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
and the value of Coulomb’s constant
is 8.98755 × 10
N · m2
Answer in units of µC
Answer This Question
More Related Questions
- Physics - Where must an object be placed to have a magniﬁcation of 3.74 ...
- physics - Hi! could any help me please, thank you. 1.An object is placed 17.8 cm...
- Physics - A lighted candle is placed 38 cm in front of a diverging lens. The ...
- Physics - two lenses,one converging with focal length 20.0 cm and one diverging ...
- physics - Please, anyone help me please. Q: An object is placed 17.8 cm from a ...
- Physics - Two converging lenses are placed 31.0 cm apart. The focal length of ...
- physics - Two converging lenses, having focal length 5 cm and 10 cm respectively...
- physics - The Yerkes refracting telescope has a 1 m diameter objective lens of ...
- Physics 2 - A converging lens with a focal length of 19 cm is used to form an ...
- Science Important - 1. A toy of height 18.4 cm is balanced in front of a ...