Answer:
total distance = 1868.478 m
Explanation:
given data
accelerate = 1.68 m/s²
time = 14.2 s
constant time = 68 s
speed = 3.70 m/s²
to find out
total distance
solution
we know train start at rest so final velocity will be after 14 .2 s is
velocity final = acceleration × time ..............1
final velocity = 1.68 × 14.2
final velocity = 23.856 m/s²
and for stop train we need time that is
final velocity = u + at
23.856 = 0 + 3.70(t)
t = 6.44 s
and
distance = ut + 1/2 × at² ...........2
here u is initial velocity and t is time for 14.2 sec
distance 1 = 0 + 1/2 × 1.68 (14.2)²
distance 1 = 169.37 m
and
distance for 68 sec
distance 2= final velocity × time
distance 2= 23.856 × 68
distance 2 = 1622.208 m
and
distance for 6.44 sec
distance 3 = ut + 1/2 × at²
distance 3 = 23.856(6.44) - 0.5 (3.70) (6.44)²
distance 3 = 76.90 m
so
total distance = distance 1 + distance 2 + distance 3
total distance = 169.37 + 1622.208 + 76.90
total distance = 1868.478 m
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)
Explanation:
Neutral carbon-12 (or any carbon atom) has 6 electrons with a total negative charge of 6e- orbiting a nucleus with a total positive charge of 6e+, so that the total net charge is zero. The nucleus is made up of 6 protons, each with a positive charge of e+, and 6 neutrons, each with zero charge.
Straight linear path
Explanation:
If the universe were completely empty with just one object which is solid moving sphere through the space at a speed of 100km/s, it will move along on a straight path.
An object will continue on a straight path or at rest unless acted upon by an external force. This is the first law of newton.
- Since there is no other mass or body in the universe, it will move along in a straight line.
- Our planet and other bodies moves in parabolic path and elliptical orbits because they are attracted by the gravitational forces of other bodies.
- Any body with mass will attract other bodies and makes it move round it.
- The body with the less mass will orbit in the vicinity of the larger mass.
- But if we have just one body out there, there is no attraction from any other body, it will continue along it's straight path.
learn more:
Gravitational force brainly.com/question/4417434
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