All categories

1 year ago

Adam has 5 barrels of different volumes. The barrel of which volume can contain the most liquid?

Engineering

1 answer:

Natalija [7]1 year ago

5 0

Answer:

Explanation:

You might be interested in

Consider a single crystal of some hypothetical metal that has the FCC crystal structure and is oriented such that a tensile stre

Naddik [55]

Answer:

imma leabe

Explanation:

leabe leave*

5 0

2 years ago

Determine the number of flipflops required to build a binary counter that count from 0 to 2043

Pie

Answer:

10 flip -flops are required to build a binary counter circuit to count to from 0 to 1023 .

Explanation:

3 0

2 years ago

For a very rough pipe wall the friction factor is constant at high Reynolds numbers. For a length L1 the pressure drop over the

Goshia [24]

Answer:

$\Delta p_{2} = 2\cdot \Delta p_{1}$

Explanation:

The pressure drop is directly proportional to the length of the pipe. Then, the new pressure drop is two time the previous one.

$\Delta p_{2} = 2\cdot \Delta p_{1}$

8 0

2 years ago

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

2 years ago

How would I find the Voltage across the open circuit

Nimfa-mama [501]

Answer:

Vab = 80V

Explanation:

The only current flowing in the circuit is supplied by the 100 V source. Its only load is the 40+60 ohm series circuit attached, so the current in that loop is (100V)/(40+60Ω) = 1A. That means V1 = (1A)(60Ω) = 60V.

Vab will be the sum of voltages around the right-side "loop" between terminals 'a' and 'b'. It is (working clockwise from terminal 'b') ...

Vab = -10V +60V +(0A×10Ω) +30V

Vab = 80V

3 0

1 year ago