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2 years ago

alguien me ayuda con una tarea? Está en mi perfil de matemática, porfavorrr regalaré corona pero porfavor

Engineering

1 answer:

tensa zangetsu [6.8K]2 years ago

5 0

Answer:

HUH?

Explanation:

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You can change lanes during a turn long as there’s no traffic and you driving slowly

Vanyuwa [196]

Your allowed to switch lanes as long as the road is clear and you use signals.

5 0

2 years ago

The compressed-air tank has an inner radius r and uniform wall thickness t. The gage pressure inside the tank is p and the centr

Sedaia [141]

Answer:

Explanation:

Given that:

The Inside pressure (p) = 1402 kPa

= 1.402 × 10³ Pa

Force (F) = 13 kN

= 13 × 10³ N

Thickness (t) = 18 mm

= 18 × 10⁻³ m

Radius (r) = 306 mm

= 306 × 10⁻³ m

Suppose we choose the tensile stress to be (+ve) and the compressive stress to be (-ve)

Then;

the state of the plane stress can be expressed as follows:

$(\sigma_ x) = \dfrac{Pd}{4t}+ \dfrac{F}{2 \pi rt}$

Since d = 2r

Then:

$(\sigma_ x) = \dfrac{Pr}{2t}+ \dfrac{F}{2 \pi rt}$

$(\sigma_ x) = \dfrac{1402 \times 306 \times 10^3}{2(18)}+ \dfrac{13 \times 10^3}{2 \pi \times 306\times 18 \times 10^{-3} \times 10^{-3}}$

$(\sigma_ x) = \dfrac{429012000}{36}+ \dfrac{13000}{34607.78467}$

$(\sigma_ x) = 11917000.38$

$(\sigma_ x) = 11.917 \times 10^6 \ Pa$

$(\sigma_ x) = 11.917 \ MPa$

$\sigma_y = \dfrac{pd}{2t} \\ \\ \sigma_y = \dfrac{pr}{t} \\ \\ \sigma _y = \dfrac{1402\times 10^3 \times 306}{18} \ N/m^2 \\ \\ \sigma _y = 23.834 \times 10^6 \ Pa \\ \\ \sigma_y = 23.834 \ MPa$

When we take a look at the surface of the circular cylinder parabolic variation, the shear stress is zero.

Thus;

$\tau _{xy} =0$

3 0

3 years ago

Copper spheres of 20-mm diameter are quenched by being dropped into a tank of water that is maintained at 280 K . The spheres ma

Ivenika [448]

Answer:

The height of the water is 1.25 m

Explanation:

copper properties are:

Kc=385 W/mK

D=20x10^-3 m

gc=8960 kg/m^3

Cp=385 J/kg*K

R=10x10^-3 m

Water properties at 280 K

pw=1000 kg/m^3

Kw=0.582

v=0.1247x10^-6 m^2/s

The drag force is:

$F_{D} =\frac{1}{2} Co*p_{w} A*V^{2}$

The bouyancy force is:

$F_{B} =V*p_{w} *g$

The weight is:

$W=V*p_{c} *g$

Laminar flow:

$v_{T} =\frac{p_{c}-p_{w}*g*D^{2} }{18*u} =\frac{(8960-1000)*9.8*(20x10^{-3})^{2} }{18*0.00143} =1213.48 m/s$

Reynold number:

$Re=\frac{1000*1213.48*20x10^{-3} }{0.00143} \\Re>>1$

Not flow region

For Newton flow region:

$v_{T} =1.75\sqrt{(\frac{p_{c}-p_{w} }{p_{w} })gD }=1.75\sqrt{(\frac{8960-1000}{1000} )*9.8*20x10^{-3} } =2.186m/s$

$Re=\frac{1000*2.186*20x10^{-3} }{0.00143} =30573.4$

$Pr=\frac{\frac{u}{p} }{\frac{K}{pC_{p} } } =\frac{u*C_{p} }{k} =\frac{0.0014394198}{0.582} =10.31$

$Nu=2+(0.4Re^{1/2} +0.06Re^{2/3} )Pr^{2/5} (u/us)^{1/4} \\Nu=2+(0.4*30573.4^{1/2}+0.06*30573.4^{2/3} )*10.31^{2/5} *(0.00143/0.00032)^{1/4} \\Nu=476.99$

$Nu=\frac{h*d}{K_{w} } \\h=\frac{476.99*0.582}{20x10^{-3} } =13880.44W/m^{2} K$

$\frac{T-T_{c} }{T_{w}-T_{c} } =e^{-t/T} \\T=\frac{m_{c}C_{p} }{hA_{c} } =\frac{8960*10x10^{-3}*385 }{13880.44*3} =0.828 s$

$e^{-t/0.828} =\frac{320-280}{360-280} \\t=0.573\\heightofthewater=2.186*0.573=1.25m$

8 0

3 years ago

700.0 liters of a gas are prepared at 760.0 mmHg and 100.0 °C. The gas is placed into a tank under high pressure. When the tank

ololo11 [35]

Answer:

The volume of the gas is 11.2 L.

Explanation:

Initially, we have:

V₁ = 700.0 L

P₁ = 760.0 mmHg = 1 atm

T₁ = 100.0 °C

When the gas is in the thank we have:

V₂ =?

P₂ = 20.0 atm

T₂ = 32.0 °C

Now, we can find the volume of the gas in the thank by using the Ideal Gas Law:

$PV = nRT$

$V_{2} = \frac{nRT_{2}}{P_{2}}$ (1)

Where R is the gas constant

With the initials conditions we can find the number of moles:

$n = \frac{P_{1}V_{1}}{RT_{1}}$ (2)

By entering equation (2) into (1) we have:

$V_{2} = \frac{P_{1}V_{1}}{RT_{1}}*\frac{RT_{2}}{P_{2}} = \frac{1 atm*700.0 L*32.0 ^{\circ}}{100.0 ^{\circ}*20.0 atm} = 11.2 L$

Therefore, When the gas is placed into a tank the volume of the gas is 11.2 L.

I hope it helps you!

5 0

3 years ago

Derive an expression for the specific heat difference of a substance whose equation of state is 1 2 ( ) RT a P b b T ν ν ν = − −

sergij07 [2.7K]

Answer:

Given data:

Equation of the state $p=\frac{RT}{v-b}-\frac{a}{v(v+b) T^{1/2} }$

Where p = pressure of fluid, pα

T = Temperature of fluid, k

V = Specific volume of fluid $m^{3} / k g$

R = gas constant , $j/k g k$

a, b = Constants

Solution:

Specific heat difference, $\begin{array}{c}c_{p}-c_{v}=-T\left(\frac{\partial v}{\partial T}\right)^{2} p \\\left(\frac{\partial P}{\partial v}\right)_{r}\end{array}$

According to cyclic reaction

$\left(\frac{\ dv}{\ dT}\right)_{p}=-\frac{\left(\frac{\ d P}{\ d T}\right)_{v}}{\left(\frac{\ d P}{\ d v}\right)_{v}}$

Hence specific heat difference is

$c_{p}-c_{v}=\frac{-T\left(\frac{\ d v}{\ d T}\right)_{p}^{2}}{\left(\frac{\ d p}{\ dv}\right)_{v}}$

Equation of state, $p=\frac{R T}{v-b}-\frac{a}{v(v+b)^{\ 1/2}}$

Differentiating the equation of state with respect to temperature at constant volume,

$\left(\frac{\ d P}{\ d T}\right)_{v}=\frac{R}{v-b}-\frac{1}{2}- \frac{a}{v(v+b)^} T^{\frac{-1}{2}}$

$\begin{aligned}&\left(\frac{\ dP}{\ dT}\right)_{V}=\frac{R}{v-b}+\frac{a}{2 v(v+b) T^{3 / 2}}\end{aligned}$

Differentiating the equation of the state with respect to volume at constant temperature.

$\left(\frac{\ dP}{\ dv}\right)_{\gamma}=+(-1) \times R T(v-b)^{-1-1}+\frac{a}{b T^{1 / 2}}\left(\frac{1}{v^{2}}-\frac{1}{(v+b)^{2}}\right)\)\\\(\left(\frac{\ dP}{\ dv}\right)_{r}=-\frac{R T}{(v-b)^{2}}+\frac{a}{T^{1 / 2}}\left(\frac{2 v+b}{v^{2}(v+b)^{2}}\right)$

Substituting both eq (3) and eq (4) in eq (2)

We get,

${cp{} - } c_{v}=\frac{T\left(\frac{R}{v-b}+\frac{a}{2 v(v+b) T^{3 / 2}}\right)^{2}}{\left(\frac{R T}{(v-b)^{2}}-\frac{a(2 v+b)}{T^{1 / 2} v^{2}(v+b)^{2}}\right)}$

Specific heat difference equation,

$c_{p} -c_{v}}=\frac{T\left(\frac{R}{v-b}+\frac{a}{2 v(v+b)^{T}^{3 / 2}}\right)^{2}}{\left(\frac{R T}{(v-b)^{2}}-\frac{a(2 v+b)}{T^{1 / 2} v^{2}(v+b)^{2}}\right)}$

7 0

3 years ago

alguien me ayuda con una tarea? Está en mi perfil de matemática, porfavorrr regalaré corona pero porfavor​

alguien me ayuda con una tarea? Está en mi perfil de matemática, porfavorrr regalaré corona pero porfavor