Affordability is most concerned with:

yulyashka [42]

Answer:you need a 3d printer

Explanation:

5 0

3 years ago

An electrochemical cell is composed of pure nickel and pure iron electrodes immersed in solutions of their divalent ions. If the

xenn [34]

Answer:C 0.12 V

Explanation:

Given

Concentration of $Fe^{2+} M_1=0.40 M$

Concentration of $Ni^{2+} M_2=0.002 M$

Standard Potential for Ni and Fe are $V_2=-0.25 V$ and $V_1=-0.44 V$

$\Delta V=V_2-V_1-\frac{0.0592}{2}\log (\frac{M_1}{M_2})$

$\Delta V=-0.25-(-0.44)-\frac{0.0592}{2}\log (\frac{0.4}{0.002})$

$\Delta V=0.12\ V$

7 0

3 years ago

Pipe (2) is supported by a pin at bracket C and by tie rod (1). The structure supports a load P at pin B. Tie rod (1) has a diam

Galina-37 [17]

Answer:

P_max = 25204 N

Explanation:

Given:

- Rod 1 : Diameter D = 12 mm , stress_1 = 110 MPa

- Rod 2: OD = 48 mm , thickness t = 5 mm , stress_2 = 65 MPa

- x_1 = 3.5 mm ; x_2 = 2.1 m ; y_1 = 3.7 m

Find:

- Maximum Force P_max that this structure can support.

Solution:

- We will investigate the maximum load that each Rod can bear by computing the normal stress due to applied force and the geometry of the structure.

- The two components of force P normal to rods are:

Rod 1 : P*cos(Q)

Rod 2: - P*sin(Q)

where Q: angle subtended between x_1 and Rod 1 @ A. Hence,

Q = arctan ( y_1 / x_1)

Q = arctan (3.7 / 2.1 ) = 60.422 degrees.

- The normal stress in each Rod due to normal force P are:

Rod 1 : stress_1 = P*cos(Q) / A_1

Rod 2: stress_2 = - P*sin(Q) / A_2

- The cross sectional Area of both rods are A_1 and A_2:

A_1 = pi*D^2 / 4

A_2 = pi*(OD^2 - ID^2) / 4

- The maximum force for the given allowable stresses are:

Rod 1: P_max = stress_1 * A_1 / cos(Q)

P_max = (110*10^6)*pi*0.012^2 / 4*cos(60.422)

P_max = 25203.61848 N

Rod 2: P_max = stress_2 * A_2 / sin(Q)

P_max = (65*10^6)*pi*(0.048^2 - 0.038^2) / 4*sin(60.422)

P_max = 50483.4 N

- The maximum force that the structure can with-stand is governed by the member of the structure that fails first. In our case Rod 1 with P_max = 25204 N.

8 0

3 years ago

Air at 293k and 1atm flow over a flat plate at 5m/s. The plate is 5m wide and 6m long. (a) Determine the boundary layer thicknes

loris [4]

Answer:

a). 8.67 x $10^{-3}$ m

b).0.3011 m

c).0.0719 m

d).0.2137 N

e).1.792 N

Explanation:

Given :

Temperature of air, T = 293 K

Air Velocity, U = 5 m/s

Length of the plate is L = 6 m

Width of the plate is b = 5 m

Therefore Dynamic viscosity of air at temperature 293 K is, μ = 1.822 X $10^{-5}$ Pa-s

We know density of air is ρ = 1.21 kg / $m^{3}$

Now we can find the Reyonld no at x = 1 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 1}{1.822\times 10^{-5} }$

Re = 332052.6

Therefore the flow is laminar.

Hence boundary layer thickness is

δ = $\frac{5.x}{\sqrt{Re}}$

= $\frac{5\times 1}{\sqrt{332052.6}}$

= 8.67 x $10^{-3}$ m

a). Boundary layer thickness at x = 1 is δ = 8.67 X $10^{-3}$ m

b). Given Re = 100000

Therefore the critical distance from the leading edge can be found by,

Re = $\frac{\rho .U.x}{\mu }$

100000 = $\frac{1.21\times5\times x}{1.822 \times10^{-5}}$

x = 0.3011 m

c). Given x = 3 m from the leading edge

The Reyonld no at x = 3 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 3}{1.822\times 10^{-5} }$

Re = 996158.06

Therefore the flow is turbulent.

Therefore for a turbulent flow, boundary layer thickness is

δ = $\frac{0.38\times x}{Re^{\frac{1}{5}}}$

= $\frac{0.38\times 3}{996158.06^{\frac{1}{5}}}$

= 0.0719 m

d). Distance from the leading edge upto which the flow will be laminar,

Re = $\frac{\rho \times U\times x}{\mu }$

5 X $10^{5}$ = $\frac{1.21 \times 5\times x}{1.822\times 10^{-5}}}$

x = 1.505 m

We know that the force acting on the plate is

$F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

and $C_{D}$ at x= 1.505 for a laminar flow is = $\frac{1.328}{\sqrt{Re}}$

= $\frac{1.328}{\sqrt{5\times10 ^{5}}}$

= 1.878 x $10^{-3}$

Therefore, $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

= $\frac{1}{2}\times 1.878\times 10^{-3}\times 1.21\times (5\times 1.505)\times 5^{2}$

= 0.2137 N

e). The flow is turbulent at the end of the plate.

Re = $\frac{\rho \times U\times x}{\mu }$

= $\frac{1.21 \times 5\times 6}{1.822\times 10^{-5} }$

= 1992316

Therefore $C_{D}$ = $\frac{0.072}{Re^{\frac{1}{5}}}$

= $\frac{0.072}{1992316^{\frac{1}{5}}}$

= 3.95 x $10^{-3}$

Therefore $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho\times A\times U^{2}$

= $\frac{1}{2}\times 3.95\times 10^{-3}\times 1.21\times (5\times 6)\times 5^{2}$

= 1.792 N

3 0

3 years ago

Before taking off a plane travels at a speed of 1/4 km per second. The runaway is 5 km. How many seconds does it take the plane

Vikentia [17]

Answer:

1 5segundos

Explanation:

8 0

3 years ago