posted by mark .
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?
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.
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
and the value of Coulomb’s constant
is 8.98755 × 10
N · m2
Answer in units of µC