# Final Exam Questions

1. May 8, 2006

2. May 8, 2006

### Staff: Mentor

Draw the diagram showing the force on the valve stem due to gravity. What is the angle between the force on the valve stem and the radius from the axle to the valve stem? Do you see why the answer to question 5 also involves this angle?

3. May 8, 2006

### daveb

If you know the equation for torque (a vector quantity), then all you have to do is find the magnitude.

4. May 8, 2006

### Alt+F4

Last edited: May 8, 2006
5. May 8, 2006

6. May 8, 2006

7. May 8, 2006

### Staff: Mentor

Your logic is faulty. Think about this: during period A you earned $100/hour; during period B you earned$1/hour. Are you sure that your average earning rate is (101)/2 = \$50.50/hour? What if period A lasted 1 hour but period B lasted 10 years?

The right way to find average velocity is displacement/time.

8. May 8, 2006

### Alt+F4

I want to thank you for all your help, I aced all the exams, just this final and i am done. Thanks

9. May 8, 2006

### Staff: Mentor

10. May 8, 2006

### Alt+F4

Horizontal Velocity = 12 Cos 65 = 5.0714

Acc = -9.8

What is confusin me is the Distance factor The Hoop is 3 meters what do i do with that

11. May 8, 2006

### Staff: Mentor

Here's one way to look at it. To get across as fast as possible, just swim straight across (with respect to the water, which will drag you downstream, but who cares?). How long does it take the water to drag the swimmer the given distance downstream? Since he swims across the river in that same time, what must be his swimming speed (with respect to the water)?

12. May 8, 2006

### Staff: Mentor

Right (in m/s).

Wrong! (What's the horizontal acceleration?)

What you do with it is ignore it. You don't need it. Focus on the horizontal component of the motion. You have the speed and the distance.

13. May 8, 2006

### Alt+F4

am i making it harder by assuming Vectors and Just figuring it out there cause the triangle is 45 45 90, i can find the angle then from there i can find the speed

14. May 8, 2006

### Staff: Mentor

You can certainly use vectors, but that's more work than needed. I don't see a 45-45-90 triangle.

15. May 8, 2006

### Alt+F4

16. May 9, 2006

### Staff: Mentor

17. May 9, 2006

### Alt+F4

18. May 9, 2006

### Staff: Mentor

Looks good to me. The person's speed with respect to the water is 2.5 m/s. (If you wanted to find the person's speed with respect to the shore, you'd have to add the cross-stream (2.5 m/s) and down-stream (1.8 m/s) components--using vector addition--to find his total speed. But you weren't asked that.)

19. May 9, 2006

### Alt+F4

20. May 9, 2006

### Alt+F4

So like (1.8^2) + (2.5^2) = X^2