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torisob [31]
3 years ago
5

It is known that the kinetics of recrystallization for some alloy obeys the Avrami equation, andthat the value of n in the expon

ential is 5.0. If, at some temperature, the fraction recrystallized is0.30 after 100 min, determine the rate of recrystallization at this temperature.
Physics
1 answer:
trapecia [35]3 years ago
3 0

Answer:8.76\times 10^{-3} min^{-1}

Explanation:

Given

n=5

0.3 fraction recrystallize after 100 min

According to Avrami equation

y=1-e^{-kt^n}

where y=fraction Transformed

k=constant

t=time

0.3=1-e^{-k(100)^5}

e^{-k(100)^5} =0.7

Taking log both sides

-k\cdot (10^{10}=\ln 0.7

k=3.566\times 10^{-11}

At this Point we want to compute t_{0.5}\ i.e.\ y=0.5

0.5=1-e^{-kt^n}

0.5=e^{-kt^n}

0.5=e^{-3.566\times 10^{-11}\cdot (t)^5}

taking log both sides

\ln 0.5=-3.566\times 10^{-11}\cdot (t)^5

t^5=1.943\times 10^{10}

t=114.2 min

Rate of Re crystallization at this temperature

t^{-1}=8.76\times 10^{-3} min^{-1}

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Georgia [21]

Explanation:

velocity is distance divided by time.

so

the average speed of the ball is 10m/20s

= 0.5 m/s

8 0
2 years ago
A circular sign has a diameter of 40 cm and is subjected to normal winds up to 150 km/h at 10°C and 100 kPa. Determine the drag
Marat540 [252]

Answer:

147.7 N

221.55 Nm

Explanation:

P = Pressure = 100000 Pa

R_s = Mass-specific gas constant = 287.015 J/kg k

T = Temperature = 10+273 = 283 K

C = Drag coefficient = 1.1

A = Area

r = Radius = 0.2 m

v = Speed of wind = \frac{150}{3.6}\ m/s

L = Length of pole

Density

\rho=\frac{P}{R_sT}\\\Rightarrow \rho=\frac{100000}{287.058\times 283}\\\Rightarrow \rho=1.2309\ kg/m^3

Drag force

F=\frac{1}{2}\rho CAv^2\\\Rightarrow F=\frac{1}{2}\times 1.2309\times 1.1\times \left(\pi \times 0.2^2\right)\times \left(\frac{150}{3.6}\right)^2\\\Rightarrow F=147.7\ N

Force on the circular sign is 147.7 N

M=F\times L\\\Rightarrow M=147.7\times 1.5\\\Rightarrow M=221.55\ Nm

Bending moment at the bottom of the pole is 221.55 Nm

6 0
3 years ago
An infant's toy has a 120 g wooden animal hanging from a spring. If pulled down gently, the animal oscillates up and down with a
Morgarella [4.7K]

Answer:

0.37 m

Explanation:

The angular frequency, ω, of a loaded spring is related to the period, T,  by

\omega = \dfrac{2\pi}{T}

The maximum velocity of the oscillation occurs at the equilibrium point and is given by

v = \omega A

A is the amplitude or maximum displacement from the equilibrium.

v = \dfrac{2\pi A}{T}

From the the question, T = 0.58 and A = 25 cm = 0.25 m. Taking π as 3.142,

v = \dfrac{2\times3.142\times0.25\text{ m}}{0.58\text{ s}} = 2.71 \text{ m/s}

To determine the height we reached, we consider the beginning of the vertical motion as the equilibrium point with velocity, v. Since it is against gravity, acceleration of gravity is negative. At maximum height, the final velocity is 0 m/s. We use the equation

v_f^2 = v_i^2+2ah

v_f is the final velocity, v_i is the initial velocity (same as v above), a is acceleration of gravity and h is the height.

h = \dfrac{v_f^2 - v_i^2}{2a}

h = \dfrac{0^2 - 2.71^2}{2\times-9.81} = 0.37 \text{ m}

3 0
3 years ago
Why might an electromagnet be used to pick up old cars in junkyards?
siniylev [52]
B) Hope it helps ,Have a nice day :)
7 0
3 years ago
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HEY YALL ANSA DIS PLS !!!
Rudik [331]

Answer:

A. 8.19 × 10^-11

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7 0
3 years ago
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