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Pepsi [2]
2 years ago
14

1. Consider the potential energy diagram shown below. This graph shows the chemical potential energy in a reaction system over t

ime. The y-axis is potential energy in kilojoules. The x-axis is the reaction progress or time.
Does this graph represent an endothermic or an exothermic reaction? Explain your answer.
What is the enthalpy change, ΔH, for this reaction? Show your work.
What is the activation energy, Ea, for this reaction? Show your work.
THE GRAPH IS BELOW
Physics
1 answer:
Gwar [14]2 years ago
4 0

150 kJ of energy was given out over the course of the reaction and 200kJ is required for the reactants to be converted into products.

<h3>What is the potential energy graph?</h3>

We know that the combination of the reactants is what leads to the formation of products. The interaction of the reactants and the products is what is shown on the potential energy curve. The hump at the highest point of the graph is the activation energy while the difference between the energy of the reactant and the products is the enthalpy of the reaction.

Now; the enthalpy change is given as;

ΔH = 200 kJ - 50 kJ = 150 kJ

Ea = 250 kJ - 50 kJ = 200kJ

We can see that 150 kJ of energy was given out over the course of the reaction and 200kJ is required for the reactants to be converted into products.

Learn more about reaction profile:brainly.com/question/23528085

#SPJ1

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The block in the figure below has a mass of 5.1 kg and it rests on an incline of angle . You pull on the rope with a force F = 3
viktelen [127]

42.9°

Explanation:

Let's assume that the x-axis is aligned with the incline and the positive direction is up the incline. We can then apply Newton's 2nd law as follows:

x:\;\;\;\;F - mg\sin{\theta} = 0\;\;\;\;

\Rightarrow mg\sin{\theta} = F

Note that the net force is zero because the block is moving with a constant speed when the angle of the incline is set at \theta. Solving for the angle, we get

\sin{\theta} = \dfrac{F}{mg}

or

\theta = \sin^{-1}\left(\dfrac{F}{mg}\right)

\;\;\;=  \sin^{-1}\left[\dfrac{34\:\text{N}}{(5.1\:\text{kg})(9.8\:\text{m/s}^2)}\right]

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2 years ago
Will Mark Brainliest if Correct PLZ!!!!! A bullet is shot at some angle above the horizontal at an initial velocity of 87m/s on
qaws [65]

Answer:

≅50°

Explanation:

We have a bullet flying through the air with only gravity pulling it down, so let's use one of our kinematic equations:

Δx=V₀t+at²/2

And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:

Δx=(V₀cosθ)t+at²/2

Now luckily we are given everything we need to solve (or you found the info before posting here):

  • Δx=760 m
  • V₀=87 m/s
  • t=13.6 s
  • a=g=-9.8 m/s²; however, at 760 m, the acceleration of the bullet is 0 because it has already hit the ground at this point!

With that we can plug the values in to get:

760=(87)(cos\theta )(13.6)+\frac{(0)(13.6^{2}) }{2}

760=(1183.2)(cos\theta)

cos\theta=\frac{760}{1183.2}

\theta=cos^{-1}(\frac{760}{1183.2})\approx50^{o}

3 0
3 years ago
A baseball is batted from a height of 1.09 m with a speed of
kobusy [5.1K]

(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

(c) The distance off course the ball would be carried is 0.38 m.

(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

<h3>Horizontal and vertical components of the ball's velocity</h3>

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

<h3>Maximum height reached by the ball</h3>

H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m

Maximum height above ground = 7.51 + 1.09 = 8.6 m

<h3>Distance off course after 2 second </h3>

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s

<h3>Resultant speed of the ball and crosswind</h3>

V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s

<h3>Distance off course the ball would be carried</h3>

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

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