Answer:
time will elapse before it return to its staring point is 23.6 ns
Explanation:
given data
speed u = 2.45 × 
m/s
uniform electric field E = 1.18 × 
N/C
to find out
How much time will elapse before it returns to its starting point
solution
we find acceleration first by electrostatic force that is
F = Eq
here
F = ma by newton law
so
ma = Eq
here m is mass , a is acceleration and E is uniform electric field and q is charge of electron
so
put here all value
9.11 × 
kg ×a = 1.18 × × 1.602 × 
a = 20.75 × 
m/s²
so acceleration is 20.75 × m/s²
and
time required by electron before come rest is
use equation of motion
v = u + at
here v is zero and u is speed given and t is time so put all value
2.45 × = 0 + 20.75 × (t)
t = 11.80 × 
s
so time will elapse before it return to its staring point is
time = 2t
time = 2 ×11.80 ×
time is 23.6 × s
time will elapse before it return to its staring point is 23.6 ns
Wave speed = (wavelength) x (frequency)
= (45 meters) x (9 per second)
= 405 meters per second .
The path of the raction occurs on the basis of mass of the nuclei involved in reaction.
In case of nuclear fusion, two or more nuclei having less mass fuse(combine, join) together to form a new nuclei(heavier mass but it is relatively stable). During fusion, matter is not conserved because some of the matter is converted into energy(light). This reaction evolves a huge amount of energy and there comes Einstein's famous Energy-mass equivalence formula E=mc^2! :D. The nuclear reaction occuring in stars(including our sun ) is "fusion".
Fission occurs with heavier nuclei such as that of Uranium-235. Which splits into smaller subatomic particles like gamma, neutrons and enormous amount of energy.
Both, Fission and Fusion releases enormous amount of energy and modern nuclear weapons works on the principle of nuclear fission.
Answer:
Explanation:
Given that
B(y, t) = k y ³t²
To find the total flux over the loop we have to integrate over the loop
Given that loop is square,so
B(y, t) = k y ³t²
We know that emf given as
So
True because the speed of a mass can change it's gravitational potential energy in the gravitational field of another massive object.
