Posted by **mark** on Wednesday, April 11, 2012 at 10:35pm.

Where must an object be placed to have a

magniﬁcation of 3.74 in front of a converging

lens of focal length 14.1 cm?

- physics -
**drwls**, Wednesday, April 11, 2012 at 10:51pm
Let do be the object distance and di be the image distance.

di/do = 3.74, the magnification.

1/do + 1/di = 1/do + 1/(3.74 do)

= 4.74/(3.74 do)

= 1.267/do = 1/14.1

Solve for do.

- Physics -
**John**, Thursday, April 12, 2012 at 8:48pm
Three identical small Styrofoam balls of mass

1.53 g are suspended from a ﬁxed point by

three nonconducting threads, each with a

length of 56.4 cm and negligible mass. At

equilibrium the three balls form an equilateral triangle with sides of 31 cm.

What is the common charge carried by

each ball? The acceleration of gravity is

9.8 m/s

2

and the value of Coulomb’s constant

is 8.98755 × 10

9

N · m2

/C

2

.

Answer in units of µC

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