We have that the electric field at the center of the metal ball due only to the charges on the surface of the metal ball is
From the question we are told that
A solid metal ball of radius 1.5 cm
bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing
uniformly distributed charge of -7 nC
The distance between the centers of the balls is 9 cm
Generally the equation for the electric field is mathematically given as
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#1). Anthony does the same amount of work as Angel, with <em>more power</em>.
#2). Power = (Work)/(Time) = 41,000 J / 500 s = <em>82 watts .</em>
#3). Power = (Work) / (Time) = 83 J / 3 sec = <em>27.7 watts</em>
Density = (mass) divided by (volume)
We know the mass (2.5 g). We need to find the volume.
The penny is a very short cylinder.
The volume of a cylinder is (π · radius² · height).
The penny's radius is 1/2 of its diameter = 9.775 mm.
The 'height' of the cylinder is the penny's thickness = 1.55 mm.
Volume = (π) (9.775 mm)² (1.55 mm)
= (π) (95.55 mm²) (1.55 mm)
= (π) (148.1 mm³)
= 465.3 mm³
We know the volume now. So we could state the density of the penny,
but nobody will understand what we have. Here it is:
mass/volume = 2.5 g / 465.3 mm³ = 0.0054 g/mm³ .
Nobody every talks about density in units of ' gram/(millimeter)³ ' .
It's always ' gram / (centimeter)³ '.
So we have to convert our number for the volume.
(0.0054 g/mm³) x (10 mm / cm)³
= (0.0054 x 1,000) g/cm³
= 5.37 g/cm³ .
This isn't actually very close to what the US mint says for the density
of a penny, but it's in a much better ball park than 0.0054 was.
Question 18: a
question 19: b
question 20: c
