To make an even better electrical junction, what should you do?

How do you calculate the rate of heat transfer between two bodies with a small area of physical contact.

gavmur [86]

Answer:

So the rate of heat transfer to an object is equal to the thermal conductivity of the material the object is made from, multiplied by the surface area in contact, multiplied by the difference in temperature between the two objects, divided by the thickness of the material.

3 0

3 years ago

If you've had anything to drink and can't get a ride home from sober friends or family, __________.

laiz [17]

Answer:

The answer is D

Explanation:

3 0

3 years ago

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Consider a mild steel specimen with yield strength of 43.5 ksi and Young's modulus of 29,000 ksi. It is stretched up to a point

mezya [45]

Answer:

0.05%

Explanation:

From the question, we have;

The yield strength of the mild steel, $\sigma _c$ = 43.5 ksi

Young's modulus of elasticity, ∈ = 29,000 ksi

The total strain, $\epsilon _c$ = 0.2% = 0.002

The inelatic strain $\epsilon_c^{in}$ is given as follows;

$\epsilon_c^{in}$ = $\epsilon _c$ - $\sigma _c$ /∈

Therefore, we have;

$\epsilon_c^{in}$ = 0.002 - 43.5/(29,000) = 0.0005

Therefore, the inelastic strain, $\epsilon_c^{in}$ = 0.0005 = 0.05%

Taking the inelastic strain as the residual strain, we have;

The residual strain = 0.05%

7 0

3 years ago

To measure an object accurately, what point on the ruler would you align with the object edge

Firlakuza [10]

Answer:

Along the zero to measure an object on a ruler

3 0

3 years ago

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Find values of the intrinsic carrier concentration n for silicon at –70° 0° 20° C, 100° C, and C. At 125° each temperature, what

Dominik [7]

Answer:

Part (i) at –70° C, intrinsic carrier concentration of silicon is 2.865 x 10⁵ carriers/cm³ and fraction of the atoms ionized is 5.37 x 10⁻¹⁸

Part (ii) at 0° C, intrinsic carrier concentration of silicon is 1.533 x 10⁹ carriers/cm³ and fraction of the atoms ionized is 3.067 x 10⁻¹⁴

Part (iii) at 20° C, intrinsic carrier concentration of silicon is 8.652 x 10⁹ carriers/cm³ and fraction of the atoms ionized is 1.731 x 10⁻¹³

Part (iv) at 100° C, intrinsic carrier concentration of silicon is 1.444 x 10¹² carriers/cm³ and fraction of the atoms ionized is 2.889 x 10⁻¹¹

Part (iv) at 125° C, intrinsic carrier concentration of silicon is 4.754 x 10¹² carriers/cm³ and fraction of the atoms ionized is 9.508 x 10⁻¹¹

Explanation:

$ni^2 = BT^3(e^{\frac{-E_g}{KT}})\\\\ni = \sqrt{ BT^3(e^{\frac{-E_g}{KT}})}$

where;

B = 5.4 x 10⁻³¹

Eg = 1.12 ev

K = 8.62 x 10⁻⁵ eV/K

T = (273 + ⁰C) K

Number of atoms in silicon crystal = 5 x 10²² atoms/cm³

Part (i) For –70° C, T = (273 -70 ⁰C)K = 203 K

$ni = \sqrt{ 5.4*10^{31}*203^3(e^ \ {\frac{-1.12}{8.62*10^{-5}*203}})}} \ =2.685*10^5 \ carriers/cm^3$

$Fraction \ of \ atoms \ ionized = \frac{2.685*10^5}{5 *10^{22}} = 5.370 *10^{-18}$

Part (ii) For 0° C, T = (273 +0 ⁰C)K = 273 K

$ni = \sqrt{ 5.4*10^{31}*273^3(e^ \ {\frac{-1.12}{8.62*10^{-5}*273}})}} \ =1.533*10^9 \ carriers/cm^3$

$Fraction \ of \ atoms \ ionized = \frac{1.533*10^9}{5 *10^{22}} = 3.067 *10^{-14}$

Part (iii) For 20° C, T = (273 + 20 ⁰C)K = 293 K

$ni = \sqrt{ 5.4*10^{31}*293^3(e^ \ {\frac{-1.12}{8.62*10^{-5}*293}})}} \ =8.652*10^9 \ carriers/cm^3$

$Fraction \ of \ atoms \ ionized = \frac{8.652*10^9}{5 *10^{22}} = 1.731 *10^{-13}$

Part (iv) For 100° C, T = (273 + 100 ⁰C)K = 373 K

$ni = \sqrt{ 5.4*10^{31}*373^3(e^ \ {\frac{-1.12}{8.62*10^{-5}*373}})}} \ =1.444*10^{12} \ carriers/cm^3$

$Fraction \ of \ atoms \ ionized = \frac{1.444*10^{12}}{5 *10^{22}} = 2.889 *10^{-11}$

Part (v) For 125° C, T = (273 + 125 ⁰C)K = 398 K

$ni = \sqrt{ 5.4*10^{31}*398^3(e^ \ {\frac{-1.12}{8.62*10^{-5}*398}})}} \ =4.754*10^{12} \ carriers/cm^3$

$Fraction \ of \ atoms \ ionized = \frac{4.754*10^{12}}{5 *10^{22}} = 9.508 *10^{-11}$

7 0

3 years ago