SolutionsPracticeExamQuestions

# SolutionsPracticeExamQuestions - Solutions to Physics 100...

• Test Prep
• 11
• 100% (2) 2 out of 2 people found this document helpful

This preview shows page 1 - 10 out of 11 pages.

Solutions to Physics 100 Practice Exam Questions Answer is c. Using kinematics (v f 2 - v i 2 =2ah), one finds the height proportional to the square of the (initial) speed. Thus, doubling the speed quadruples the height.

Subscribe to view the full document.

Subscribe to view the full document.

Answers: (a), (b), (d)
Answer: Acceleration is down, and its magnitude is 0.5 m/s 2 . Use F net = n W = ma, where the convention is positive up. Since n = 570 and W = 600 N, a is negative, indicating down. Use total charge = (Number of ions)*(charge of one ion) = current * time. When open: A > D = E > B = C When closed: E > A > B = C > D

Subscribe to view the full document.

Since all the light bulbs have the same current, to compare the power it is enough to compare the current running through each. When switch is open, no current runs through it. When closed, no current runs through C. Answer: 15 W Use P=I 2 R for each resistor; add their contributions. Same current runs through them. The latter is determined by using the voltage across resistor one, and its resistance. Total current is 3 A. From R 1 and I 1 you can find V 1 . The latter is the same as V 2 . Use V 2 to find I 2 . The total current is the sum of I 1 and I 2 .

Subscribe to view the full document.

(a) When sled + brother are accelerating, you is still pushing. Thus, as indicated on the right.

Subscribe to view the full document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern