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3 years ago

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What chemicals must be cycled through a Martian environment in order for the planet to be habitable for life? Describe how scien

tists intend to make those resources available in such a harsh environment.

1 answer:

photoshop1234 [79]3 years ago

4 0

The scientists intend to make chemicals

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In a distillation operation, the heat added to the boiler a) increases with increase in reflux ratio b) decreases with increase

Leni [432]

Answer:

a) increases with increase in reflux ratio

Explanation:

If we increase the reflux ratio, which can de defined as : L/D = Condesate back to column /distillate discarded from column, it means that we increase the amount of condensate back to column. So the reboiler load will increase and be maximum in case of Total reflux ratio. And as vapor load will be increased, diameter of column will be increased too.

7 0

3 years ago

Find the de Broglie wavelength lambda for an electron moving at a speed of 1.00 \times 10^6 \; {\rm m/s}. (Note that this speed

masya89 [10]

(A) $7.28\cdot 10^{-10} m$

The De Broglie wavelength of an electron is given by

$\lambda=\frac{h}{p}$ (1)

where

h is the Planck constant

p is the momentum of the electron

The electron in this problem has a speed of

$v=1.00\cdot 10^6 m/s$

and its mass is

$m=9.11\cdot 10^{-31} kg$

So, its momentum is

$p=mv=(9.11\cdot 10^{-31} kg)(1.00\cdot 10^6 m/s)=9.11\cdot 10^{-25}kg m/s$

And substituting into (1), we find its De Broglie wavelength

$\lambda=\frac{6.63\cdot 10^{-34}Js}{9.11\cdot 10^{-25} kg m/s}=7.28\cdot 10^{-10} m$

(B) $1.16\cdot 10^{-34}m$

In this case we have:

m = 0.143 kg is the mass of the ball

v = 40.0 m/s is the speed of the ball

So, the momentum of the ball is

$p=mv=(0.143 kg)(40.0 m/s)=5.72 kg m/s$

And so, the De Broglie wavelength of the ball is given by

$\lambda=\frac{h}{p}=\frac{6.63\cdot 10^{-34} Js}{5.72 kg m/s}=1.16\cdot 10^{-34}m$

(C) $9.02\cdot 10^{-9}m$

The location of the first intensity minima is given by

$y=\frac{L\lambda}{a}$

where in this case we have

$y=0.492 cm = 4.92\cdot 10^{-3} m$

L = 1.091 is the distance between the detector and the slit

$a=2.00\mu m=2.00\cdot 10^{-6}m$ is the width of the slit

Solving the formula for $\lambda$ , we find the wavelength of the electrons in the beam:

$\lambda=\frac{ya}{L}=\frac{(4.92\cdot 10^{-3}m)(2.00\cdot 10^{-6} m)}{1.091 m}=9.02\cdot 10^{-9}m$

(D) $7.35\cdot 10^{-26}kg m/s$

The momentum of one of these electrons can be found by re-arranging the formula of the De Broglie wavelength:

$p=\frac{h}{\lambda}$

where here we have

$\lambda=9.02\cdot 10^{-9}m$ is the wavelength

Substituting into the formula, we find

$p=\frac{6.63\cdot 10^{-34}Js}{9.02\cdot 10^{-9}m}=7.35\cdot 10^{-26}kg m/s$

7 0

3 years ago

Which is NOT part of the definition of matter

qaws [65]

Answer:

Everything that has mass and takes up space is matter. Every day, you find something usual that either does't have mass or else don't take up space. Those things are non-matter. Basically, any type of energy or any abstract concept is an example of something that does not have matter.

Explanation:

7 0

3 years ago

Read 2 more answers

Use the ruler to determine the length of this object. Record your answer to the nearest tenth. The object is ____ long.

VLD [36.1K]

The object is 2.7 cm long

7 0

3 years ago

What is the main greenhouse gas responsible for influencing the world climate

cluponka [151]

Answer:

carbon dioxide

Explanation:

6 0

3 years ago