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3 years ago

10

Major processing methods for fiberglass composited include which of the following? Mark all that apply) a)- Open Mold b)- Closed

Mold c)- Preforming d)- Postforming e)- None of the above f)- All the above

1 answer:

Novay_Z [31]3 years ago

4 0

Answer:

it is f all of the above

Explanation:

let me know if im right

im not positive if im right but i should be right

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At a certain location, wind is blowing steadily at 10 m/s. Determine the mechanical energy of air per unit mass and the power ge

ale4655 [162]

Answer:

<em>a) 50 J/kg</em>

<em>b) 721 67 KW</em>

<em></em>

Explanation:

The velocity of the wind v = 10 m/s

diameter of the blades d = 70 m

efficiency of the turbine η = 30%

density of air ρ = 1.25 kg/m^3

The area of the blade A = $\pi d^2/4$

A = $\frac{3.142 * 70^2}{4}$ = 3848.95 m^2

The mechanical energy air per unit mass is gives as

e = $v^2/2$ = $\frac{10^2}{2}$ = <em>50 J/kg</em>

<em></em>

Theoretical Power of the turbine P = ρAve

where

ρ is the density of air

A is the area of the blade

v is the velocity of the wind

e is the energy per unit mass

substituting values, we have

P = 1.25 x 3848.95 x 10 x 50 = 2405593.75 W

Actual power = ηP

where η is the efficiency of the turbine

P is the theoretical power of the turbine

Actual power = 0.3 x 2405593.75 = 721678.1 W

==> <em>721 67 KW</em>

7 0

3 years ago

Which of the following is not caused by alcohol?

Brrunno [24]

Answer:

Inhibited comprehension

4 0

2 years ago

Read 2 more answers

Air at 40C flows over a 2 m long flat plate with a free stream velocity of 7 m/s. Assume the width of the plate (into the paper)

mezya [45]

Complete Question

Air at 40C flows over a 2 m long flat plate with a free stream velocity of 7m/s. Assume the width of the plate (into the paper) is 0.5 m. If the plate is at a co temperature of 100C,find:

The total heat transfer rate from the plate to the air

Answer:

$q=1.7845$

Explanation:

From the question we are told that:

Air Temperature $T_1=40c$

Length $l=2m$

Velocity $v=7m/s$

Width $w=0.5$

Constant temperature $T_t= 100C$

Generally the equation for Total heat Transfer is mathematically given by

$q=hA(T_s-T_\infty)$

Where

h=Convective heat transfer coefficient

$h=29.9075w/m^2k$

Therefore

$q=h(L*B)(T_s-T_\infty)$

$q=29.9075*(2*0.5)(100+273-(40+273))$

$q=1794.45w$

$q=1.7845$

5 0

3 years ago

Consider a single crystal of nickel oriented such that a tensile stress is applied along a [001] direction. If slip occurs on a

Elena L [17]

Answer:

$\mathbf{\tau_c =5.675 \ MPa}$

Explanation:

Given that:

The direction of the applied tensile stress =[001]

direction of the slip plane = [ $\bar 1$ 01]

normal to the slip plane = [111]

Now, the first thing to do is to calculate the angle between the tensile stress and the slip by using the formula:

$cos \lambda = \Big [\dfrac{d_1d_2+e_1e_2+f_1f_2}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_2^2+e_2^2+f_2^2) }} \Big]$

where;

$[d_1\ e_1 \ f_1]$ = directional indices for tensile stress

$[d_2 \ e_2 \ f_2]$ = slip direction

replacing their values;

i.e $d_1$ = 0 , $e_1$ = 0 $f_1$ = 1 & $d_2$ = -1 , $e_2$ = 0 , $f_2$ = 1

$cos \lambda = \Big [\dfrac{(0\times -1)+(0\times 0) + (1\times 1) }{\sqrt{(0^2+0^2+1^2)+((-1)^2+0^2+1^2) }} \Big]$

$cos \ \lambda = \dfrac{1}{\sqrt{2}}$

Also, to find the angle $\phi$ between the stress [001] & normal slip plane [111]

Then;

$cos \ \phi = \Big [\dfrac{d_1d_3+e_1e_3+f_1f_3}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_3^2+e_3^2+f_3^2) }} \Big]$

replacing their values;

i.e $d_1$ = 0 , $e_1$ = 0 $f_1$ = 1 & $d_3$ = 1 , $e_3$ = 1 , $f_3$ = 1

$cos \ \phi= \Big [ \dfrac{ (0 \times 1)+(0 \times 1)+(1 \times 1)} {\sqrt {(0^2+0^2+1^2)+(1^2+1^2 +1^2)} } \Big]$

$cos \phi= \dfrac{1} {\sqrt{3} }$

However, the critical resolved SS(shear stress) $\mathbf{\tau_c}$ can be computed using the formula:

$\tau_c = (\sigma )(cos \phi )(cos \lambda)$

where;

applied tensile stress $\sigma =$ 13.9 MPa

∴

$\tau_c =13.9\times ( \dfrac{1}{\sqrt{2}} )( \dfrac{1}{\sqrt{3}})$

$\mathbf{\tau_c =5.675 \ MPa}$

3 0

3 years ago

Calculate and plot the radial and circumferential stress distribution in the left ventricle at the end of systole (p 5 80 mmHg;

alexandr402 [8]

Answer:

62990.08 N/M

Explanation:

Circumferential stress, or hoop stress, a normal stress in the tangential (azimuth) direction. axial stress, a normal stress parallel to the axis of cylindrical symmetry. radial stress, a stress in directions coplanar with but perpendicular to the symmetry axis.

See attachment for the step by step solution of the given problem..

8 0

3 years ago