The final velocity is a vector quantity that measures the speed and direction of a moving body after it has reached its maximum acceleration
C 1.6 amps hipe this helps
Answer:
The smallest distance the student that the student could be possibly be from the starting point is 6.5 meters.
Explanation:
For 2 quantities A and B represented as
and 
The sum is represented as
For the the values given to us the sum is calculated as
Now the since the uncertainity inthe sum is 
The closest possible distance at which the student can be is obtained by taking the negative sign in the uncertainity
Thus closest distance equals 
meters
You asked a question. I'm about to answer it.
Sadly, I can almost guarantee that you won't understand the solution.
This realization grieves me, but there is little I can do to change it.
My explanation will be the best of which I'm capable.
Here are the Physics facts I'll use in the solution:
-- "Apparent magnitude" means how bright the star appears to us.
-- "Absolute magnitude" means the how bright the star WOULD appear
if it were located 32.6 light years from us (10 parsecs).
-- A change of 5 magnitudes means a 100 times change in brightness,
so each magnitude means brightness is multiplied or divided by ⁵√100 .
That's about 2.512... .
-- Increasing magnitude means dimmer.
Decreasing magnitude means brighter.
+5 is 10 magnitudes dimmer than -5 .
-- Apparent brightness is inversely proportional to the square
of the distance from the source (just like gravity, sound, and
the force between charges).
That's all the Physics. The rest of the solution is just arithmetic.
____________________________________________________
-- The star in the question would appear M(-5) at a distance of
32.6 light years.
-- It actually appears as a M(+5). That's 10 magnitudes dimmer than M(-5),
because of being farther away than 32.6 light years.
-- 10 magnitudes dimmer is ( ⁵√100)⁻¹⁰ = (100)^(-2) .
-- But brightness varies as the inverse square of distance,
so that exponent is (negative double) the ratio of the distances,
and the actual distance to the star is
(32.6) · (100)^(1) light years
= (32.6) · (100) light years
= approx. 3,260 light years . (roughly 1,000 parsecs)
I'll have to confess that I haven't done one of these calculations
in over 50 years, and I'm not really that confident in my result.
If somebody's health or safety depended on it, or the success of
a space mission, then I'd be strongly recommending that you get
a second opinion.
But, quite frankly, I do feel that mine is worth the 5 points.
Explanation:
According to Rydberg's formula, the wavelength of the balmer series is given by:
R is Rydberg constant for an especific hydrogen-like atom, we may calculate R for hydrogen and deuterium atoms from:
Here, 
is the "general" Rydberg constant, 
is electron's mass and M is the mass of the atom nucleus
For hydrogen, we have, 
:
Now, we calculate the wavelength for hydrogen:
![\frac{1}{\lambda}=R_H(\frac{1}{2^2}-\frac{1}{3^2})\\\lambda=[R_H(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.0967*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5646*10^{-7}m=656.46nm](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Clambda%7D%3DR_H%28%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B3%5E2%7D%29%5C%5C%5Clambda%3D%5BR_H%28%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B3%5E2%7D%29%5D%5E%7B-1%7D%5C%5C%5Clambda%3D%5B1.0967%2A10%5E7m%5E%7B-1%7D%28%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B3%5E2%7D%29%5D%5E%7B-1%7D%5C%5C%5Clambda%3D6.5646%2A10%5E%7B-7%7Dm%3D656.46nm)
For deuterium, we have 
:
![R_D=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{2*1.67*10^{-27}kg})}\\R_D=1.09707*10^7m^{-1}\\\\\lambda=[R_D(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.09707*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5629*10^{-7}=656.29nm](https://tex.z-dn.net/?f=R_D%3D%5Cfrac%7B1.09737%2A10%5E7m%5E%7B-1%7D%7D%7B%281%2B%5Cfrac%7B9.11%2A10%5E%7B-31%7Dkg%7D%7B2%2A1.67%2A10%5E%7B-27%7Dkg%7D%29%7D%5C%5CR_D%3D1.09707%2A10%5E7m%5E%7B-1%7D%5C%5C%5C%5C%5Clambda%3D%5BR_D%28%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B3%5E2%7D%29%5D%5E%7B-1%7D%5C%5C%5Clambda%3D%5B1.09707%2A10%5E7m%5E%7B-1%7D%28%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B3%5E2%7D%29%5D%5E%7B-1%7D%5C%5C%5Clambda%3D6.5629%2A10%5E%7B-7%7D%3D656.29nm)
