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°
Answer:
L = a 1,929 10⁴ m
a = 0.1 mm = 0.1 10⁻³ m, L = 1,929 m
Explanation:
The diffraction phenomenon is described by the expression
a sin θ = m λ
Let's use trigonometry to find the breast
tan θ = y / x
As the angles are very small
tan θ = sin θ/ cos θ = sin θ = y / x
We replace
a y / L = m λ
L = a y / m λ
The red light has a wavelength of Lam = 700 nm = 700 10⁻⁹ m, in the third pattern it is m = 3
L = a 4.05 10⁻² / (3 700 10⁻⁹)
L = a 1,929 10⁴ m
To give a specific value we must know the width of the slit, suppose a value of a = 0.1 mm = 0.1 10⁻³ m
L = 1,929 m
A peak = A Rms x Sq root 2
Therefore 3.6 x sq root of 2
A peak = 5.09
Stars having less mass collapses early than those with more mass. This can be explained by Einstein's equation E=mc².
According to this equation, mass of stars is converted into light due to thermonuclear reactions occuring in the core of star which acts as engine of the stars. This thermonuclear reactions keeps star alive. Thermonuclear reactions occurs slowly in massive stars hence massive stars live more than light stars.
Answer:
²₁H + ³₂He —> ⁴₂He + ¹₁H
Explanation:
From the question given above,
²₁H + ³₂He —> __ + ¹₁H
Let ⁿₐX be the unknown.
Thus the equation becomes:
²₁H + ³₂He —> ⁿₐX + ¹₁H
We shall determine, n, a and X. This can be obtained as follow:
For n:
2 + 3 = n + 1
5 = n + 1
Collect like terms
n = 5 – 1
n = 4
For a:
1 + 2 = a + 1
3 = a + 1
Collect like terms
a = 3 – 1
a = 2
For X:
n = 4
a = 2
X =?
ⁿₐX => ⁴₂X => ⁴₂He
Thus, the balanced equation is
²₁H + ³₂He —> ⁴₂He + ¹₁H
