{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

qz2sp09sol - PHY2054 Discussion-Spring 09 Quiz 2(Chapter...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
January 27, 2009 PHY2054 Discussion-Spring ‘09 Quiz 2 (Chapter 16.1-16.8) Name: UFID: **1. (2.5pts) On planet Tehar, the free-fall acceleration is the same as that on Earth, but there is also a strong vertical electric field that is uniform close to the planet's surface. A 1.00-kg ball with a charge of -5.00×10 -4 C is thrown straight up at a speed of 18.0 m/s from the ground. It hits the ground 3.00 s later. What is the potential difference between the starting point and the top point of the trajectory? (Calculate V top -V bottom .) When the ball hits the ground, the vertical displacement is zero. Thus we have Δy = v 0 t+(1/2)at 2 0 = v 0 t+(1/2)at 2 a = -2 v 0 /t = -2×18/3 = -12.0 m/s 2 Using Newton’s second law, we calculate the electric field. ma = qE-mg E = (g+a)/q = (9.8-12)/(-5×10 -4 ) = 4.40×10 3 N/C The velocity of the ball zero at the peak of the motion. Therefore, the maximum height is v 2 = v 0 2 +2aΔy 0 = v 0 2 +2aΔy Δy = - v 0 2 /2a = -18 2 /2×(-12) = 13.5 m. The potential difference is ΔV = -E y Δy = -4.4×10 3 ×13.5 = -5.94×10 4 V *2. (2.5pts) Three charges are situated at corners of a rectangles shown as below. How much energy
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern