EBK ASTRONOMY TODAY
EBK ASTRONOMY TODAY
9th Edition
ISBN: 9780134583662
Author: Chaisson
Publisher: PEARSON CO
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Chapter 2, Problem 11D
To determine

To explain: The Newton's laws of motion and the gravity.

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Review Conceptual Example 3 and the drawing as an aid in solving this problem. A conducting rod slides down between two frictionless vertical copper tracks at a constant speed of 3.9 m/s perpendicular to a 0.49-T magnetic field. The resistance of th rod and tracks is negligible. The rod maintains electrical contact with the tracks at all times and has a length of 1.4 m. A 1.1-Q resistor is attached between the tops of the tracks. (a) What is the mass of the rod? (b) Find the change in the gravitational potentia energy that occurs in a time of 0.26 s. (c) Find the electrical energy dissipated in the resistor in 0.26 s.
A camera lens used for taking close-up photographs has a focal length of 21.5 mm. The farthest it can be placed from the film is 34.0 mm. (a) What is the closest object (in mm) that can be photographed? 58.5 mm (b) What is the magnification of this closest object? 0.581 × ×
Given two particles with Q = 4.40-µC charges as shown in the figure below and a particle with charge q = 1.40 ✕ 10−18 C at the origin. (Note: Assume a reference level of potential V = 0 at r = ∞.) Three positively charged particles lie along the x-axis of the x y coordinate plane.Charge q is at the origin.Charge Q is at (0.800 m, 0).Another charge Q is at (−0.800 m, 0).(a)What is the net force (in N) exerted by the two 4.40-µC charges on the charge q? (Enter the magnitude.) N(b)What is the electric field (in N/C) at the origin due to the two 4.40-µC particles? (Enter the magnitude.) N/C(c)What is the electrical potential (in kV) at the origin due to the two 4.40-µC particles? kV(d)What If? What would be the change in electric potential energy (in J) of the system if the charge q were moved a distance d = 0.400 m closer to either of the 4.40-µC particles?
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