Recent content by chengbin

Thank you guys. @RandomGuy88, this is so true. This is why I'm asking this equation in my other thread here https://www.physicsforums.com/showthread.php?p=3197469#post3197469 @HallsofIvy, The non-linear equation ends up involving tanh and stuff, which I didn't learn yet. From Google...

If a person has reached terminal velocity (say 50m/s), and he opens his parachute, and I want to find his speed 2 seconds after he opens his parachute, can I do this? Solve the equation m dv/dt = mg - kv^2 and substitute k with 1/2 pCdA and substitute the value for p, Cd, and A to find...

What is the difference between these two equations. m dv/dt = mg - kv and m dv/dt = mg - kv^2 as the equation modeling a falling body with air resistance?

The chute was open, and they were airborne with the chute open for 2 seconds before impact. I hope someone can answer my question of stoke's drag and if the variable b is the same as drag coefficient. If not, what is b, and how can I find a value of b for a parachute?

Could you please explain how? Drag is not taught in grade 11 physics, so I'm not too familiar with its calculation. I want to calculate if someone is falling at 50m/s, and he releases his parachute at t = 0, what is his velocity after 2 seconds. I'm guessing the drag coefficient has to be...

So for stokes drag calculation, would the drag coefficient be the "b" variable?

I'm trying to prove why it is impossible for James Bond to open the parachute 2 seconds before impact and still land relatively unharmed. My argument is that there is no way a parachute can slow Bond from 190km/h to about 40-50km/h in 2 seconds. I tried searching the web for information on...

Homework Statement A dragster reaches 350km/h from rest to 6.2s. If the car is 800kg and generates a driving force of 14000N, find the force of friction acting on the car. Homework Equations The Attempt at a Solution 350km/h = 97.2m/s a=\frac{97.2m/s - 0}{6.2s} = 15.7m/s^2...

Thank you! That solved the problem.

Homework Statement A puck of mass 30g slides across rough ice, experiencing a frictional force of 0.2N. If it was moving at 10km/h when it hit the ice patch, how long did it take to stop? How long was the ice patch? Homework Equations The Attempt at a Solution I don't think this question is...

Thanks. Apparently I can't use my calculator properly. I guess 170g is probably a typo in the book.

Homework Statement A net force of 200N is applied to an object, causing its velocity to change from 30km/h to 20km/h in 2.3s. Homework Equations What is the object's acceleration? What is its mass? The Attempt at a Solution I got the correct answer for acceleration (-1.2m/s^2)...

I know how to find the answer, but I need to show it, because that's half the marks.

The hypotenuse is 2 m/s, the horizontal side (water) is 1.5 m/s, and the vertical side is the velocity relative to the ground. \sqrt {(2.0)^2 - (1.3)^2} = 1.3 To find the angle, \sin \theta = \frac {1.5}{2} \theta = 49^{\circ}

Actually you do. The answer is actually 1.3 m/s at 49 degrees. Our teacher said that's a classic problem, and most people think the answer is 37 degrees, while it is actually 49. Therefore I need to show notation to show my understanding of how it works.