Ask question

Ask question

All categories

3 years ago

10

Calculate the value of the diffusion coefficient D (in m2/s) at 567°C for the diffusion of some species in a metal; assume that

the values of D0 and Qd are 5.60 x 10-5 m2/s and 177 kJ/mol, respectively.

1 answer:

timofeeve [1]3 years ago

3 0

Explanation:

Below is an attachment containing the solution.

You might be interested in

How long must a pendulum be to have a period of 4.6 5 on Neptune, where g = 11.15 m/s?

Anon25 [30]

Answer:

5.98 m

Explanation:

$T=2\pi \sqrt\frac{L}{g}$

$T^2=4\pi ^2L/g\\L = gT^2/(4\pi^2)$ = 11.15*4.6²/(4π²) = 5.98 m

5 0

2 years ago

I NEED HELP PLEASE THANKS! :)

Nady [450]

Answer:

0.962 s

Explanation:

Speed = frequency × wavelength

v = fλ

2.60 m/s = f (2.50 m)

f = 1.04 Hz

Period = 1 / frequency

T = 1/f

T = 1 / (1.04 Hz)

T = 0.962 s

7 0

3 years ago

The mechanical force of contraction is generated by

maria [59]

it is generated by two objects pushing against each other with equal/almost equal force, therefore contracting against each other

4 0

4 years ago

A 19.0-kg cart is moving with a velocity of 7.20 m/s down a level hallway. A constant force of -13.0 N acts on the cart and its

gulaghasi [49]

Answer:

a. -369.36J

b. -123.9J

c. 9.52m

Explanation:

From the expression for kinetic energy

K. E=1/2mv^2

Since the mass is constant, but the velocity changes. Hence the change in kinetic energy is

K.E=1/2*19(3.6²-7.2²)

K.E= -369.36J

b. to determine the workdone by the force,we determine the distance moved.

But the acceleration is from

F=ma ,

a=f/m

a=-13/19

0.68m/s²

the distance moved is

s=v²/2a

s=3.6²/2*0.68

s=9.52m

Hence the work done is

W=force * distance

W=-13*9.52

W=-123.9J

d. the distance moved is

s=v²/2a

s=3.6²/2*0.68

s=9.52m

4 0

4 years ago

Read 2 more answers

Two spherical asteroids have the same radius R. Asteroid 1 has mass M and asteroid 2 has mass 1.97·M. The two asteroids are rele

nekit [7.7K]

Answer:

$0.536\sqrt{\frac{GM}{R}}$

Explanation:

We are given that

Mass of one asteroid 1, $m_1=M$

Mass of asteroid 2, $m_2=1.97 M$

Initial distance between their centers,d=13.63 R

Radius of each asteroid=R

d'=R+R=2R

Initial velocity of both asteroids

$u=0$

We have to find the speed of second asteroid just before they collide.

According to law of conservation of momentum

$(m_1+m_2)u=m_1v_1+m_2v_2$

$(M+1.97 M)\times 0=Mv_1+1.97Mv_2$

$Mv_1=-1.97 Mv_2$

$v_1=-1.97v_2$

According to law of conservation of energy

$Gm_1m_2(\frac{1}{d'}-\frac{1}{d})=\frac{1}{2}m_1v^2_1+\frac{1}{2}m_2v^2_2$

$GM(1.97M)(\frac{1}{2R}-\frac{1}{13.63R})=\frac{1}{2}M(-1.97v_2)^2+\frac{1}{2}(1.97M)v^2_2$

$1.97M^2G(\frac{13.63-2}{27.26R})=\frac{1}{2}Mv^2_2(3.8809+1.97)$

$1.97MG(\frac{11.63}{27.26 R})=\frac{1}{2}(5.8509)v^2_2$

$v^2_2=\frac{1.97GM\times11.63\times 2}{27.26R\times 5.8509}$

$v_2=\sqrt{\frac{1.97GM\times11.63\times 2}{27.26R\times 5.8509}}$

$v_2=0.536\sqrt{\frac{GM}{R}}$

Hence, the speed of second asteroid = $0.536\sqrt{\frac{GM}{R}}$

8 0

3 years ago