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melamori03 [73]

3 years ago

5

A patient has swollen feet, ankles, and legs, and has pain in his lower back. Tests show the level of urea, a waste product, in

the patient's blood has suddenly risen. Which body system is most likely not functioning properly?

1 answer:

myrzilka [38]3 years ago

3 0

Answer: B Excretory

Explanation: Hope this helps :)

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What process is used to remove collodal and dissolved organic matter in waste water

Juli2301 [7.4K]

Answer:

Aerobic biological treatment process

Explanation:

Aerobic biological treatment process in which micro-organisms, in the presence of oxygen, metabolize organic waste matter in the water, thereby producing more micro-organisms and inorganic waste matter like CO₂, NH₃ and H₂O.

3 0

3 years ago

A shaft made of stainless steel has an outside diameter of 42 mm and a wall thickness of 4 mm. Determine the maximum torque T th

fiasKO [112]

Answer:

Explanation:

Using equation of pure torsion

$\frac{T}{I_{polar} }=\frac{t}{r}$

where

T is the applied Torque

$I_{polar}$ is polar moment of inertia of the shaft

t is the shear stress at a distance r from the center

r is distance from center

For a shaft with

$D_{0} =$ Outer Diameter

$D_{i} =$ Inner Diameter

$I_{polar}=\frac{\pi (D_{o} ^{4}-D_{in} ^{4}) }{32}$

Applying values in the above equation we get

$I_{polar} =\frac{\pi(0.042^{4}-(0.042-.008)^{4})}{32}\\I_{polar}= 1.74$ x $10^{-7} m^{4}$

Thus from the equation of torsion we get

$T=\frac{I_{polar} t}{r}$

Applying values we get

$T=\frac{1.74X10^{-7}X100X10^{6} }{.021}$

T =829.97Nm

7 0

3 years ago

A sinusoidal voltage source produces the waveform, v t = 1 + cos 2πft. Design a system with v t as its input such that an LED wi

DerKrebs [107]

Answer:

See explaination

Explanation:

LM358 is the useful IC which works as buffer. It enables circuit to remove overloading effect on each other. Image is in attachment.

We can define a light-emitting diode (LED) as a semiconductor light source that emits light when current flows through it. Electrons in the semiconductor recombine with electron holes, releasing energy in the form of photons

See attached file for detailed solution of the given problem.

3 0

3 years ago

While walking across campus one windy day, an engineering student speculates about using an umbrella as a "sail" to propel a bic

makvit [3.9K]

Answer:

$Given data:\\While walking across campus one windy day\\Frontal area, \(A=0.3 m ^{2}\)\\Wind speed \(V=24 Km / hr\)\\The drag coefficient \(C_{D, b}=1.2\)\\The combined mass \(m=75 kg\)\\Umbrella diameter, \(D=1.22 m\)\\Velocity of wind \(V=24 \frac{ km }{ hr }\)\\The rolling resistance \(C_{R}=0.75 \%\)$

Solution:

Note: Refer the diagram

$Basic equation:\\'s law of motion: \(\sum F_{x}=m a_{x}\)\\Lift coefficient, \(C_{L}=\frac{F_{L}}{\frac{1}{2} \rho V^{2} A_{p}}\)\\Drag coefficient, \(C_{D}=\frac{F_{D}}{\frac{1}{2} \rho V^{2} A_{p}}\)$

$From force balance equation:\\\(\sum F_{x}=F_{D}-F_{R}=0\)\\But \(F_{D}=\left(C_{D, \alpha} A_{u}+C_{D, B} A_{b}\right) \frac{1}{2} \rho\left(V_{\nu}-V_{b}\right)^{2}\)\(F_{R}=C_{R} m g\)\\Area of the Umbrella \(A_{u}=\frac{\pi D_{u}^{2}}{4}\)\(A_{x}=\frac{\pi \times 1.22^{2}}{4} m ^{2}\)\(A_{v}=1.17 m ^{2}\)$

Drag coefficient data for selected objects table at

Hemisphere (open end facing flow), $C_{D, x}=1.42$

Substituting all parameters,

$\begin{aligned}&F_{R}=0.0075 \times 75 \times 9.81\\&F_{R}=5.52 N\end{aligned}$

Then,

$\begin{aligned}&V_{b}=V_{w}-\left[\frac{2 F_{R}}{\rho\left(C_{D, w} A_{w}+C_{D, B} A_{b}\right)}\right]^{\frac{1}{2}} \dots\\&V_{w}=24 \times 1000 \times \frac{1}{3600}\\&V_{w}=6.67 \frac{ m }{ s }\end{aligned}$

And the equation becomes,

$\begin{aligned}&V_{b}=6.67-\left[\frac{2 \times 5.52}{1.23(1.42 \times 1.17+1.2 \times 0.3)}\right]^{\frac{1}{2}}\\&V_{b}=6.67-2.11\\&V_{b}=4.56 \frac{ m }{ s }\end{aligned}$

Thus the floyds travels at $68.3^{\circ}$ wind speed.

7 0

4 years ago

A light bar AD is suspended from a cable BE and supports a 20-kg block at C. The ends A and D of the bar are in contact with fri

babymother [125]

Answer:

Tension in cable BE= 196.2 N

Reactions A and D both are 73.575 N

Explanation:

The free body diagram is as attached sketch. At equilibrium, sum of forces along y axis will be 0 hence

$T_{BE}-W=0$ hence

$T_{BE}=W=20*9.81=196.2 N$

Therefore, tension in the cable, $T_{BE}=196.2 N$

Taking moments about point A, with clockwise moments as positive while anticlockwise moments as negative then

$196.2\times 0.125- 196.2\times 0.2+ D_x\times 0.2=0$

$24.525-39.24+0.2D_x=0$

$D_x=73.575 N$

Similarly,

$A_x-D_y=0$

$A_x=73.575 N$

Therefore, both reactions at A and D are 73.575 N

7 0

3 years ago