Answer:
Yes , it is possible for two different atoms of carbon to have different numbers of neutrons in their nuclei .
Explanation:
Isotopes -
Atoms of the element with same number of the electrons and protons , but differ in the number of neutrons , are called as the isotopes .
The isotopes of the element have the same number of protons and electrons , hence have the same chemical and physical properties .
Many isotopes occurs naturally .
In case of Carbon ,
Carbon too have isotopes ,
i.e.
Carbon - 13 and Carbon - 14
carbon - 13 , have seven neutrons
Carbon - 14 , have eight neutrons .
it is just a matter of integration and using initial conditions since in general dv/dt = a it implies v = integral a dt
v(t)_x = integral a_{x}(t) dt = alpha t^3/3 + c the integration constant c can be found out since we know v(t)_x at t =0 is v_{0x} so substitute this in the equation to get v(t)_x = alpha t^3 / 3 + v_{0x}
similarly v(t)_y = integral a_{y}(t) dt = integral beta - gamma t dt = beta t - gamma t^2 / 2 + c this constant c use at t = 0 v(t)_y = v_{0y} v(t)_y = beta t - gamma t^2 / 2 + v_{0y}
so the velocity vector as a function of time vec{v}(t) in terms of components as[ alpha t^3 / 3 + v_{0x} , beta t - gamma t^2 / 2 + v_{0y} ]
similarly you should integrate to find position vector since dr/dt = v r = integral of v dt
r(t)_x = alpha t^4 / 12 + + v_{0x}t + c let us assume the initial position vector is at origin so x and y initial position vector is zero and hence c = 0 in both cases
r(t)_y = beta t^2/2 - gamma t^3/6 + v_{0y} t + c here c = 0 since it is at 0 when t = 0 we assume
r(t)_vec = [ r(t)_x , r(t)_y ] = [ alpha t^4 / 12 + + v_{0x}t , beta t^2/2 - gamma t^3/6 + v_{0y} t ]
Answer: a) vcar= 7 m/s ; b) a train= 0.65 m/s^2
Explanation: By using the kinematic equation for the car and the train we can determine the above values of the car velocity and the acceletarion of the train, respectively.
We have for the car
distance = v car* t, considering the length of train (81.1 m) travel by the car during the first 11.6 s
the v car = distance/time= 81.1 m/11.6s= 7 m/s
In order to calculate the acceleration we have to use the kinematic equation for the train from the rest
distance train = (a* t^2)/2
distance train : distance travel by the car at constant speed
so distance train= (vcar*36.35)m=421 m
the a traiin= (2* 421 m)/(36s)^2=0.65 m/s^2
Bergeron–Findeisen Process.
