{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Hwk4Solns

# Hwk4Solns - Assignment 4 Solutions Chpt 26 1(a The charge...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Assignment 4 Solutions Chpt 26 1. (a) The charge that passes through any cross section is the product of the current and time. Since t = 4.0 min = (4.0 min)(60 s/min) = 240 s, q = it = (5.0 A)(240 s) = 1.2 × 10 3 C. (b) The number of electrons N is given by q = Ne , where e is the magnitude of the charge on an electron. Thus, N = q / e = (1200 C)/(1.60 × 10 –19 C) = 7.5 × 10 21 . 23. The resistance of conductor A is given by 2 A A r L R π ρ = , where r A is the radius of the conductor. If r o is the outside diameter of conductor B and r i is its inside diameter, then its cross-sectional area is π ( r o 2 – r i 2 ), and its resistance is ) ( 2 2 i B r r L R − = π ρ The ratio is . 3 ) 50 . ( ) 50 . ( ) . 1 ( 2 2 2 2 2 2 = − = − = mm mm mm r r r R R A i B A . 26. Let be the radius of the kite string and 2.00 mm r = 0.50 mm t = be the thickness of the water layer. The cross-sectional area of the layer of water is ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

Hwk4Solns - Assignment 4 Solutions Chpt 26 1(a The charge...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online