By Newton's second law, the net vertical force acting on the object is 0, so that
<em>n</em> - <em>w</em> = 0
where <em>n</em> = magnitude of the normal force of the surface pushing up on the object, and <em>w</em> = weight of the object. Hence <em>n</em> = <em>w</em> = <em>mg</em> = 196 N, where <em>m</em> = 20 kg and <em>g</em> = 9.80 m/s².
The force of static friction exerts up to 80 N on the object, since that's the minimum required force needed to get it moving, which means the coefficient of <u>static</u> friction <em>µ</em> is such that
80 N = <em>µ</em> (196 N) → <em>µ</em> = (80 N)/(196 N) ≈ 0.408
Moving at constant speed, there is a kinetic friction force of 40 N opposing the object's motion, so that the coefficient of <u>kinetic</u> friction <em>ν</em> is
40 N = <em>ν</em> (196 N) → <em>ν</em> = (40 N)/(196 N) ≈ 0.204
And so the closest answer is C.
(Note: <em>µ</em> and <em>ν</em> are the Greek letters mu and nu)
When a force causes a body to move, work is done on the object by the force. Work is the measure of the energy transfer when a force 'F' moves an object through a distance 'd'. So we say that energy is transferred from one energy store to another when work is done, and therefore, energy transferred = work done.
Answer:
A. 1.64 J
Explanation:
First of all, we need to find how many moles correspond to 1.4 mg of mercury. We have:

where
n is the number of moles
m = 1.4 mg = 0.0014 g is the mass of mercury
Mm = 200.6 g/mol is the molar mass of mercury
Substituting, we find

Now we have to find the number of atoms contained in this sample of mercury, which is given by:

where
n is the number of moles
is the Avogadro number
Substituting,
atoms
The energy emitted by each atom (the energy of one photon) is

where
h is the Planck constant
c is the speed of light
is the wavelength
Substituting,

And so, the total energy emitted by the sample is

Speed is different from velocity. Velocity is a vector quantity and has a direction. Speed is a scalar quantity and does not require a direction. The answer must be D).