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
Answer:
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
Explanation:
The orbital period of a planet around a star can be expressed mathematically as;
T = 2π√(r^3)/(Gm)
Where;
r = radius of orbit
G = gravitational constant
m = mass of the star
Given;
Let R represent radius of earth orbit and r the radius of planet orbit,
Let M represent the mass of sun and m the mass of the star.
r = 4R
m = 16M
For earth;
Te = 2π√(R^3)/(GM)
For planet;
Tp = 2π√(r^3)/(Gm)
Substituting the given values;
Tp = 2π√((4R)^3)/(16GM) = 2π√(64R^3)/(16GM)
Tp = 2π√(4R^3)/(GM)
Tp = 2 × 2π√(R^3)/(GM)
So,
Tp/Te = (2 × 2π√(R^3)/(GM))/( 2π√(R^3)/(GM))
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
The image formed by a plane mirror is virtual, upright and the same size with the actual object. The upright image of an object in a plane mirror is can be found on the other side of the mirror which is why it is also virtual.
