Where do I buy a 1997 MK4 Toyota Supra twin turbo manual for cheap

I need help with this question

Ad libitum [116K]

Answer:

LOL where is the question, that u need help with?

Explanation:

5 0

3 years ago

Harmonic excitation of motion is represent as

Gennadij [26K]

Harmonic excitation refers to a sinusoidal external force of a certain frequency applied to a system. ... Resonance occurs when the external excitation has the same frequency as the natural frequency of the system. It leads to large displacements and can cause a system to exceed its elastic range and fail structurally.

6 0

3 years ago

Steam at a pressure of 100 bar and temperature of 600 °C enters an adiabatic nozzle with a velocity of 35 m/s. It leaves the noz

butalik [34]

Answer:

Exit velocity $V_2=1472.2$ m/s.

Explanation:

Given:

At inlet:

$P_1=100 bar,T_=600°C,V_1=35m/s$

Properties of steam at 100 bar and 600°C

$h_1=3624.7\frac{KJ}{Kg}$

At exit:Lets take exit velocity $V_2$

We know that if we know only one property inside the dome then we will find the other property by using steam property table.

Given that dryness or quality of steam at the exit of nozzle is 0.85 and pressure P=80 bar.So from steam table we can find the other properties.

Properties of saturated steam at 80 bar

$h_f= 1316.61\frac{KJ}{Kg} ,h_g= 2757.8\frac{KJ}{Kg}$

So the enthalpy of steam at the exit of turbine

$h_2=h_f+x(h_g-h_f)\frac{KJ}{Kg}$

$h_2=1316.61+0.85(2757.8-1316.61)\frac{KJ}{Kg}$

$h_2=2541.62\frac{KJ}{Kg}$

Now from first law for open system

$h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w$

In the case of adiabatic nozzle Q=0,W=0

$3624.7+\dfrac{35^2}{2000}+0=2541.62+\dfrac{(V_2)^2}{2000}+0$

$V_2=1472.2$ m/s

So Exit velocity $V_2=1472.2$ m/s.

4 0

3 years ago

At an operating frequency of 5 GHz, a 50 lossless coaxial line with insulating material having a relative permittivity r = 2.25

lapo4ka [179]

Answer:

The answer is "150 $\Omega$ ".

Explanation:

Its line length must be converted into wavelengths for the Smith chart to be used.

$\to \beta = \frac{2 \pi }{\lambda } \\\\\to u_p = \frac{\omega }{\beta}\\\\\lambda =\frac{2 pi}{\beta} =\frac{2 \pi U_p}{\omega}=\frac{c}{\sqrt{\varepsilon_r f}}$

$= \frac{3 \times 10^8}{ 2.25 \times (5 \times 10^9)}\\\\= \frac{3}{ 2.25 \times 5\times 10 }\\\\= \frac{3}{ 22.5 \times 5 }\\\\= \frac{3}{ 112.5}\\\\= 0.04 \ m.$