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Lubov Fominskaja [6]
2 years ago
9

A system samples a sinusoid of frequency 230 Hz at a rate of 175 Hz and writes the sampled signal to its output without further

modification. Determine the frequency that the sampling system will generate in its output.
a. 120
b. 55
c. 175
d. 230
Engineering
1 answer:
Irina-Kira [14]2 years ago
6 0
B I got it right so (;
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Establishes general guidelines concerning licensing and vehicle
gavmur [86]

Answer:

A. National Highway Safety Act

Explanation:

The National Highway Safety Act establishes general guidelines concerning licensing, vehicle registration and inspection, and traffic laws for state regulations. The act was made in 1966 to reduce the amount of death on the highway as a result of increase in deaths by 30% between 1960 and 1965

National Traffic and Motor Vehicle Safety Act regulates vehicle manufacturers  by ensuring national safety standards and issuance recalls for defective vehicles

Uniform Traffic Control Devices Act  defines shapes, colors and locations for road signs, traffic signals, and road markings

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What causes the charging system warning lamp to go out when the engine starts up?
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A thick oak wall initially at 25°C is suddenly exposed to gases for which T =800°C and h =20 W/m2.K. Answer the following questi
Schach [20]

Answer:

a) What is the surface temperature, in °C, after 400 s?

   T (0,400 sec) = 800°C

b) Yes, the surface temperature is greater than the ignition temperature of oak (400°C) after 400 s

c) What is the temperature, in °C, 1 mm from the surface after 400 s?

   T (1 mm, 400 sec) = 798.35°C

Explanation:

oak initial Temperature = 25°C = 298 K

oak exposed to gas of temp = 800°C = 1073 K

h = 20 W/m².K

From the book, Oak properties are e=545kg/m³   k=0.19w/m.k   Cp=2385J/kg.k

Assume: Volume = 1 m³, and from energy balance the heat transfer is an unsteady state.

From energy balance: \frac{T - T_{\infty}}{T_i - T_{\infty}} = Exp (\frac{-hA}{evCp})t

Initial temperature wall = T_i

Surface temperature = T

Gas exposed temperature = T_{\infty}

6 0
4 years ago
Water flows steadily through the pipe as shown below, such that the pressure at section (1) and at section (2) are 300 kPa and 1
steposvetlana [31]

Answer:

The velocity at section is approximately 42.2 m/s

Explanation:

For the water flowing through the pipe, we have;

The pressure at section (1), P₁ = 300 kPa

The pressure at section (2), P₂ = 100 kPa

The diameter at section (1), D₁ = 0.1 m

The height of section (1) above section (2), D₂ = 50 m

The velocity at section (1), v₁ = 20 m/s

Let 'v₂' represent the velocity at section (2)

According to Bernoulli's equation, we have;

z_1 + \dfrac{P_1}{\rho \cdot g} + \dfrac{v^2_1}{2 \cdot g} = z_2 + \dfrac{P_2}{\rho \cdot g} + \dfrac{v^2_2}{2 \cdot g}

Where;

ρ = The density of water = 997 kg/m³

g = The acceleration due to gravity = 9.8 m/s²

z₁ = 50 m

z₂ = The reference = 0 m

By plugging in the values, we have;

50 \, m + \dfrac{300 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{(20 \, m/s)^2}{2 \times 9.8 \, m/s^2} = \dfrac{100 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}50 m + 30.704358 m + 20.4081633 m = 10.234786 m + \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}

50 m + 30.704358 m + 20.4081633 m - 10.234786 m = \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}

90.8777353 m = \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}

v₂² = 2 × 9.8 m/s² × 90.8777353 m

v₂² = 1,781.20361 m²/s²

v₂ = √(1,781.20361 m²/s²) ≈ 42.204308 m/s

The velocity at section (2), v₂ ≈ 42.2 m/s

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