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3 years ago

Three-dimensional measuring references all of these EXCEPT:

Engineering

1 answer:

seraphim [82]3 years ago

8 0

Answer:

Radius. It is only used for Spheres

Explanation:

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A budding electronics hobbyist wants to make a simple 1.0-nF capacitor for tuning her crystal radio, using two sheets of aluminu

bazaltina [42]

Answer:

a. 8 sheets of paper is needed between her plates to get the proper capacitance

b. Area of Aluminum Foil needed = 0.45m²

c. To keep a 1.0-nF, a larger area of Teflon is required.

Explanation:

First, we need to calculate the distance between two plates.

This is given by

d = Kε0A/C

Where

K = 3

ε0 = Physical Constant = 8.854 * 10^-12 C²/Nm²

A = Area = 22 * 28 = 616cm² = 0.0616m²

C = 1.0-nF = 1 * 10^-12F

So, d = (3 * 8.854 * 10^-12 C²/Nm² * 0.0616) / (1 * 10^-12F)

d = 1.64 * 10^-3m

d = 1.64mm

Now, that the distance has been solved.

The Number of Sheets, N is given by

N = d/d,sheet where d, sheet =the sheet thickness = 0.2mm

N = 1.64/0.2

N = 8.2

N = 8 sheets --- Approximated

Here, she's changed the diameter of the sheets to 12mm

Well make use of the formula in (a) above

Using d = Kε0A/C

Where

d = 12 * 10^-3m

Other constraints remain unchanged

Make A the subject of formula

A = dC/Kε0

A = (12 * 10^-3m * 1 * 10^-12F)/(3 * 8.854 * 10^-12 C²/Nm²)

A= 0.45m²

c. From (b) above

A ∝ 1/K

As the dielectric constant increase, the area decreases

The dielectric constant of a Teflon is 2.1

This means that if she used a Teflon instead, the area will be larger.

So, to keep a 1.0-nF, a larger area of Teflon is required.

7 0

4 years ago

A soil is at a void ratio e = 0.90 with a specific gravity of the solid particles Gs = 2.70.

Alexus [3.1K]

Answer:

The correct answers are:

a. % w = 33.3%

b. mass of water = 45g

Explanation:

First, let us define the parameters in the question:

void ratio e = $\frac{V_v}{V_s}$ = $\frac{\left\begin{array}{ccc}volume&of&void\end{array}\right}{\left\begin{array}{ccc}volume&of&solid\end{array}\right}------ (1)$

Specific gravity $G_{s}$ = $\frac{P_s}{P_w} =$ $\frac{\left\begin{array}{ccc}density&of&soil\end{array}\right}{\left\begin{array}{ccc}density&of&water\end{array}\right}------ (2)$

% Saturation S = $\frac{V_w}{Vv}$ × $\frac{100}{1}$ = $\frac{\left\begin{array}{ccc}volume&of&water\end{array}\right}{\left\begin{array}{ccc}volume&of&void\end{array}\right}$ × $\frac{100}{1}--------(3)$

water content w = $\frac{M_w}{M_s}$ = $\frac{\left\begin{array}{ccc}mass&of&water\end{array}\right}{\left\begin{array}{ccc}mass&of&solid\end{array}\right} ------(4)$

a) To calculate the lower and upper limits of water content:

when S = 100%, it means that the soil is fully saturated and this will give the upper limit of water content.

when S < 100%, the soil is partially saturated, and this will give the lower limit of water content.

Note; S = 0% means that the soil is perfectly dry. Hence, when s = 1 will give the lowest limit of water content.

To get the relationship between water content and saturation, we will manipulate the equations above;

w = $\frac{M_w}{Ms}$

Recall; mass = Density × volume

w = $\frac{V_wP_w}{V_sP_s} ------(5)$

From eqn. (2) $G_{s}$ = $\frac{P_s}{P_w}$

∴ $\frac{1}{G_s} = \frac{P_w}{P_s} ------(6)$

putting eqn. (6) into (5)

w = $\frac{V_w}{V_sG_s} -----(7)$

Again, from eqn (1)

$V_s = \frac{V_v}{e}$

substituting into eqn. (7)

$w = \frac{V_w}{\frac{V_v}{e}{G_s} } = \frac{V_w e}{V_vG_s} \\ but \frac{V_w}{V_v} = S$

∴ $w = \frac{Se}{G_s} -----(8)$

With eqn. (7), we can calculate

upper limit of water content

when S = 100% = 1

Given, $G_{s} = 2.7, e= 0.9$

∴ $w= \frac{0.9*1}{2.7} = 0.333$

∴ %w = 33.3%

Lower limit of water content

when S = 1% = 0.01

$w= \frac{0.01*0.9 }{2.7} = 0.0033$

∴ % w = 0.33%

b) Calculating mass of water in 100 cm³ sample of soil ( $P_w=\frac{1_g}{cm^{3} }$ )

Given, $V_{s} = 100 cm^{3 }$ , S = 50% = 0.5

%S = $\frac{V_w}{V_v}$ × $\frac{100}{1}$ = $\frac{V_w}{eV_s}$ × $\frac{100}{1}$

0.50 = $\frac{V_w}{0.9* 100} = 45cm^{3}$

mass of water = $P_wV_w= 1 * 45 = 45_{g}$

7 0

4 years ago

Line.

Veronika [31]

Air supplied to a pneumatic system is supplied through the C. Actuator

Explanation

Pneumatic systems are like hydraulic systems, it is just that these systems uses compressed air rather than hydraulic fluid. Pneumatic systems are used widely across the industries. these pneumatic systems needs a constant supply of compressed air to operate. This is provided by an air compressor. The compressor sucks in air at a very high rate from the environment and stores it in a pressurized tank. the Air is supplied thereafter with the help of a actuator valve that is a more sophisticated form of a valve.

From the above statement it is clear that Air supplied to a pneumatic system is supplied through the Actuator

7 0

4 years ago

An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standar

Dennis_Churaev [7]

Answer:

a) the heat exchanger area required for the evaporator is 11178.236 m²

b) the required flow rate is 1993630.38 kg/s

Explanation:

Given the data in the question;

Water temperature near the surface = 300 K

temperature at reasonable depths ( cold ) = 280 K

power plant output W' = 2 MW

efficiency η = 3% = 0.03

we know that; efficiency η = W' $_{power-out$ / Q $_{supplied$

we substitute

0.03 = 2 / Q $_{supplied$

Q $_{supplied$ = 2 / 0.03

Q $_{supplied$ = 66.667 MW = 66.667 × 10⁶ Watt

T $h_{in$ = 300 K T $h_{out$ = 292 K

T $c_{in$ = 290 K T $c_{out$ = 290 K

Now, Heat transfer in evaporator;

Q = UA( LMTD )

LMTD = (ΔT₁ - ΔT₂) / ln( ΔT₁ / ΔT₂ )

first we get ΔT₁ and ΔT₂

ΔT₁ = T $h_{in$ - T $c_{out$ = 300 - 290 = 10 K

ΔT₂ = T $h_{out$ - T $c_{in$ = 292 - 290 = 2 K

so we substitute into our equation;

LMTD = (10 - 2) / ln( 10 / 2 )

LMTD = 8 / ln( 5 )

LMTD = 8 / 1.6094379

LMTD = 4.97

a) Heat transfer Area will be;

Q $_H$ = UA( LMTD )

we substitute

66.667 × 10⁶ = 1200 × A × 4.97

66.667 × 10⁶ = 5964 × A

A = (66.667 × 10⁶) / 5964

A = 11178.236 m²

Therefore, the heat exchanger area required for the evaporator is 11178.236 m²

b) Flow rate

we know that;

Q $_H$ = m'C $_P$ ( $T_{in$ - $T_{out$ )

specific heat capacity of water Cp = 4.18 (kJ/kg∙°C)

we substitute

66.667 × 10⁶ = m' × 4.18 × ( 300 - 292 )

66.667 × 10⁶ = m' × 33.44

m' = ( 66.667 × 10⁶ ) / 33.44

m' = 1993630.38 kg/s

Therefore, the required flow rate is 1993630.38 kg/s

7 0

3 years ago

Calculate the value of ni for gallium arsenide (GaAs) at T = 300 K. The constant B = 3. 56 times 1014 9cm -3 K-3/2) and the band

meriva

This question is incomplete, the complete question is;

Calculate the value of ni for gallium arsenide (GaAs) at T = 300 K.

The constant B = 3.56×10¹⁴ (cm⁻³ K^-3/2) and the bandgap voltage E = 1.42eV.

Answer: the value of ni for gallium arsenide (GaAs) is 2.1837 × 10⁶ cm⁻³

Explanation:

Given that;

T = 300k

B = 3.56×10¹⁴ (cm⁻³ K^-3/2)

Eg = 1.42 eV

we know that, the value of Boltzmann constant k = 8.617×10⁻⁵ eV/K

so to find the ni for gallium arsenide;

ni = B×T^(3/2) e^ ( -Eg/2kT)

we substitute

ni = (3.56×10¹⁴)(300^3/2) e^ ( -1.42 / (2× 8.617×10⁻⁵ 300))

ni = (3.56×10¹⁴)(5196.1524)e^-27.4651

ni = (3.56×10¹⁴)(5196.1524)(1.1805×10⁻¹²)

ni = 2.1837 × 10⁶ cm⁻³

Therefore the value of ni for gallium arsenide (GaAs) is 2.1837 × 10⁶ cm⁻³

4 0

3 years ago