Passband to baseband conversion: Consider the following passband signal,

A jet impinges directly on to a plate that is oriented normal to the axis of the jet. The mass flow rate of the jet is 50 kg/min

PilotLPTM [1.2K]

Answer:

166.67 N

Explanation:

Given:

Mass flow rate = 50 kg/min = 50 kg / 60 seconds = 0.833 kg/s

Initial velocity = 200 m/s

after striking the normal board the water will flow in the normal direction, thus the final velocity in the direction of the initial flow will be zero

therefore,

Force = (change in momentum)

or

Force = Initial momentum - final momentum

or

Force = 0.833 × 200 - 0.833 × 0

or

Force = 166.67 N

3 0

2 years ago

Find the equivalent impedance Zeq seen by the source when Vs = 2 cos (5t) v, C = 0.2 F, R = 1 Ω and L = 0.1 H. (Give angles in d

Yanka [14]

Answer:

0.89 cos (st +116.57°v)

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

6 0

2 years ago

Consider the series solution, Equation 5.42, for the plane wall with convection. Calculate midplane (x* = 0) and surface (x* = 1

VashaNatasha [74]

Answer:

We conclude that the approximate series solution (with only one eigein value) provides systematically high results but by less than 1.5%, for the biot number range from 0.11 to 10. See attached image.

Explanation:

8 0

3 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

Suppose within your web browser you click on a link to obtain a web page. The IP address for the associated URL is already cache

aleksandrvk [35]

Answer:

All the detailed steps are mentioned in pictures.

Explanation:

See attached pictures.

8 0

2 years ago