If the gear ratio is 3:5 does it increase torque or speed?

Parity Detect of Three Inputs 20 pts. (EVEN) a) Draw a function table showing the three data lines and the output function F. 4

DIA [1.3K]

Answer:

See attached picture.

Explanation:

4 0

3 years ago

2- A 2-m3 insulated tank containing ammonia at -20 C, 80% quality, is connected by a valve to a line flowing ammonia at 2 MPa, 6

alekssr [168]

Answer:

The valve should be closed at 1081 kPa ⇒ 11 bar

Explanation:

$Q_{C_V} + m_ih_i = m_2u_2 - m_1u_1 + W_{C_V}$

$Q_{C_V} = W_{C_V} = 0$

$m_1 = \frac{V}{v_1} = \frac{2}{0.49927} = 4.006 kg$

$m_i = m_2 - m_1 = 15 - 4.006 = 10.994 kg$

$u_1 = 1057.5, h_i = 1509.9$

$u_2 = m_ih_i + m_1u_1$

$m_2 = \frac{[10.994*1509.9] + [4.006*1057.5]}{15} = 1389.1 kJ/kg$

$v_2 = \frac{V}{m_2} = \frac{2}{15} = 0.1333 m^3/kg$

Therefore, v₂, u₂ fix state 2.

By trial and error, P₂ = 1081 kPa & T₂ = 50.4° C

1081 kPa ⇒ 11 bar

8 0

4 years ago

dentify a semiconducting material and provide the value of its band gap) that could be used in: (a) (1 point) red LED (b) (1 poi

OLga [1]

Answer:

(a) Aluminum Indium Gallium Phosphide (AlInGaP). Band gap = 1.81eV ≈ 2eV

(b) Gallium Nitride (GaN). Band Gap = 3.4eV

(c) Aluminium Gallium Arsenide (AlGaA). Band Gap = 1.42eV ≈ 2.16eV

(d) Zinc Selenide (ZnSe). Band Gap = 2.82eV

(e) Gallium Phosphide (GaP). Band Gap = 2.24eV

Explanation:

LED's are semi-conducting materials that convert electrical energy to light energy. The light color emitted from the LED depends on the semi-conducting material and other compositions.

The band gap of the semi-conductor determines its wavelength. High band gap semi-conductors emit lower wavelengths which means greater power(UV semi-conducting macterials fall under this category).

5 0

3 years ago

2.

den301095 [7]

Jae pain seems the most off

4 0

3 years ago

A polymeric extruder is turned on and immediately begins producing a product at a rate of 10 kg/min. An operator realizes 20 min

hodyreva [135]

Answer:

The plot of the function production rate m(t) (in kg/min) against time t (in min) is attached to this answer.

The production rate function M(t) is:

$m(t)=[H(t)\cdot10+H(t-20)\cdot5-H(t-80)\cdot14+H(t-81)\cdot9]kg/min$ (1)

The Laplace transform of this function is:

$\displaystyle m(s)=[\frac{10+5e^{-20s}-14e^{-80s}+9e^{-81s}}{s}]kg/min$ (2)

Explanation:

The function of the production rate can be considered as constant functions by parts in the domain of time. To make it a continuous function, we can use the function Heaviside (as seen in equation (1)). To join all the constant functions, we consider at which time the step for each one of them appears and sum each function multiply by the function Heaviside.

For the Laplace transform we use the following rules:

$\mathcal{L}[f(x)+g(x)]=\mathcal{L}[f(x)]+\mathcal{L}[g(x)]=F(s)+G(s)$ (3)

$\mathcal{L}[aH(x-b)]=\displaystyle\frac{ae^{-bs}}{s}$ (4)

4 0

4 years ago