Answer:
1.776 x 10^-19 m
Explanation:
Energy, E = 7 TeV
Let λ be the wavelength.
Energy = h c / λ
Where, h is the Planks,s constant and c be the velocity of light
h = 6.63 x 10-^-34 Js
c = 3 x 10^8 m/s
Convert TeV into J
1 TeV = 1.6 x 10^-7 J
So, E = 7 x 1.6 x 10^-7 = 11.2 x 10^-7 J
11.2 x 10^-7 = (6.63 x 10^-34 x 3 x 10^8) / λ
λ = 1.776 x 10^-19 m
Answer:
Explanation:
According to the law of conservation of energy, the total energy of the system can be expresed as the sum of the potential energy and kinetic energy:
When the spring is in its equilibrium position, that is , the object speed its maximum. So, we have:
In order to double its maximum speed, that is . We have:
Answer:
Explanation:
a ) It is given that bomb was at rest initially , so , its momentum before the explosion was zero.
b ) We shall apply law of conservation of momentum along x and y direction separately because no external force acts on the bomb.
If v be the velocity of the third part along a direction making angle θ
with x axis ,
x component of v = vcosθ
So momentum along x axis after explosion of third part = mv cosθ
= 10 v cosθ
Momentum along x of first part = - 5 x 42 m/s
momentum of second part along x direction =0
total momentum along x direction before explosion = total momentum along x direction after explosion
0 = - 5 x 42 + 10 v cosθ
v cosθ = 21
Similarly
total momentum along y direction before explosion = total momentum along y direction after explosion
0 = - 5 x 38 + 10 v sinθ
v sinθ= 21
squaring and and then adding the above equation
v² cos²θ +v² sin²θ = 21² +19²
v² = 441 + 361
v = 28.31 m/s
Tanθ = 21 / 19
θ = 48°
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