Answer:
3.57 MJ
Explanation:
ASSUMING it's fresh water with density of 1000 kg/m³
W = ΔPE = mgΔh = 14.0(1000)(9.81)(26.0) = 3,570,840 J
Salt water would require more.
Answer:
a) 
For this case we know the following values:




So then if we replace we got:

b) 
With 
And replacing we have:

And then the scattered wavelength is given by:

And the energy of the scattered photon is given by:

c) 
Explanation
Part a
For this case we can use the Compton shift equation given by:
For this case we know the following values:
So then if we replace we got:
Part b
For this cas we can calculate the wavelength of the phton with this formula:
With
And replacing we have:
And then the scattered wavelength is given by:
And the energy of the scattered photon is given by:
Part c
For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:
We make a graphic of this problem to define the angle.
The angle we can calculate through triangle relation, that is,

With this function we should only calculate the derivate in function of c

That is the rate of change of
.
b) At this point we need only make a substitution of 0 for c in the equation previously found.

Hence we have finally the rate of change when c=0.
Answer:
1531 m
Explanation:
The motion of the jet ski is an uniformly accelerated motion, so we can find the distance travelled by using the following suvat equation:

where
s is the distance
u is the initial velocity
t is the time
a is the acceleration
For the jet ski in this problem,

t = 35 s
u = 0 (it starts from rest)
Solving for s, we find the distance travelled:

Answer:
4.7 is 10 as much as the number 0.47.
If you multiply 0.47 x 10 it will equal 4.7
Explanation: