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Alexeev081 [22]

3 years ago

7

What is not a good inslator ??

1 answer:

Dahasolnce [82]3 years ago

5 0

I was gonna say your mom but the correct awnser is body movement

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

trapecia [35]

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

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

3 0

3 years ago

You push on a cart (18.0kg) at a 30 degree below horizontal angle. The coefficient of kinetic friction between the chair and the

podryga [215]

Given that force is applied at an angle of 30 degree below the horizontal

So let say force applied if F

now its two components are given as

$F_x = Fcos30$

$F_y = Fsin30$

Now the normal force on the block is given as

$N = Fsin30 + mg$

$N = 0.5F + (18\times 9.8)$

$N = 0.5F + 176.4$

now the friction force on the cart is given as

$F_f = \mu N$

$F_f = 0.625(0.5F + 176.4)$

$F_f = 110.25 + 0.3125F$

now if cart moves with constant speed then net force on cart must be zero

so now we have

$F_f + F_x = 0$

$Fcos30 - (110.25 + 0.3125F) = 0$

$0.866F - 0.3125F = 110.25$

$F = 199.2 N$

so the force must be 199.2 N

8 0

3 years ago

A disk-shaped grindstone of mass 3.0 kg and radius 8.0 cm is spinning at 600 rev/min. After the power is shut off, a man continu

kolbaska11 [484]

Answer:

τ=0.060 N.m

Explanation:

By kinematics:

$\omega f = \omega o-\alpha*t$

Solving for α:

$\alpha=\frac{\omega o-\omega f}{t}$

where ωo = 600*2*π/60; ωf = 0; t=10s

$\alpha=6.283rad/s^2$

The sum of torque is:

$\tau=I*\alpha$

$\tau=M*R^2/2*\alpha$

$\tau=0.060 N.m$

8 0

3 years ago

Some of the energy the car gained as it was pulled was lost because of heat from friction. The rest of the energy was transforme

Lisa [10]

Answer:

454,320 joules

Explanation:

The work done on an object is equal to its change in kinetic energy: Change in KE = F × d.

Plug the values for F and d into the formula and solve:

Change in KE = 2,524 × 180

= 454,320 joules

The roller coaster gains 454,320 joules of energy from the work done on it by the chain.

7 0

3 years ago

Read 2 more answers

_________ is a layer of the earth that is classified not by composition

Sidana [21]

Answer:

Asthenosphere

Explanation:

The asthenosphere is a part of the upper mantle just below the lithosphere that is involved in plate tectonic movement and isostatic adjustments.

7 0

3 years ago