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vovangra [49]
4 years ago
7

The diffusion coefficients for iron in nickel are given at two temperatures:T (K)D (m2/s)12739.4 × 10–1614732.4 × 10–14(a) Deter

mine the values of D0 and the activation energy Qd.(b) What is the magnitude of D at 1100°C (1373 K)?

Engineering
1 answer:
hram777 [196]4 years ago
5 0

Answer:

The diffusion coefficients for iron in nickel are given at two temperatures:

T (K)        1273          1473

D (m^{2}/s) 9.4 × 10^{-16}    2.4 × 10^{-14}

(a) Determine the values of Do and the activation energy Qd.

(b) What is the magnitude of D at 1100°C (1373 K)?

<em>A </em>

<em>The pre-exponential factor Do = 2.1 x </em>10^{-16}<em></em>

<em>The activation energy Qd = 252,609 J/mol</em>

<em>B</em>

<em>The diffusion coefficient D= 5.14 x </em>10^{-15}<em></em>

Explanation:

The full explanation is contained in the attached images;

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8 0
3 years ago
An experiment compares the initial speed of bullets fired from two handguns: a 9 mm and a 0.44 caliber. The guns are fired into
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Answer:

1.176

Explanation:

When the bullets impact the mass they become embedded on it, it is a plastic collision, therefore momentum is conserved.

v2 * (M + mb) = v1 * mb

Where

v1: muzzle velocity of the bullet

M: mass of the bob

mb: mass of the bullet

v2: mass of the bob with the bullet after being hit

v2 = v1 * mb / (M + mb)

Upon being impacted the bob will acquire speed v2, this implies a kinetic energy. The bob will then move and raise a height h. Upon acheiving the maximum height it will have a speed of zero. At that point all kinetic energy will be converted into potential energy.

Ek = 1/2 (M + mb) * v2^2

Ep = (M + mb) * g * h

Ek = Ep

1/2 (M + mb) * v2^2 = (M + mb) * g * h

1/2 * (v1 * mb / (M + mb))^2 = g * h

1/2 * v1^2 * mb^2 / (M + mb)^2 = g * h

v1^2 = g *h * (M+ mb)^2 / (1/2 * mb^2)

v2 = \sqrt{\frac{g *h * (M+ mb)^2}{\frac{1}{2} * mb^2}}

The height h that it reaches is related to the length L of the pendulum arm and the angle it forms with the vertical.

h = L * (1 - cos(a))

v2 = \sqrt{\frac{g * L * (1 - cos(a)) * (M+ mb)^2}{\frac{1}{2} * mb^2}}

For the 9 mm:

v2 = \sqrt{\frac{9.81 * L * (1 - cos(4.3)) * (10+ 0.006)^2}{\frac{1}{2} * 0.006^2}} = \sqrt{L} * 391

For the 0.44 caliber:

v2 = \sqrt{\frac{9.81 * L * (1 - cos(10.1)) * (10+ 0.012)^2}{\frac{1}{2} * 0.012^2}} = \sqrt{L} * 460

The ratio is 460 / 391 = 1.176

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3 years ago
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Thepotemich [5.8K]

Answer:

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Answer:

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Answer:

Explanation:

the solution to the problem is given in the pictures attached. (b) is answered first then (a). I hope the explanation helps you.Thank you

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