Answer:
rotation
Explanation:
I just answered this!
Which option identifies the type of engineer described in the following scenario?
Sean is an engineer whose current project is a skyscraper in Richmond, VA. He relies heavily on geometry in his research of building design.
Material
Civil
Mechanical
Chemical
Answer:
civil
Explanation:
mark be branilist
Answer:
Civil
Explanation:
civil engineering – the application of planning, designing, constructing, maintaining, and operating infrastructure while protecting the public and environmental health, as well as improving existing infrastructure that may
Implement the following Matlab code:
x=zeros(10,128);
t1=[0:1/128:1-1/128];
z=cos(2*pi*2*t1);
x(1,:)=z;
x=reshape(x,1,1280);
figure(1);
plot(x)
Examine Figure 1 – what does it “look like”? Zoom in. Take the FFT of x and explain what you see.
Manipulate the FFT and perform the IFFT to create a signal which is a continuous (i.e. no interpolating zeroes) sinusoid in the “time” domain. Explain the amplitude of the sinusoid.
The amplitude of the sinusoid is 28384 *x soít cos.
What is amplitude?Amplitude is defined as the greatest deviation from equilibrium of a point on a vibrating body or wave in terms of displacement or distance traveled. In most cases, amplitude is calculated by looking at a wave graph and determining the height of the wave from rest. The strength or intensity of the wave is gauged by its amplitude.
Sinusoid is defined as a signal with sine wave characteristics. In the liver, spleen, and bone marrow, sinusoids and irregular tubules transport blood in place of venules and capillaries. The sine or cosine functions from trigonometry form the foundation of sinusoidal signals, which are periodic functions.
Thus the amplitude of the sinusoid is 28384 *x soít cos.
To learn more about amplitude, refer to the link below:
https://brainly.com/question/8662436
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A sample of soil has a volume of 0.45 ft^3 and a weight of 53.3 lb. After being dried inan oven, it has a weight of 45.1 lb. It has a specific gravity of solids of 2.70. Compute its moisture content and degree of saturation before it was placed in the oven.
Answer:
a) the moisture content before it was placed in the oven is 18.18%
b) degree of saturation for soil is 72.19%
Explanation:
Given the data in the question;
Moisture Content = [(Weight of soil before dry - dry weight) / dry weight] × 100
so we substitute
Moisture content = [(53.3 - 45.1) / 45.1 ] × 100
= (8.2/45.1) × 100
= 18.18%
Therefore the moisture content before it was placed in the oven is 18.18%
Dry Unit Weight = dry weight / volume
Dry Unit Weight = 45.1 lb / 0.45 ft³
Dry Unit Weight = 100.22 lb/ft³
we know that;
dry unit weight = (Specific gravity × unit weight of water) / (1 + e)
we also know that; unit weight of water is 62.43 lbf/ft³
so we substitute
e = (2.70×62.43 / 100.22) - 1
e = 1.68 - 1
e = 0.68
so void ratio e = 0.68
Now we determine the degree of saturation using the equation;
degree of saturation = (Moisture content × specific gravity) / void ratio
we substitute
degree of saturation = ( 18.18% × 2.7) / 0.68
= 0.49086 / 0.68
= 0.7219 ≈ 72.19%
Therefore degree of saturation for soil is 72.19%
Technician a says that diesel engines can produce more power because air in fuel or not mix during the intake stroke. Technician be says that diesel engines produce more power because they use excess air to burn feel who is correct
Answer:
Technician be says that diesel engines produce more power because they use excess air to burn feel who is correct
Explanation:
He is correct as many engines are run by diesel. It produces more power as that is how cars produce more power.
An NMOS transistor with kn = 1mA/V2 and Vt = 1V is operated with VGS = 2.5V. At what value of VDS does the transistor enter the saturation region? What value of ID is obtained in saturation?
Answer:
a) the point at which the transistor enters the saturation region is 1.5 v
b) the value of ID is obtained in saturation is 1.125 mA
Explanation:
Given the data in the question;
for an NMOS, the condition for the saturation is;
V_DS ≥ V_GS - V_t
V_GS is 2.5 v and V_t is 1 v
so we substitute
V_DS ≥ 2.5 - 1
V_DS = 1.5 v
so the point at which the transistor enters the saturation region is 1.5 v
The drain current I_d in the saturation region;
I_d = 1/2×μₙ×Cₐₓ×W/L×( V_GS - V_t)²
= 1/2Kₙ ( V_GS - V_t)²
our Kₙ is 1 mA/V², V_GS is 2.5v and V_t is 1 v
so we substitute
I_d = 1/2(1 mA/V²)( 2.5 - 1 )²
= 1/2(1 mA/V²)( 2.25)
= 1.125 mA
therefore, the value of ID is obtained in saturation is 1.125 mA
