1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
schepotkina [342]
3 years ago
14

Calculate the fraction of methyl isonitrile molecules that have an energy of 160.0 kJ or greater at 510 K .

Physics
1 answer:
soldi70 [24.7K]3 years ago
3 0

Answer:

4.0932672025\times 10^{-17}

Explanation:

E_a = Activation energy = 160 kJ

T = Temperature = 510 K

R = Universal gas constant = 8.314 J/mol K

The fraction of energy is given by

f=e^{-\dfrac{E_a}{RT}}\\\Rightarrow f=e^{-\dfrac{160000}{8.314\times 510}}\\\Rightarrow f=4.0932672025\times 10^{-17}

The fraction of energy is 4.0932672025\times 10^{-17}

You might be interested in
There is the more role of the moon than the sun to occur tides in the oceans,why?
Nana76 [90]

Answer:

The ocean tides on earth are caused by both the moon's gravity and the sun's gravity. ... Even though the sun is much more massive and therefore has stronger overall gravity than the moon, the moon is closer to the earth so that its gravitational gradient is stronger than that of the sun.

8 0
2 years ago
A rectangular block of steel measures 4cm*2cm*1.5cm*. its mass is 93.6g
nata0808 [166]

Answer:

b

Explanation:

5 0
3 years ago
An infinite line of charge with linear density λ1 = 8.2 μC/m is positioned along the axis of a thick insulating shell of inner r
bixtya [17]

1) Linear charge density of the shell:  -2.6\mu C/m

2)  x-component of the electric field at r = 8.7 cm: 1.16\cdot 10^6 N/C outward

3)  y-component of the electric field at r =8.7 cm: 0

4)  x-component of the electric field at r = 1.15 cm: 1.28\cdot 10^7 N/C outward

5) y-component of the electric field at r = 1.15 cm: 0

Explanation:

1)

The linear charge density of the cylindrical insulating shell can be found  by using

\lambda_2 = \rho A

where

\rho = -567\mu C/m^3 is charge volumetric density

A is the area of the cylindrical shell, which can be written as

A=\pi(b^2-a^2)

where

b=4.7 cm=0.047 m is the outer radius

a=2.7 cm=0.027 m is the inner radius

Therefore, we have :

\lambda_2=\rho \pi (b^2-a^2)=(-567)\pi(0.047^2-0.027^2)=-2.6\mu C/m

 

2)

Here we want to find the x-component of the electric field at a point at a distance of 8.7 cm from the central axis.

The electric field outside the shell is the superposition of the fields produced by the line of charge and the field produced by the shell:

E=E_1+E_2

where:

E_1=\frac{\lambda_1}{2\pi r \epsilon_0}

where

\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m is the linear charge density of the wire

r = 8.7 cm = 0.087 m is the distance from the axis

And this field points radially outward, since the charge is positive .

And

E_2=\frac{\lambda_2}{2\pi r \epsilon_0}

where

\lambda_2=-2.6\mu C/m = -2.6\cdot 10^{-6} C/m

And this field points radially inward, because the charge is negative.

Therefore, the net field is

E=\frac{\lambda_1}{2\pi \epsilon_0 r}+\frac{\lambda_2}{2\pi \epsilon_0r}=\frac{1}{2\pi \epsilon_0 r}(\lambda_1 - \lambda_2)=\frac{1}{2\pi (8.85\cdot 10^{-12})(0.087)}(8.2\cdot 10^{-6}-2.6\cdot 10^{-6})=1.16\cdot 10^6 N/C

in the outward direction.

3)

To find the net electric field along the y-direction, we have to sum the y-component of the electric field of the wire and of the shell.

However, we notice that since the wire is infinite, for the element of electric field dE_y produced by a certain amount of charge dq along the wire there exist always another piece of charge dq on the opposite side of the wire that produce an element of electric field -dE_y, equal and opposite to dE_y.

Therefore, this means that the net field produced by the wire along the y-direction is zero at any point.

We can apply the same argument to the cylindrical shell (which is also infinite), and therefore we find that also the field generated by the cylindrical shell has no component along the y-direction. Therefore,

E_y=0

4)

Here we want to find the x-component of the electric field at a point at

r = 1.15 cm

from the central axis.

We notice that in this case, the cylindrical shell does not contribute to the electric field at r = 1.15 cm, because the inner radius of the shell is at 2.7 cm from the axis.

Therefore, the electric field at r = 1.15 cm is only given by the electric field produced by the infinite wire:

E=\frac{\lambda_1}{2\pi \epsilon_0 r}

where:

\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m is the linear charge density of the wire

r = 1.15 cm = 0.0115 m is the distance from the axis

This field points radially outward, since the charge is positive . Therefore,

E=\frac{8.2\cdot 10^{-6}}{2\pi (8.85\cdot 10^{-12})(0.0115)}=1.28\cdot 10^7 N/C

5)

For this last part we can use the same argument used in part 4): since the wire is infinite, for the element of electric field dE_y produced by a certain amount of charge dq along the wire there exist always another piece of charge dq on the opposite side of the wire that produce an element of electric field -dE_y, equal and opposite to dE_y.

Therefore, the y-component of the electric field is zero.

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

4 0
3 years ago
An 89 kg man drops from rest on a diving board −3.1 m above the surface of the water and comes to rest 0.5 s after reaching the
OLga [1]

To solve this problem we will use the linear motion kinematic equations, for which the change of speed squared with the acceleration and the change of position. The acceleration in this case will be the same given by gravity, so our values would be given as,

m= 89 kg\\x = 3.1 m\\t = 0.5s\\a = g = 9.8m/s^2

Through the aforementioned formula we will have to

v_f^2-v_i^2 = 2ax

The particulate part of the rest, so the final speed would be

v_f^2 = 2gx

v_f=\sqrt{2(9.8)(3.1)}

v_f = 7.79m/s

Now from Newton's second law we know that

F = ma

Here,

m = mass

a = acceleration, which can also be written as a function of velocity and time, then

F = m\frac{dv}{dt}

Replacing we have that,

F = (89)\frac{7.79}{0.5}

F = 1386.62N

Therefore the force that the water exert on the man is 1386.62

3 0
3 years ago
The inventor of the electric cell was:<br> O Coulomb<br> Franklin<br> o Gilbert<br> O Volta
Tomtit [17]

Answer:

The inventor of the electric cell was:

Alessandro Volta (in other words, Volta)

Explanation:

3 0
3 years ago
Read 2 more answers
Other questions:
  • Can a black light be used to charge a solar powered product
    14·1 answer
  • 2. The stage choreography for a play requires an
    7·1 answer
  • An apple falls out of a tree from a height of 2.3m. what is the impact speed of the apple
    6·1 answer
  • What type of simple machine is a slide shovel broom screwdriver
    14·2 answers
  • How long is a pendulum with a period of 1.0 s on a planet with twice the gravity of the earth?
    14·1 answer
  • . A telescope is constructed with two lenses separated by a distance of 25 cm. The focal length of the objective is 20 cm. The f
    9·1 answer
  • PLS ANSWER DUE LATER TODAY!!!
    9·1 answer
  • Pls help i need an answer for this ill mark u brianlist or anything u want
    11·1 answer
  • 29. Jorge is conducting an investigation into perfectly inelastic collisions using equipment where two carts collide with
    11·1 answer
  • If an asteroid were going to impact Earth, would you prefer to know in advance or let the NEO remain a mystery until impact? Why
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!