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vampirchik [111]

2 years ago

6

An FCC iron-carbon alloy initially containing 0.55 wt% C is exposed to an oxygen-rich and virtually carbon-free atmosphere at 13

25 K. Under these circumstances, the carbon diffuses from the alloy and reacts at the surface with the oxygen in the atmosphere -- that is, the carbon concentration at the surface position is maintained essentially at 0 wt% C. This process of carbon depletion is called decarburization. At what position will the carbon concentration be 0.20 wt% after 10 h treatment?

1 answer:

lukranit [14]2 years ago

7 0

Answer:

See attached pictures.

Explanation:

See attached pictures for explanation.

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A 2.00-kg box is suspended from the end of a light vertical rope. A time-dependent force is applied to the upper end of the rope

Alex

Answer:

T = 27.92 N

Explanation:

For this exercise let's use Newton's second law

T - W = m a

The weight

W = mg

The acceleration can be found by derivatives

a = dv / dt

v = 2 t + 0.6 t²

a = 2 + 0.6 t

We replace

T - mg = m (2 + 0.6t)

T = m (g + 2 + 0.6 t) (1)

Let's look for the time for the speed of 15 m / s

15 = 2 t + 0.6 t²

0.6 t² + 2 t - 15 = 0

We solve the second degree equation

t = [-2 ±√(4 - 4 0.6 (-15))] / 2 0.6

t = [-2 ±√40] / 1.3 = [-2 ± 6.325] / 1.2

We take the positive time

t = 3.6 s

Let's calculate from equation 1

T = 2.00 (9.8 + 2 + 0. 6 3.6)

T = 27.92 N

4 0

2 years ago

Two children hang by their hands from the same tree branch. The branch is straight, and grows out from the tree trunk at an angl

vazorg [7]

Answer:879.29 N-m

Explanation:

Given

mass of first child $m_1=44 kg$

distance of first child from tree is $r_1=1 m$

tree is inclined at an angle of $\theta =27^{\circ}$

mass of second child $m_1=27 kg$

distance of second child from tree is $r_2=2.1 m$

Weight of first child $=m_1g=431.2 kg$

Weight of second child $=m_2g=264.6 kg$

Torque of first child weight $=m_1g\cos \theta \cdot r_1$

$T_1=44\times 9.8\times \cos 27\times 1=384.202 N-m$

Torque of second child weight $=m_2g\cos \theta \cdot r_2$

$T_2=27\times 9.8\times \cos 27\times 2.1=495.096 N-m$

Net torque $T_{net}=T_1+T_2=384.202+495.096=879.29 N-m$

6 0

3 years ago

A 10 g bullet is fired into, and embeds itself in, a 2 kg block attached to a spring with a force constant of 19.6 n/m and whose

dmitriy555 [2]

Answer:

$x=0.478\ m$ is the compression in the spring

Explanation:

Given:

mass of the bullet, $m=10\ g=0.01\ kg$

mass of block, $M=2\ kg$

stiffness constant of the spring, $k=19.6\ N.m^{-1}$

initial velocity of the spring just before it hits the block, $u=300\ m.s^{-1}$

<u>Now since the bullet-mass gets embed into the block, we apply the conservation of momentum as:</u>

$m.u=(M+m).v$

$0.01\times 300=(2+0.01)\times v$

$v=1.4925\ m.s^{-1}$

Now this kinetic energy of the combined mass gets converted into potential energy of the spring.

$\rm Kinetic\ energy=Spring\ potential\ energy$

$\frac{1}{2} (M+m).v^2=\frac{1}{2} k.x^2$

$\frac{1}{2} \times (2+0.01)\times 1.4925^2=\frac{1}{2} \times 19.6\times x^2$

$x=0.478\ m$ is the compression in the spring

4 0

3 years ago

Read 2 more answers

Protons and neutrons in the nucleus are held together by what

True [87]

They are held together because of Strong Nuclear Force.

4 0

3 years ago

In a hydroelectric power plant, 65 m3 /s of water flows from an elevation of 90-m to a turbine-generator, where electricity is g

Mariulka [41]

The electric output of the plant is 48.19 MW

First we need to calculate the water power, it is given by the formula

WP=ρQgh

Here, ρ=1000 kg/m3 is density of water,Q is the flow rate, g is the gravity, and h is the water head

Therefore, WP=1000*65*9.81*90=57388500 W=57.38 MW

Now the overall efficiency of the hydroelectric power plant is given as

η= $\frac{electric power}{water power}$

Plugging the values in the above equation

0.84=EP/57.38

EP=48.19 MW

Therefore, the electric output of the plant is 48.19 MW.

3 0

2 years ago