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2 years ago

7

This woman is riding a bicycle down a hill at a constant speed and in a straight line. Which change will increase the speed of t

he bicycle?
A. An added force of 20 N down the hill
B. Added forces of 30 N up the hill and 30 N down the hill
C. Added forces of 30 N up the hill and 20 N down the hill
D. An added force of 20 N to the side of the hill

1 answer:

Lubov Fominskaja [6]2 years ago

5 0

Answer:

The answer is A. An added force of 20 N to the side of the hill

Explanation:

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Under equilibrium condition, what is the probability of an electron state being occupied if it is located at the Fermi level

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3 years ago

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2 years ago

A flash distillation chamber operating at 101.3 kpa is separating an ethanol water mixture the feed mixture contains z weight et

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Answer:

The answer is $[\frac{LX_1}{46} + \frac{L(1-x)}{18}]\times0.454$ mol/hr

$[\frac{Vy_1}{46}+\frac{V(1-y)}{18}]\times0.454$ mol/hr

Explanation:

For flash distillation

F = V+L

$\frac{V}{F} + \frac{L}{F} = 1$

$\frac{F}{V} -\frac{L}{V} = 1$

Fz = Vy+Lx

Y = $\frac{F}{V}\times Z - \frac{L}{V}\times X$ let, $\frac{V}{F} = F$

$y = \frac{Z}{F} -[ \frac{1}{F} -1]\times X$

Highlighted reading

F = 299; $\frac{V}{F}$ = 0.85 ; z = 0.36

y = $\frac{0.36}{0.85} - (-0.15)\times X$

= 0.423 + 0.15x ------------(i)

$y^{*}$ = -43.99713 $x^{6}$ + 148.27274 $x^{5}$ - 195.46 $x^{4}$ +127.99 $x^{3}$ -43.3 $x^{2}$ + 7.469 $x^{}$ + 0.02011

At equilibrium, $y^{*}$ = y

0.423+0.15 $x^{}$ = $y^{*}$

-43.99713 $x^{6}$ + 148.27274 $x^{5}$ - 195.46 $x^{4}$ +127.99 $x^{3}$ -43.3 $x^{2}$ + 7.319 $x^{}$ -0.403

F(x) for Newton's Law

Let $x_{0} = 0$

$x_{1}$ = $\frac{0-[{-0.403}]}{7.319}$

= 0.055

$x_{2}$ = $\frac{{0.055}-{f(0.055)} {{{{{{{}}}}}}}}{f^{'} (0.055)}$

= $\frac{{(0.055)}-(-0.11)}{3.59}$

= 0.085

$x^{3}$ = $\frac{{0.085}-(0.024)}{2.289}$

= 0.095

$x^{4}$ = $\frac{{0.095}-(-0.0353)}{-1.410}$

= 0.07

From This x and y are found from equation (i) and L and V are obtained from $\frac{V}{F}$ and F values

$[\frac{LX_1}{46} + \frac{L(1-x)}{18}]\times0.454$ mol/hr

$[\frac{Vy_1}{46}+\frac{V(1-y)}{18}]\times0.454$ mol/hr

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3 years ago