E=hc/l
E=
<span><span>E=<span>(6.626 x 10-34 J s)(3.0 x 108m/s )</span><span>=2.88 x 10-19J</span></span><span>6.90 x 10-7m</span></span>
It is natural and u can't by it
Answer:
Explanation:
In general, an increase in pressure (decrease in volume) favors the net reaction that decreases the total number of moles of gases, and a decrease in pressure (increase in volume) favors the net reaction that increases the total number of moles of gases.
Δn= b - a
Δn= moles of gaseous products - moles of gaseous reactants
Therefore, <u>after the increase in volume</u>:
- If Δn= −1 ⇒ there are more moles of gaseous reactants than gaseous products. The equilibrium will be shifted towards the products, that is, from left to right, and K>Q.
- If Δn= 0 ⇒ there is the same amount of gaseous moles, both in products and reactants. The system is at equilibrium and K=Q.
- Δn= +1 ⇒ there are more moles of gaseous products than gaseous reactants. The equilibrium will be shifted towards the reactants, that is, from right to left, and K<Q.
The statement is true in this situation is C. The size of Ffric is the same as the size of Fapp:
From the diagram, since the body is in equilibrium, the sum of vertical forces equals zero. Also, the sum of horizontal forces equal zero.
So, ∑Fx = 0 and ∑Fy = 0
Since Fapp acts in the negative x - direction and Ffric acts in the positive x - direction,
∑Fx = -Fapp + Ffric = 0
-Fapp + Ffric = 0
Fapp = Ffric
Also, since Fgrav acts in the negative y - direction and Fnorm acts in the positive y - direction,
∑Fy = Fnorm + (-Fgrav) = 0
Fnorm - Fgrav = 0
Fnorm = Fgrav
So, we see that the size of Fapp <u>equals</u> size of Ffric and the size of Fnorm <u>equals</u> the size of Fgrav.
So, the correct option is C
The statement which is true in this situation is C. The size of Ffric is the same as the size of Fapp.
Learn more about equilibrium of forces here:
brainly.com/question/12980489
