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2 years ago

Two solid yellow center lines on a two-lane highway indicate:

Engineering

2 answers:

Vladimir79 [104]2 years ago

4 0

That you can pass the person in front of you while using the other lane

ikadub [295]2 years ago

4 0

Answer:

tttl8

Explanation:

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This problem demonstrates aliasing. Generate a 512-point waveform consisting of 2 sinusoids at 200 and 400-Hz. Assume a sampling

aalyn [17]

Answer and Explanation:

clear all; close all;

N=512;

t=(1:N)/N;

fs=1000;

f=(1:N)*fs/N;

x= sin(2*pi*200*t) + sin(2*pi*400*t);

y= sin(2*pi*200*t) + sin(2*pi*900*t);

for n = 1:20

a(n) = (2/N)*sum(x.*(cos(2*pi*n*t)))

b(n) = (2/N)*sum(x.*(sin(2*pi*n*t)))

c(n) = sqrt(a(n).^2+b(n).^2)

theta(n) =-(360/(2*pi))*atan(b(n)./a(n));

end

plot(f(1:20),c(1:20),'rd');

disp([a(1:4),b(1:4),c(1:4),theta(1:4)])

8 0

3 years ago

A heat recovery system (HRS) is used to conserve heat from the surroundings and supply it to the Mars Rover. The HRS fluid loop

blagie [28]

Answer:

0.304 L of Freon is needed

Explanation:

Q = mCT

Q is quantity of energy that must be removed = 47 BTU = 47×1055.06 = 49587.82 J

C is specific heat of Freon = 74 J/mol.K = 74 J/mol.K × 1 mol/120 g = 0.617 J/g.K

T is temperature in the area of Mars = 189 K

m = Q/CT = 49587.82/(0.617×189) = 452.23 g = 452.24/1000 = 0.45223 kg

Density of Freon = specific gravity of Freon × density of water = 1.49 × 1000 kg/m^3 = 1490 kg/m^3

Volume of Freon = mass/density = 0.45223/1490 = 0.000304 m^3 = 0.000304×1000 = 0.304 L

7 0

3 years ago

The toggle (t) flip-flop has one input, clk, and one output, q. on each rising edge of clk, q toggles to the complement of its p

jeyben [28]

Answer:

See attached

Explanation:

The next state of a toggle flip-flop is the inverse of the present state. This behavior can be produced using a D flip-flop that has its input connected to the inverse of its output.

A schematic is attached.

7 0

2 years ago

You are using a Jupyter Notebook to explore data in a DataFrame named productDF. You want to write some inline SQL by using the

defon

You need to explain it more simple as everyone is clueless

7 0

3 years ago

An inductor has a 50.0-Ω reactance when connected to a 60.0-Hz source. The inductor is removed and then connected to a 45.0-Hz s

nignag [31]

Given:

$X_{L} = 50.0 \ohm$

frequency, f = 60.0 Hz

frequency, f' = 45.0 Hz

$V_rms} = 85.0 V$

Solution:

To calculate max current in inductor, $I_{L(max)$ :

At f = 60.0 Hz

$X_{L} = 2\pi fL$

$50.0 = 2\pi\times 60.0\times L$

L = 0.1326 H

Now, reactance $X_{L}$ at f' = 45.0 Hz:

$X'_{L} = 2\pi f'L$

$X'_{L} = 2\pi\times 45.0\times 0.13263 = 37.5\ohm$

Now, $I_{L(max)$ is given by:

$I_{L(max) = \sqrt {\frac{2V_{rms}}{X'_{L}}}$

$I_{L(max) = \sqrt {\frac{2\times 85.0}{37.5}} = 2.13 A$

Therefore, max current in the inductor, $I_{L(max)$ = 2.13 A

7 0

3 years ago