The main cause of this is Friction. The more oil that is laid down, the less friction there is between the ball and the lane surface. The less friction, the harder it is for the bowler to send the ball in a curved path imparted by the spin that the bowler puts on the ball at the instant of release.
Work done = force * distance moved (in direction of the force)
force= mass* acceleration
force=58.1N
58.1*(5.8*10^4)
=3,369,800 J
Answer:
Approximately 1.62 × 10⁻⁴ V.
Explanation:
The average EMF in the coil is equal to
,
Why does this formula work?
By Faraday's Law of Induction, the EMF
induced in a coil (one loop) is equal to the rate of change in the magnetic flux
through the coil.
.
Finding the average EMF in the coil is similar to finding the average velocity.
.
However, by the Fundamental Theorem of Calculus, integration reverts the action of differentiation. That is:
.
Hence the equation
.
Note that information about the constant term in the original function will be lost. However, since this integral is a definite one, the constant term in
won't matter.
Apply this formula to this question. Note that
, the magnetic flux through the coil, can be calculated with the equation
.
For this question,
is the strength of the magnetic field.
is the area of the coil.
is the number of loops in the coil.
is the angle between the field lines and the coil. - At
, the field lines are parallel to the coil,
. - At
, the field lines are perpendicular to the coil,
.
Initial flux:
.
Final flux:
.
Average EMF, which is the same as the average rate of change in flux:
.
A) 
The energy of an x-ray photon used for single dental x-rays is

The energy of a photon is related to its wavelength by the equation

where
is the Planck constant
is the speed of light
is the wavelength
Re-arranging the equation for the wavelength, we find

B) 
The energy of an x-ray photon used in microtomography is 2.5 times greater than the energy of the photon used in part A), so its energy is

And so, by using the same formula we used in part A), we can calculate the corresponding wavelength:
