There is no <span>radioactive decay</span>
Answer:
A u = 0.36c B u = 0.961c
Explanation:
In special relativity the transformation of velocities is carried out using the Lorentz equations, if the movement in the x direction remains
u ’= (u-v) / (1- uv / c²)
Where u’ is the speed with respect to the mobile system, in this case the initial nucleus of uranium, u the speed with respect to the fixed system (the observer in the laboratory) and v the speed of the mobile system with respect to the laboratory
The data give is u ’= 0.43c and the initial core velocity v = 0.94c
Let's clear the speed with respect to the observer (u)
u’ (1- u v / c²) = u -v
u + u ’uv / c² = v - u’
u (1 + u ’v / c²) = v - u’
u = (v-u ’) / (1+ u’ v / c²)
Let's calculate
u = (0.94 c - 0.43c) / (1+ 0.43c 0.94 c / c²)
u = 0.51c / (1 + 0.4042)
u = 0.36c
We repeat the calculation for the other piece
In this case u ’= - 0.35c
We calculate
u = (0.94c + 0.35c) / (1 - 0.35c 0.94c / c²)
u = 1.29c / (1- 0.329)
u = 0.961c
The velocity when function p(t)=11 is 8 .
According to the question
The position of a car at time t represented by function :
Now,
When function p(t) = 11 , t will be
11 = t²+2t-4
0 = t² + 2t - 15
or
t² +2t-15 = 0
t² +(5-3)t-15 = 0
t² +5t-3t-15 = 0
t(t+5)-3(t+5) = 0
(t-3)(t+5) = 0
t = 3 , -5
as t cannot be -ve as given ( t≥0)
so,
t = 3
Now,
the velocity when p(t)=11
As we know velocity =
therefore to get the value of velocity from function p(t)
we have to differentiate the function with respect to time
v(t) = 2t + 2
where v(t) = velocity at that time
as t = 3 for p(t)=11
so ,
v(t) = 2t + 2
v(t) = 2*3 + 2
v(t) = 8
Hence, the velocity when function p(t)=11 is 8 .
To know more about function here:
brainly.com/question/12431044
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Explanation :
When an electron jumps from one energy level to another, the energy of atom gets changed.
If a photon gets absorbed, the electron will move to higher energy levels and then fall back to the lower energy levels. Then each time a photon will be absorbed whose energy is given by difference between the initial and final energy levels i.e
In Balmer series, the transition is from higher energy levels to n = 2.
So, the necessary condition for Balmer series is that the electron should be at first excited state or n = 2 level as shown in figure.