarF said:
They are balanced...greg_rack said:
I mean you would get that f=mg, but then how can we find out m?Delta2 said:
What do you mean? Be more precise.OscarF said:
You don't need the mass. Think in terms of forces and forget ##F=mg## for now.OscarF said:
We must take into account weight and air resistance (drag)greg_rack said:
That's right. So, in order to have that parachutist traveling at terminal velocity, you must have the weight equal in magnitude to the drag. Right?OscarF said:
yes... I still don't understand though how the weight is found out from that being that we are only given speed. Although, I guess that's the next step - I'm just stumped.greg_rack said:
The graph just tells you the intensity of the resistive force acting, depending on the speed at which the body is traveling.OscarF said:
Would that value be 600N?greg_rack said:
Which case are you referring to?OscarF said:
When the parachutist is traveling at 5/msgreg_rack said:
Yes, when the parachutist is traveling at 5m/s, the resistive force acting is ABOUT 600N(you can't tell exactly since you have no precise reference points/intersections in the graph).OscarF said:
so then if the weight were 600N, that point would be terminal velocity. I still don't understand why 600N MUST be the parachutist's weight though.greg_rack said:
That's exact.OscarF said:
but why is her weight 600N, if we had done this at a different point we would just be concluding that her weight is 800N for example.greg_rack said:
The given of the problem is that his particular skydiver's terminal velocity is 5 m/s. From this information we derive the skydivers weight as 600 N.OscarF said:
The statement is telling you that her terminal velocity is 5m/s, so her weight MUST be 600N because of how this problem is formulated.OscarF said:
oh ok, makes sense. Just quickly for question c, would the terminal velocities be 400N and 1000Ngreg_rack said:
For point c) you are being given the weights of two other parachutists.OscarF said:
great, makes sense! And finally, part d, is it basically just the same graph but with the drag values doubled?greg_rack said:
Exactly, you will have that graph vertically dilated:OscarF said:
Terminal velocity is the maximum speed that an object can reach when falling through a fluid, such as air or water. It occurs when the force of gravity pulling the object down is equal to the resistance of the fluid pushing against it.
The formula for calculating terminal velocity is: Vt = √(2mg/ρAC), where Vt is the terminal velocity, m is the mass of the object, g is the acceleration due to gravity, ρ is the density of the fluid, A is the cross-sectional area of the object, and C is the drag coefficient.
The weight of an object does not directly affect its terminal velocity. However, a heavier object will have a greater force of gravity pulling it down, which means it will need a larger force of resistance from the fluid to reach terminal velocity.
No, an object cannot have a terminal velocity in a vacuum because there is no fluid present to provide resistance. In a vacuum, an object will continue to accelerate until it reaches the maximum speed allowed by its environment, such as the escape velocity of a planet.
Air resistance, also known as drag, is the force that opposes the motion of an object through the air. As an object falls, the force of air resistance increases until it reaches a point where it is equal to the force of gravity, resulting in terminal velocity. The shape and size of the object, as well as the density and viscosity of the fluid, can all affect the amount of air resistance an object experiences.
