Random question, does anyone here use Lego, do not answer unless that is a yes

Driving Distraction Brainstorming Session

Leto [7]

texting, phone calls, putting on makeup, brushing hair, movies playing in car, loud music, children, and that's pretty much all I could think of

please give <u>BRAINLIEST ANSWER └[T‸T]┘</u>

5 0

3 years ago

An ideal Otto cycle has a compression ratio of 9.2 and uses air as the working fluid. At the beginning of the compression proces

Allushta [10]

Answer:

(a) The amount of heat transferred to the air, $q_{out}$ is 215.5077 kJ/kg

(b) The net work output, $W_{net}$ , is 308.07 kJ/kg

(c) The thermal efficiency is 58.8%

(d) The Mean Effective Pressure, MEP, is 393.209 kPa

Explanation:

(a) The assumptions made are;

$c_p$ = 1.005 kJ/(kg·K), $c_v$ = 0.718 kJ/(kg·K), R = 0.287 kJ/(kg·K),

Process 1 to 2 is isentropic compression, therefore;

$T_{2}= T_{1}\left (\dfrac{v_{1}}{v_{2}} \right )^{k-1} = 300.15\times 9.2^{0.4} = 729.21 \, K$

From;

$\dfrac{p_{1}\times v_{1}}{T_{1}} = \dfrac{p_{2}\times v_{2}}{T_{2} }$

We have;

$p_{2} = \dfrac{p_{1}\times v_{1}\times T_{2}}{T_{1} \times v_{2}} = \dfrac{98\times 9.2\times 729.21}{300.15 } = 2190.43 \, kPa$

Process 2 to 3 is reversible constant volume heating, therefore;

$\dfrac{p_3}{T_3} =\dfrac{p_2}{T_2}$

p₃ = 2 × p₂ = 2 × 2190.43 = 4380.86 kPa

$T_3 = \dfrac{p_3 \times T_2}{p_2} =\dfrac{4380.86 \times 729.21}{2190.43} = 1458.42 \, K$

Process 3 to 4 is isentropic expansion, therefore;

$T_{3}= T_{4}\left (\dfrac{v_{4}}{v_{3}} \right )^{k-1}$

$1458.42= T_{4} \times \left (9.2 \right )^{0.4}$

$T_4 = \dfrac{1458.42}{(9.2)^{0.4}} = 600.3 \, K$

$q_{out} = m \times c_v \times (T_4 - T_1) = 0.718 \times (600.3 - 300.15) = 215.5077 \, kJ/kg$

The amount of heat transferred to the air, $q_{out}$ = 215.5077 kJ/kg

(b) The net work output, $W_{net}$ , is found as follows;

$W_{net} = q_{in} - q_{out}$

$q_{in} = m \times c_v \times (T_3 - T_2) = 0.718 \times (1458.42 - 729.21) = 523.574 \, kJ/kg$

$\therefore W_{net} = 523.574 - 215.5077 = 308.07 \, kJ/kg$

(c) The thermal efficiency is given by the relation;

$\eta_{th} = \dfrac{W_{net}}{q_{in}} \times 100= \dfrac{308.07}{523.574} \times 100= 58.8\%$

(d) From the general gas equation, we have;

$V_{1} = \dfrac{m\times R\times T_{1}}{p_{1}} = \dfrac{1\times 0.287\times 300.15}{98} =0.897\, m^{3}/kg$

The Mean Effective Pressure, MEP, is given as follows;

$MEP =\dfrac{W_{net}}{V_1 - V_2} = \dfrac{W_{net}}{V_1 \times (1- 1/r)}= \dfrac{308.07}{0.897\times (1- 1/9.2)} = 393.209 \, kPa$

The Mean Effective Pressure, MEP = 393.209 kPa.

3 0

3 years ago

A tensile test was operated to test some important mechanical properties. The specimen has a gage length = 1.8 in and diameter =

oee [108]

Answer:

a) 60000 psi

b) 1.11*10^6 psi

c) 112000 psi

d) 30.5%

e) 30%

Explanation:

The yield strength is the load applied when yielding behind divided by the section.

yield strength = Fyield / A

A = π/4 * D^2

A = 0.5 in^2

ys = Fy * A

y2 = 30000 * 0.5 = 60000 psi

The modulus of elasticity (E) is a material property that is related to the object property of stiffness (k).

k = E * L0 / A

And the stiffness is related to change of length:

Δx = F / k

Then:

Δx = F * A / (E * L0)

E = F * A / (Δx * L0)

When yielding began (approximately the end of the proportional peroid) the force was of 30000 lb and the change of length was

Δx = L - L0 = 1.8075 - 1.8 = 0.0075

Then:

E = 30000 * 0.5 / (0.0075 * 1.8) = 1.11*10^6 psi

Tensile strength is the strees at which the material breaks.

The maximum load was 56050 lb, so:

ts = 56050 / 0.5 = 112000 psi

The percent elongation is calculated as:

e = 100 * (L / L0)

e = 100 * (2.35 / 1.8 - 1) = 30.5 %

If it necked with and area of 0.35 in^2 the precent reduction in area was:

100 * (1 - A / A0)

100 * (1 - 0.35 / 0.5) = 30%

5 0

3 years ago

Can you answer what is attached.

kirill [66]

29.4 bro I hope that helps

3 0

3 years ago

For each of the following combinations of parameters, determine if the material is a low-loss dielectric, a quasi-conductor, or

Alborosie

Answer:

Glass: Low-Loss dielectric

α = 8.42*10^-11 Np/m

β = 468.3 rad/m

λ = 1.34 cm

up = 1.34*10^8 m/s

ηc = 168.5 Ω

Tissue: Quasi-Conductor

α = 9.75 Np/m

β = 12.16 rad/m

λ = 51.69 cm

up = 0.52*10^8 m/s

ηc = 39.54 + j 31.72 Ω

Wood: Good conductor

α = 6.3*10^-4 Np/m

β = 6.3*10^-4 Np/m

λ = 10 km

up = 0.1*10^8 m/s

ηc = 6.28*( 1 + j )

Explanation:

Given:

Glass with µr = 1, εr = 5, and σ = 10−12 S/m at 10 GHz

Animal tissue with µr = 1, εr = 12, and σ = 0.3 S/m at 100 MHz.

Wood with µr = 1, εr = 3, and σ = 10−4 S/m at 1 kHz

Find:

Determine if the material is a low-loss dielectric, a quasi-conductor, or a good conductor, and then calculate α, β, λ, up, and ηc:

Solution:

- We need to determine the loss tangent to determine category of the medium as follows:

σ / w*εr*εo

Where, w is the angular speed of wave

εo is the permittivity of free space = 10^-9 / 36*pi

- Now we classify as follows:

$Glass = \frac{10^-^1^2 }{2*\pi * 10*10^9 * \frac{5*10^-^9}{36\pi } } = 3.6*10^-^1^3\\\\Tissue = \frac{0.3 }{2*\pi * 100*10^6 * \frac{12*10^-^9}{36\pi } } = 4.5\\\\Wood = \frac{10^-^4 }{2*\pi * 1*10^3 * \frac{3*10^-^9}{36\pi } } = 600\\$

- For σ / w*εr*εo < 0.01 --- Low-Loss dielectric and σ / w*εr*εo > 100 --- Good conducting material.

Glass: Low-Loss dielectric

Tissue: Quasi-Conductor

Wood: Good conductor

- Now we will use categorized material base equations from Table 17-1 as follows:

Glass: Low-Loss dielectric

α = (σ / 2)*sqrt(u / εr*εo) = (10^-12 / 2)*sqrt( 4*pi*10^-7/5*8.85*10^-12)

α = 8.42*10^-11 Np/m

β = w*sqrt (u*εr*εo) = 2pi*10^10*sqrt (4*pi*10^-7*5*8.85*10^-12)

β = 468.3 rad/m

λ = 2*pi / β = 2*pi / 468.3

λ = 1.34 cm

up = λ*f = 0.0134*10^10

up = 1.34*10^8 m/s

ηc = sqrt ( u / εr*εo ) = sqrt( 4*pi*10^-7/12*8.85*10^-12)

ηc = 168.5 Ω

Tissue: Quasi-Conductor

α = (σ / 2)*sqrt(u / εr*εo) = (0.3 / 2)*sqrt( 4*pi*10^-7/12*8.85*10^-12)

α = 9.75 Np/m

β = w*sqrt (u*εr*εo) = 2pi*100*10^6*sqrt (4*pi*10^-7*12*8.85*10^-12)

β = 12.16 rad/m

λ = 2*pi / β = 2*pi / 12.16

λ = 51.69 cm

up = λ*f = 0.5169*100*10^6

up = 0.52*10^8 m/s

ηc = sqrt ( u / εr*εo )*( 1 - j (σ / w*εr*εo))^-0.5

ηc = sqrt (4*pi*10^-7*12*8.85*10^-12)*( 1 - j 4.5)^-0.5

ηc = 39.54 + j 31.72 Ω

Wood: Good conductor

α = sqrt (pi*f*σ u) = sqrt( pi* 10^3 *4*pi* 10^-7 * 10^-4 )

β = α = 6.3*10^-4 Np/m

λ = 2*pi / β = 2*pi / 6.3*10^-4

λ = 10 km

up = λ*f = 10,000*1*10^3

up = 0.1*10^8 m/s

ηc = α*( 1 + j ) / б = 6.3*10^-4*( 1 + j ) / 10^-4

ηc = 6.28*( 1 + j )

8 0

3 years ago