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2 years ago

7

What term refers to the part of a spacecraft that is occupied by the crew for takeoff and landing?

1 answer:

Semenov [28]2 years ago

7 0

Command module ✅

service module

lunar module

annum module

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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

2. A large pack of snow of mass m sits at the top of a trail at a ski resort. Friction keeps the snow pack in place, but when it

Aleks [24]

Answer:

tanΘ

Explanation:

Let gravitational acceleration be g. When the avalanche starts to occur, the gravity force that is parallel to the slope is the same as friction force.

Gravity force that is parallel to the slope can be written as:

G = mgsinΘ

The friction force is the product of normal force and coefficient:

$F_f = N\mu$

where normal force N is the gravity in the direction perpendicular to the slope

$F_f = \mu mgcos\theta$

As stated before, gravity force that is parallel to the slope is the same as friction force:

$G = F_f$

$mgsin\theta = \mu mgcos\theta$

$\mu = sin\theta / cos\theta = tan\theta$

8 0

3 years ago

People benefit by participating in the market because

solniwko [45]

Answer:

Market participation allows individuals to specialize and, with trade, ultimately consume more.

3 0

3 years ago

Please help me with this!

Neporo4naja [7]

hope this helps good luck

5 0

3 years ago

The length of the grassy park where the aunt lives is 1 kilometer. How many times longer than the aunt is the park? Explain your

Oksi-84 [34.3K]

Answer:ssundee

Explanation:

6 0

3 years ago