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3 years ago

PLEASE HURRY IM BEING TIMED

1 answer:

8 0

A compound machine is one that consists of a couple simple machines working together, like a wheelbarrow contains the lever, axle, and wheel. While a simple machine would just be those individually, or anything else that can be used for applying a force.<span />

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Carbon is allowed to diffuse through a steel plate 9.7-mm thick. The concentrations of carbon at the two faces are 0.664 and 0.3

cupoosta [38]

Answer:

844°C

Explanation:

The problem can be easily solve by using Fick's law and the Diffusivity or diffusion coefficient.

We know that Fick's law is given by,

$J = - D \frac{\Delta c}{\Delta x}$

Where $\frac{\Delta c}{\Delta x}$ is the concentration of gradient

D is the diffusivity coefficient

and J is the flux of atoms.

In the other hand we have, that

$D= D_0 e^{\frac{E_d}{RT}}$

Where $D_0$ is the proportionality constant,

R is the gas constant, T the temperature and $E_d$ is the activation energy.

Replacing the value of diffusivity coefficient in Fick's law we have,

$J = -D_0 ^{\frac{E_d}{RT}}\frac{\Delta c}{\Delta x}$

Rearrange the equation to get the value of temperature,

$T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}$

We have all the values in our equation.

$\Delta c = 0.664-0.339 = 0.325 C. cm^{-1}$

$\Delta x = 9.7*10^{-3}m$

$E_d = 82000J$

$D_0 = 6.5*10^{-7}m^2/s$

$J = 3.2*10^{-9}m^2/s$

$R= 8.31Jmol^{-1}K$

Substituting,

$T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}$

$T=-\frac{-82000}{(8.31)ln(\frac{3.2*10^{-9}(9.7*10^{-3})}{6.5*10^{-7} (0.325)})}$

$T=1118.07K=844\°C$

4 0

3 years ago

The resistance of a circuit is doubled. How is the value of the current affected?

tekilochka [14]

A because it makes more sense

3 0

3 years ago

Witch will delsove faster 1 gram of sugar or 1 gram of salt

natka813 [3]

Answer:

Sugar dissolves faster than salt

4 0

3 years ago

A 36.0 kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130

konstantin123 [22]

Answer:

(a) W = 650J

(b) Wf = 529.2J

(d) W = 0J

(e) ΔK = 120.8J

(f) v2 = 2.58 m/s

Explanation:

(a) In order to find the work done by the applied force you use the following formula:

$W=Fd$ (1)

F: applied force = 130N

d: distance = 5.0m

$W=(130N)(5.0m)=650J$

The work done by the applied force is 650J

(b) The increase in the internal energy of the box-floor system is given by the work done of the friction force, which is calculated as follow:

$W_f=F_fd=\mu Mgd$ (2)

μ: coefficient of friction = 0.300

M: mass of the box = 36.0kg

g: gravitational constant = 9.8 m/s^2

$W_f=(0.300)(36.0kg)(9.8m/s^2)(5.0m)=529.2J$

The increase in the internal energy is 529.2J

(c) The normal force does not make work on the box because the normal force is perpendicular to the motion of the box.

$W = 0J$

(d) The same for the work done by the normal force. The work done by the gravitational force is zero because the motion of the box is perpendicular o the direction of the gravitational force.

(e) The change in the kinetic energy is given by the net work on the box. You use the following formula:

$\Delta K=W_T$ (3)