Let’s create a diagram that shows the chemical reaction absorb energy. Take a look at the diagram I created that is attached to this answer.
Now, the start of the graph line is pretty flat and low leveled. This is the start, which is the reactants. The beginning of a chemical reaction that absorbs energy will almost always start with reactants. I labeled this this “Start: Reactants” and colored the text light green so you can easily detect the part I’m talking about.
Moving on, we see there’s a light blue arrow pointing from the the near bottom to the top curve of the graph line. This is the where the activation energy is rising. Activation energy is the difference between the reactants and products. When the graph line begins to cure, this is when the transition state is beginning. I labeled the transition state, “Transition State” and colored the text a light orange. I colored the line a light blue and also labeled it “Activation Energy” and colored the text light blue as well. I hope it’s easy enough for you to follow along. :)
Next we see that the graph line dips and then goes flat, but it doesn’t go low leveled. It says fairly higher than where it started. This is the end product. I labeled this “End: Product” and colored the text light purple so you can easily detect it.
Okay, now that that’s clear, let’s go ahead and fill in the first blank. According to the diagram we see that the start should be a reactant, and since the graph line starts out pretty low. So, this means that the first tile should be “Low energy of reactants."
Now the second tile should definitely be “Transition State.” This is because the second part in the diagram is transition state. Again, this is mainly where the activation energy is starting to rise.
As for the last and final tile, we should chose “High energy of products.” This is because the diagram is left off on a high leveled part at the end of the graph line.
So, when you place the tiles they should look something like this.
Low energy of reactants
↓
Transition state
↓
High energy of products
- Marlon Nunez
Answer:
The equivalent capacitance will be
Explanation:
We have given two capacitance
They are connected in parallel
So equivalent capacitance
This equivalent capacitance is now connected in series with
In series combination of capacitors the equivalent capacitance is given by
So the equivalent capacitance will be
Answer:
v > 133.5 m/s
Explanation:
Let's analyze this problem a little, park run a complete circle we must know the speed of the system at the top of the circle.
Let's start by using the concepts of energy to find the velocity at the top of the circle
Initial. Top circle
Em₀ = K + U = ½ m v² + m g y
If we place the reference system at the bottom of the cycle y = 2R = L
Em₀ = ½ m v² + m g y
final. Low circle
= K = ½ m v₁²
Emo =
½ m v² + m g y = 1/2 m v₁²
v₁² = v² + (2g L)
v₁² = v² + 2 g L
The smallest value that v can have is zero, with this value the bullet + block system reaches this point and falls, with any other value exceeding it and completing the circle. Let's calculate for this minimum speed point
v₁ = √2g L
We already have the speed system at the bottom we can use the moment
Starting point before crashing
p₀ = m v₀
End point after collision at the bottom of the circle
= (m + M) v₁
The system is formed by the two bodies and therefore the forces to last before the crash are internal and the moment is conserved
p₀ =
m v₀ = (m + M) v₁
v₀ = (m + M) / m v₁
Let's replace
v₀ = (1+ M / m) √ 2g L
Let's reduce to the SI system
m = 40 g (kg / 1000g) = 0.040 kg
Let's calculate
v₀ = (1 + 1.35 / 0.040) RA (2 9.8 0.753)
v₀ = 34.75 3.8417
v₀ = 133.5 m / s
the velocity must be greater than this value
v > 133.5 m/s