Answer:
a) 2.4 mm
b) 1.2 mm
c) 1.2 mm
Explanation:
To find the widths of the maxima you use the diffraction condition for destructive interference, given by the following formula:
a: width of the slit
λ: wavelength
m: order of the minimum
for little angles you have:
y: height of the mth minimum
a) the width of the central maximum is 2*y for m=1:
b) the width of first maximum is y2-y1:
c) and for the second maximum:
Answer:
a) m=20000Kg
b) v=0.214m/s
Explanation:
We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.
For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, , were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as
, and the mass of the first and second coals as
and
respectively
We start with the transition between parts A and B, so we have:
Which means
And since we want the mass of the first coal thrown () we do:
Substituting values we obtain
For the transition between parts B and C, we can write:
Which means
Since we want the new final speed of the car () we do:
Substituting values we obtain
Answer:
Given
initial velocity (u) =27.030
Force of gravidity g) =9.8
Rtc maximum height Hmix =?