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Alex73 [517]
4 years ago
6

Polymer ropes and lines for use on water are often designed to float, to aid in their retrieval and to avoid applying a downward

s load to an object or person attached to them in the water. Excessive stretch is undesirable, so a lower limit of 0.5 GPa is also imposed on Young's modulus. Identify suitable polymers, using Young's modulus-density chart.

Engineering
1 answer:
andrew-mc [135]4 years ago
8 0

Answer:

We choose PTFE

Explanation:

Attached are the modulus density and modulus strength chart.

Due to its young modulus, the density is near 0.5 GPa, as seen in the chart and support water gliding. The PTFE density is between 1 and 10 Mg / cubic meter (see module and chart of density), and the resistance is between 10 and 100 Mpa (see module and chart of strength). Therefore, the finest ploymer will be PTFE that meets the requirements.

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a. Determine R for a series RC high-pass filter with a cutoff frequency (fc) of 8 kHz. Use a 100 nF capacitor. b. Draw the schem
Readme [11.4K]

Answer:

a) 199.04 ohms

b) attached in image

c) -0.696dB

Explanation:

We are given:

Fc = 8Khz = 8000hz

C = 100nF = 100*10^-^9F

a)Using the formula:

F_c = \frac{1}{2pie*Rc}

8000= \frac{1}{2*3.14*R*100*10^-^9}

R =\frac{1}{2*3.14*100*10^-^9*8000}

R = 199.04 ohms

b) diagram is attached

c) H(w) = \frac{V_out(w)}{Vin(w)} = \frac{1}{1-j\frac{wc}{w}}

H(F) = \frac{1}{1-j\frac{fc}{f}}

At F = 20KHz and Fc= 8KHz we have:

H(F)= \frac{1}{1-j\frac{8}{20}} = \frac{1}{1-j(0.4)}

|H(F)|= \frac{1}{\sqrt{1^2+(0.4)^2}}

=0.923

|H(F)| in dB = 20log |H(F)|

=20log0.923

= -0.696dB

5 0
3 years ago
Discuss the applications of numerical weather forecasting​
olchik [2.2K]

Numerical weather prediction (NWP) uses mathematical models of the atmosphere and oceans to predict the weather based on current weather conditions. Though first attempted in the 1920s, it was not until the advent of computer simulation in the 1950s that numerical weather predictions produced realistic results. A number of global and regional forecast models are run in different countries worldwide, using current weather observations relayed from radiosondes, weather satellites and other observing systems as inputs.
3 0
3 years ago
Two technicians are discussing a vehicle that cranks slowly when the key is turned to the crank position. The positive battery t
Norma-Jean [14]

Answer:

Both are right

Explanation:

3 0
2 years ago
A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 55 MPa √m (50 ksi √in.). If, during
astra-53 [7]

Answer:

0.024 m = 24.07 mm

Explanation:

1) Notation

\sigma_c = tensile stress = 200 Mpa

K = plane strain fracture toughness= 55 Mpa\sqrt{m}

\lambda= length of a surface crack (Variable of interest)

2) Definition and Formulas

The Tensile strength is the ability of a material to withstand a pulling force. It is customarily measured in units (F/A), like the pressure. Is an important concept in engineering, especially in the fields of materials and structural engineering.

By definition we have the following formula for the tensile stress:

\sigma_c=\frac{K}{Y\sqrt{\pi\lambda}}   (1)

We are interested on the minimum length of a surface that will lead to a fracture, so we need to solve for \lambda

Multiplying both sides of equation (1) by Y\sqrt{\pi\lambda}

\sigma_c Y\sqrt{\pi\lambda}=K   (2)

Sequaring both sides of equation (2):

(\sigma_c Y\sqrt{\pi\lambda})^2=(K)^2  

\sigma^2_c Y^2 \pi\lambda=K^2   (3)

Dividing both sides by \sigma^2_c Y^2 \pi we got:

\lambda=\frac{1}{\pi}[\frac{K}{Y\sigma_c}]^2   (4)

Replacing the values into equation (4) we got:

\lambda=\frac{1}{\pi}[\frac{55 Mpa\sqrt{m}}{1.0(200Mpa)}]^2 =0.02407m

3) Final solution

So the minimum length of a surface crack that will lead to fracture, would be 24.07 mm or more.

7 0
4 years ago
Because of ____________ people must make choices, and when they choose, they incur a(n)______________.
lbvjy [14]
There must be a photo for me to answer!
3 0
3 years ago
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