Energy released by the sun results from the process where atomic nuclei
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The only balanced equation is B.
If you look at the equation and break it down you can see that in:
→ 
Starting from the left side of the equation there are 2 Nitrogen atoms, and 2 oxygen atoms as indicated by the subscript.
To balance the equation, the number of atoms of each element in the right side equation should be equal to left. By putting the numerical coefficient of 2, you will distribute that to each element. So you will end up with 2 nitrogen atoms and 2 oxygen atoms on the left side of the equation. Thus, the equation is balanced.
The answer again, is B.
If you look at the equation and break it down you can see that in:
Starting from the left side of the equation there are 2 Nitrogen atoms, and 2 oxygen atoms as indicated by the subscript.
To balance the equation, the number of atoms of each element in the right side equation should be equal to left. By putting the numerical coefficient of 2, you will distribute that to each element. So you will end up with 2 nitrogen atoms and 2 oxygen atoms on the left side of the equation. Thus, the equation is balanced.
The answer again, is B.
Answer:
T₁ = 93.6 N , T₂ = 155.6 N , T₃ = 200 N
Explanation:
This is a balance exercise where we must apply the expressions for translational balance in the two axes
∑ F = 0
Suppose that cable t1 goes to the left and the angles are 41º with respect to the horizontal and cable t2 goes to the right with angles of 63º
decompose the tension of the two upper cables
cos 41 = T₁ₓ / T1
sin 41 = T₁y / T1
T₁ₓ = T₁ cos 41
T₁y= T₁ sin 41
for cable gold
cos 63 = T₂ / T₂
sin 63 = / T₂
We apply the two-point equilibrium equation: The junction point of the three cables and the point where the traffic light joins the vertical cable.
Let's start by analyzing the point where the traffic light meets the vertical cable
T₃ - W = 0
T₃ = W
T₃ = 200 N
now let's write the equations for the single point of the three wires
X axis
- T₁ₓ + T₂ₓ = 0
T₁ₓ = T₂ₓ
T1 cos 41 = T2 cos 63
T1 = T2 cos 63 / cos 41 (1)
y Axis
+ T_{2y} - T3 = 0
T₁ sin 41 + T₂ sin 63 = T₃ (2)
to solve the system we substitute equation 1 in 2
T₂ cos 63 / cos 41 sin 41 + T₂ sin 63 = W
T₂ (cos 63 tan 41 + sin 63) = W
T₂ = W / (cos 63 tan 41 + sin 63)
We calculate
T₂ = 200 / (cos 63 tan 41 + sin 63)
T₂ = 200 / 1,2856
T₂ = 155.6 N
we substitute in 1
T₁ = T₂ cos 63 / cos 41
T₁ = 155.6 cos63 / cos 41
T₁ = 93.6 N
therefore the tension in each cable is
T₁ = 93.6 N
T₂ = 155.6 N
T₃ = 200 N