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Sergeeva-Olga [200]
2 years ago
8

Estimate the maximum expected thermal conductivity for a Cermet that contains 58 vol% titanium carbide (TiC) particles in a coba

lt matrix. Assume thermal conductivities of 27 and 69 W/m-K for TiC and Co respectively.
Engineering
1 answer:
Alik [6]2 years ago
8 0

Answer:

The right answer is "36.32 W/mk".

Explanation:

According to the question,

TiC = 76%

or,

      = 0.76

CO = 24%

or,

     = 0.24

Thermal conductivity of TiC,

= 26 W/mk

Thermal conductivity of CO,

= 69 W/mk

On applying the rule, we get

⇒  "K" \ of \ Cermet=(K)_{TiC} (V_f)_{TiC}+(K)_{CO} (V_m)_{CO}

On putting the given values, we get

⇒                            =(26)(0.76)+(69)(0.24)

⇒                            =19.76+16.56

⇒                            =36.32 \ W/mk

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3 years ago
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g Let the charges start infinitely far away and infinitely far apart. They are placed at (6 cm, 0) and (0, 3 cm), respectively,
irina1246 [14]

Answer:

a) V =10¹¹*(1.5q₁ + 3q₂)

b) U = 1.34*10¹¹q₁q₂

Explanation:

Given

x₁ = 6 cm

y₁ = 0 cm

x₂ = 0 cm

y₂ = 3 cm

q₁ = unknown value in Coulomb

q₂ = unknown value in Coulomb

A) V₁ = Kq₁/r₁

where   r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m

V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁

V₂ = Kq₂/r₂

where   r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m

V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂

The electric potential due to the two charges at the origin is

V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)

B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows

U = Kq₁q₂/r₁₂

where

r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m

then

U = 9*10⁹q₁q₂/(3√5/100)

⇒ U = 1.34*10¹¹q₁q₂

5 0
3 years ago
1. A team of students have designed a battery-powered cooler, which promises to keep beverages at a high-drinkability temperatur
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Answer:

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Minimum battery size needed = 3.03 Amp-hr

Explanation:

Temperature of the beverages, T_L = 36^0 F = 275.372 K

Outside temperature, T_H = 100^0F = 310.928 K

rate of insulation, Q = 100 Btu/h

To get the minimum electrical power required, use the relation below:

\frac{T_L}{T_H - T_L} = \frac{Q}{W} \\W = \frac{Q(T_H - T_L)}{T_L}\\W = \frac{100(310.928 - 275.372)}{275.372}\\W = 12.91 Btu/h\\1 Btu/h = 0.293071 W\\W = 12.91 * 0.293071\\W_{min} = 3.784 Watt

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Power = IV

W_{min} = I_{min} V\\3.784 = 5I_{min}\\I_{min} = \frac{3.784}{5} \\I_{min} = 0.7568 A

If the cooler is supposed to work for 4 hours, t = 4 hours

I_{min} = 0.7568 * 4\\I_{min} = 3.03 Amp-hr

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Which kind of fracture (ductile or brittle) is associated with each of the two crack propagation mechanisms?
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dutile is the correct answer

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Answer:

i dont know

Explanation:

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