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3 years ago

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The diffusion coefficients for species A in metal B are given at two temperatures: T (°C) D (m2/s) 1020 8.01 × 10-17 1290 7.86×

10-16 (a) Determine the value of the activation energy Qd (in J/mol).

1 answer:

Solnce55 [7]3 years ago

5 0

Explanation:

Below is an attachment containing the solution.

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The nuclear fuel distribution in a nuclear reactor is chosen so that when in operation the wall temperature of the reactor is a

cestrela7 [59]

Answer:

Explanation:

We Often solve the the integral neutron transport equation using the collision probability (CP) method which usually requires flat flux (FF) approach. In this research, it has been carried out in the cylindrical nuclear fuel cell with the spatial of mesh with quadratic flux approach. This simply means that the neutron flux at any region of the nuclear fuel cell is forced to follow the pattern of a quadratic function.

Furthermore The mechanism may be referred to as the process of non-flat flux (NFF) approach. The parameters that calculated in this study are the k-eff and the distribution of neutron flux. The result shows that all parameters are in accordance with the result of SRAC.

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3 years ago

In science, Bob learns that the energy of a wave is directly proportional to the square of the waves amplitude. If the energy of

KIM [24]

The energy of a wave is directly proportional to the square of the waves amplitude. Therefore, E = A² where A is the amplitude. This therefore means when the amplitude of a wave is doubled the energy will be quadrupled, when the amplitude is tripled the energy increases by a nine fold and so on.
Thus, in this case if the energy is 4J, then the amplitude will be √4 = 2 .

4 0

3 years ago

Sound with a frequency of 1250 Hz leaves a room through a doorway with a width of 1.05 m.At what minimum angle relative to the c

kiruha [24]

Answer:

15.19°, 31.61°, 51.84°

Explanation:

We need to fin the angle for m=1,2,3

We know that the expression for wavelenght is,

$\lambda = \frac{c}{f}$

Substituting,

$\lambda = \frac{344}{1250}$

$\lambda = 0.2752m$

Once we have the wavelenght we can find the angle by the equation of the single slit difraction,

$sin\theta = \frac{m \lambda}{W}$

Where,

W is the width

m is the integer

$\lambda$ the wavelenght

Re-arrange the expression,

$\theta = sin^{-1} \frac{m\lambda}{W}$

For m=1,

$\theta = sin^{-1} \frac{1 (0.2752)}{1.05}= 15.19\°$

For m=2,

$\theta = sin^{-1} \frac{2 (0.2752)}{1.05}= 31.61\°$

For m=3,

$\theta = sin^{-1} \frac{3 (0.2752)}{1.05}= 51.84\°$

<em>The angle of diffraction is directly proportional to the size of the wavelength.</em>

5 0

3 years ago

A bicyclist in the Tour de France crests a mountain pass as he moves at 18 km/h. At the bottom, 4.0 km farther, his speed is at

Allisa [31]

We are given:

v0 = initial velocity = 18 km/h

d = distance = 4 km

v = final velocity = 75 km/h

a =?

<span>
We can solve this problem by using the formula:</span>

v^2 = v0^2 + 2 a d

75^2 = 18^2 + 2 (a) * 4

5625 = 324 + 8a

<span>a = 662.625 km/h^2</span>

6 0

3 years ago

The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east.

Elena L [17]

Answer:

The magnitude of the acceleration is $a_r = 1.50 \ m/s^2$

The direction is $\theta = 32.5 6^o$ north of east

Explanation:

From the question we are told that

The force exerted by the wind is $F_{sail} = (330 ) \ N \ north$

The force exerted by water is $F_{keel} = (210 ) \ N \ east$

The mass of the boat(+ crew) is $m_b = 260 \ kg$

Now Force is mathematically represented as

$F = ma$

Now the acceleration towards the north is mathematically represented as

$a_n = \frac{F_{sail}}{m_b}$

substituting values

$a_n = \frac{330 }{260}$

$a_n = 1.269 \ m/s^2$

Now the acceleration towards the east is mathematically represented as

$a_e = \frac{F_{keel}}{m_b }$

substituting values

$a_e = \frac{210}{260}$

$a_e =0.808 \ m/s^2$

The resultant acceleration is

$a_r = \sqrt{a_e^2 + a_n^2}$

substituting values

$a_r = \sqrt{(0.808)^2 + (1.269)^2}$

$a_r = 1.50 \ m/s^2$

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

$tan \theta = \frac{a_e}{a_n}$

$\theta = tan ^{-1} [\frac{a_e}{a_n } ]$

substituting values

$\theta = tan ^{-1} [\frac{0.808}{1.269 } ]$

$\theta = tan ^{-1} [0.636 ]$

$\theta = 32.5 6^o$

5 0

3 years ago