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3 years ago

12

How to change a fuel fiter

1 answer:

alisha [4.7K]3 years ago

8 0

Remove whatever is holding the old filter in place and remove it. Put the new filter on in the same position as the old filter. Replace whatever holds the filter in place, and make sure it's secure. Replace the fuse for the fuel pump in the fuse box.

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Very thin films are usually deposited under vacuum conditions to prevent contamination and ensure that atoms can fly directly fr

katrin [286]

Answer:

a. 9947 m

b. 99476 times

c. 2*10^11 molecules

Explanation:

a) To find the mean free path of the air molecules you use the following formula:

$\lambda=\frac{RT}{\sqrt{2}\pi d^2N_AP}$

R: ideal gas constant = 8.3144 Pam^3/mol K

P: pressure = 1.5*10^{-6} Pa

T: temperature = 300K

N_A: Avogadros' constant = 2.022*10^{23}molecules/mol

d: diameter of the particle = 0.25nm=0.25*10^-9m

By replacing all these values you obtain:

$\lambda=\frac{(8.3144 Pa m^3/mol K)(300K)}{\sqrt{2}\pi (0.25*10^{-9}m)^2(6.02*10^{23})(1.5*10^{-6}Pa)}=9947.62m$

b) If we assume that the molecule, at the average, is at the center of the chamber, the times the molecule will collide is:

$n_{collision}=\frac{9947.62m}{0.05m}\approx198952\ times$

c) By using the equation of the ideal gases you obtain:

$PV=NRT\\\\N=\frac{PV}{RT}=\frac{(1.5*10^{-6}Pa)(\frac{4}{3}\pi(0.05m)^3)}{(8.3144Pa\ m^3/mol\ K)(300K)}=3.14*10^{-13}mol\\\\n=(3.14*10^{-13})(6.02*10^{23})\ molecules\approx2*10^{11}\ molecules$

5 0

3 years ago

John has just graduated from State University. He owes $35,000 in college loans, but he does not have a job yet. The college loa

IRISSAK [1]

Answer:

The correct response is "821.88". A further explanation is given below.

Explanation:

According to the question,

The largest amount unresolved after five years would have been:

= $35000\times (\frac{F}{P}, 4 \ percent,5 )$

= $35000\times 1.216 7$

= $42584.50$

Now,

time (t) will be:

= $5\times 12$

= $60 \ monthly \ payments$

So, monthly payment will be:

= $42582.85\times (\frac{A}{P}, 0.5 \ percent,60 )$

= $42584.50\times 0.0193$

= $821.88$

6 0

3 years ago

If the specific surface energy for aluminum oxide is 0.90 J/m2 and its modulus of elasticity is (393 GPa), compute the critical

vampirchik [111]

Answer:

critical stress required for the propagation is 27.396615 × $10^{6}$ N/m²

Explanation:

given data

specific surface energy = 0.90 J/m²

modulus of elasticity E = 393 GPa = 393 × $10^{9}$ N/m²

internal crack length = 0.6 mm

to find out

critical stress required for the propagation

solution

we will apply here critical stress formula for propagation of internal crack

( σc ) = $\sqrt{\frac{2E\gamma s}{\pi a}}$ .....................1

here E is modulus of elasticity and γs is specific surface energy and a is half length of crack i.e 0.3 mm = 0.3 × $10^{-3}$ m

so now put value in equation 1 we get

( σc ) = $\sqrt{\frac{2E\gamma s}{\pi a}}$

( σc ) = $\sqrt{\frac{2*393*10^9*0.90}{\pi 0.3*10^{-3}}}$

( σc ) = 27.396615 × $10^{6}$ N/m²

so critical stress required for the propagation is 27.396615 × $10^{6}$ N/m²

6 0

3 years ago

A gas stream contains 4.0 mol % NH3 and its ammonia content is reduced to 0.5 mol % in a packed absorption tower at 293 K and 10

bagirrra123 [75]

Answer:

Explanation:

Step by step solved solution is given in the attached document.

8 0

3 years ago

3. Which of these instruments is used to measure wind speed? A. anemometer C. wind sock B. thermometer D. wind vane It is an ins

siniylev [52]

Answer:

wind vane if it can be used to show wind speed and the other is a

Explanation:

please mark 5 star if im right and brainly when ya can

5 0

3 years ago