In order to avoid slipping in the shop, your footwear should _

All of these are true about a magnesium part EXCEPT that it:

d1i1m1o1n [39]

The option that is an exception about the properties of magnesium is the option;

<u>Is typically made from </u><u>stampings</u>

The reason why the option is an exception is given as follows:

Magnesium is a nonferrous metal as it is usually combined with manganese, zinc and aluminum.

Within the group of structural metals, magnesium is the lightest, having a high strength-to-weight ratio.

Magnesium posses good dampening characteristics, and is capable of reducing vibration of mechanical parts.

<em />

<em>However</em>

Magnesium parts used for vehicle construction are die cast, and not made from stampings because magnesium is more difficult to stamp compared to aluminum.

Therefore, the correct option for the statement that is an exception from the true statements about magnesium is magnesium; <u>is typically made from stampings</u>

Learn more about the properties of magnesium here:

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8 0

2 years ago

Air enters the compressor of a cold air-standard Brayton cycle with regeneration and reheat at 100 kPa, 300 K, with a mass flow

yanalaym [24]

Answer:

a. 47.48%

b. 35.58%

c. 2957.715 KW

Explanation:

$T_2 =T_1 + \dfrac{T_{2s} - T_1}{\eta _c}$

T₁ = 300 K

$\dfrac{T_{2s}}{T_1} = \left( \dfrac{P_{2}}{P_1} \right)^{\dfrac{k-1}{k} }$

$T_{2s} = 300 \times (10) ^{\dfrac{0.4}{1.4} }$

$T_{2s}$ = 579.21 K

T₂ = 300+ (579.21 - 300)/0.8 = 649.01 K

T₃ = T₂ + $\epsilon _{regen}$ (T₅ - T₂)

T₄ = 1400 K

Given that the pressure ratios across each turbine stage are equal, we have;

$\dfrac{T_{5s}}{T_4} = \left( \dfrac{P_{5}}{P_4} \right)^{\dfrac{k-1}{k} }$

$T_{5s}$ = 1400× $\left( 1/\sqrt{10} \right)^{\dfrac{0.4}{1.4} }$ = 1007.6 K

T₅ = T₄ + ( $T_{5s}$ - T₄)/ $\eta _t$ = 1400 + (1007.6- 1400)/0.8 = 909.5 K

T₃ = T₂ + $\epsilon _{regen}$ (T₅ - T₂)

T₃ = 649.01 + 0.8*(909.5 - 649.01 ) = 857.402 K

T₆ = 1400 K

$\dfrac{T_{7s}}{T_6} = \left( \dfrac{P_{7}}{P_6} \right)^{\dfrac{k-1}{k} }$

$T_{7s}$ = 1400× $\left( 1/\sqrt{10} \right)^{\dfrac{0.4}{1.4} }$ = 1007.6 K

T₇ = T₆ + ( $T_{7s}$ - T₆)/ $\eta _t$ = 1400 + (1007.6 - 1400)/0.8 = 909.5 K

a. $W_{net \ out}$ = cp(T₆ -T₇) = 1.005 * (1400 - 909.5) = 492.9525 KJ/kg

Heat supplied is given by the relation

cp(T₄ - T₃) + cp(T₆ - T₅) = 1.005*((1400 - 857.402) + (1400 - 909.5)) = 1038.26349 kJ/kg

Thermal efficiency of the cycle = (Net work output)/(Heat supplied)

Thermal efficiency of the cycle = (492.9525 )/(1038.26349 ) =0.4748 = 47.48%

b. $bwr = \dfrac{W_{c,in}}{W_{t,out}}$

bwr = (T₂ -T₁)/[(T₄ - T₅) +(T₆ -T₇)] = (649.01 - 300)/((1400 - 909.5) + (1400 - 909.5)) = 35.58%

c. Power = 6 kg *492.9525 KJ/kg = 2957.715 KW

3 0

3 years ago

Which of the following can cause a flopping sound at the front of the engine

Irina-Kira [14]

B) timing chain too loose

3 0

2 years ago

Oil (SAE 30) at 15.6 oC flows steadily between fixed, horizontal, parallel plates. The pressure drop per unit length along the c

Nitella [24]

Answer:

(a) The volume rate of flow per meter width = 5.6*10⁻³ m²/s

(b) The shear stress acting on the bottom plate = 157.5 N/m²

(c) The velocity along the centerline of the channel = 0.93 m/s

Explanation:

(a)

Calculating the distance of plate from centre line using the formula;

h = d/2

where h = distance of plate

d = diameter of flow = 9 mm

Substituting, we have;

h = 9/2

= 4.5 mm = 4.5*10^-3 m

Calculating the volume flow rate using the formula;

Q = (2h³/3μ)* (Δp/L)

Where;

Q = volume flow rate

h = distance of plate = 4.5*10^-3 m

μ = dynamic viscosity = 0.38 N.s/m²

(Δp/L) = Pressure drop per unit length = 35 kPa/m = 35000 Pa

Substituting into the equation, we have;

Q = (2*0.0045³/3*0.38) *(35000)

= (1.8225*10⁻⁷/1.14) * (35000)

= 1.60*10⁻⁷ * 35000

= 5.6*10⁻³ m²/s

Therefore, the volume flow rate = 5.6*10⁻³ m³/s

(b) Calculating the shear stress acting at the bottom plate using the formula;

τ = h*(Δp/L)

= 0.0045* 35000

= 157.5 N/m²

(c) Calculating the velocity along the centre of the channel using the formula;

u(max) = h²/2μ)* (Δp/L)

= (0.0045²/2*0.38) * 35000

=2.664*10⁻⁵ *35000

= 0.93 m/s

7 0

3 years ago

A company intends to market a new product and it estimates that there is a 20% chance that it will be first in the market

Stells [14]

The EMV - Ending Market Value is given as:
$2,400,000.

<h3>How is the EMV Arrived At?</h3>

The EMV is given as:

BMV x (i + r); Where

BMV is the Beginning Market Value; and

r is the interest rateor percentage given.

Hence the EMV = 2,000,000 x ( 1 + 20%)

= 2,000,000 x 1.2

= $ 2, 400,000.

It is to be noted that the BMV is the Beginning Market Value which is the value of an investment at the start of the business period.

Learn more about Market Value at:

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4 0

2 years ago

In order to avoid slipping in the shop, your footwear should __