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alisha [4.7K]
3 years ago
10

At an axial load of 22 kN, a 15-mm-thick × 40-mm-wide polyimide polymer bar elongates 4.1 mm while the bar width contracts 0.15

mm. The bar is 270-mm long. At the 22-kN load, the stress in the polymer bar is less than its proportional limit. Determine Poisson’s ratio.
Engineering
1 answer:
Alenkasestr [34]3 years ago
3 0

Answer:

The Poisson's Ratio of the bar is 0.247

Explanation:

The Poisson's ratio is got by using the formula

Lateral strain / longitudinal strain

Lateral strain = elongation / original width (since we are given the change in width as a result of compession)

Lateral strain = 0.15mm / 40 mm =0.00375

Please note that strain is a dimensionless quantity, hence it has no unit.

The Longitudinal strain is the ratio of the elongation to the original length in the longitudinal direction.

Longitudinal strain = 4.1 mm / 270 mm = 0.015185

Hence, the Poisson's ratio of the bar is 0.00375/0.015185 = 0.247

The Poisson's Ratio of the bar is 0.247

Please note also that this quantity also does not have a dimension

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The settlement of foundations is typically the result of three separate occurrences that take place in the soil which provides s
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Answer:

The differences are listed below

Explanation:

The differences between consolidation and compaction are as follows:

In compaction the mechanical pressure is used to compress the soil. In consolidation, there is an application of stead pressure.

In compaction, there is a dynamic load by rapid mechanical methods like tamping, rolling, etc. In consolidation, there is static and sustained pressure applied for a long time.

In compaction, the soil volume is reduced by removing air from the void. In consolidation, the soil volume is reduced by squeezing out water from the pores.

Compaction is used for sandy soil, consolidation on the other hand, is used for clay soil.

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3 years ago
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BE-40 What is a characteristic of a catamaran hull?
Nat2105 [25]

Answer:

<em>A stable ride</em>

Explanation:

A Catamaran hull is a form of sea craft invented by the Austronesian peoples, the invention of the Catamaran hull enabled these people to sail across the sea in their expansion to the islands of the Indian and Pacific Oceans. Catamaran has multiple hulls, usually two parallel hulls of equal size. This geometric feature gives the craft an increased stability because,<em> it derives extra stability from its wide beam, in the place of a ballasted keel employed in a regular monohull sailboat. </em>A Catamaran hull will require four times the force needed to capsize it, when compared to an equivalent monohull.

8 0
3 years ago
The hot combustion gases of a furnace are separated from the ambient air and its surroundings, which are at 25 oC, by a brick wa
yanalaym [24]

Answer:

T1 = 625.54 K

Explanation:

We are given;

T_α = Tsur = 25°C = 298K

h = 20 W/m².K,

L = 0.15 m

K = 1.2 W/m.K

ε = 0.8

Ts = T2 = 100°C = 373K

T1 = ?

Assumption:

-Steady- state condition

-One- dimensional conduction

-No uniform heat generation

-Constant properties

From Energy balance equation;

E°in - E°out = 0

Thus,

q"cond – q"conv – q"rad = 0

K[(T1 - T2)/L] - h(Ts-T_α) - εσ (Ts⁴ – Tsur⁴)

Where σ is Stephan-Boltzmann constant and has a value of 5.67 x 10^(-8)

Thus;

K[(T1 - T2)/L] - h(Ts-T_α) - εσ (Ts⁴ – Tsur⁴) = 1.2[(T1 - 373)/0.15] - 20(373 - 298] - 0.8x5.67x10^(-8)[373⁴ - 298⁴] = 0

This gives;

(8T1 - 2984) - (1500) - 520.31 = 0

8T1 = 2984 + 1500 + 520.31

8T1 = 5004.31

T1 = 5004.31/8

T1 = 625.54 K

7 0
3 years ago
An important concept to aid understanding of electromagnetics is electrical length. Electrical _________ is a unitless measure t
kondor19780726 [428]

Answer: Electrical length is a unitless measure that refers to the length of a wire or device at a certain frquency. It is defined as the ratio of the physical length of the device to the wavelength of the signal frequency.

Explanation:

In nature, the maximum speed achievable for any interaction or perturbation along a medium (like a wire) can't be higher than the speed of light, otherwise it would be violating Einstein's Relativity Theory.

So, in a given circuit, the voltage and current aren't set up instantaneously, and a finite time exists since a battery is connected to a circuit till a stable current  traverses a load resistor connected to it.

Now, the electrical length, defines how important in is this delay, in the calculation of the voltage through the resistor, for instance.

We can write the electrical length (L.E.) as follows:

L.E. = L / (v/f) = L / λ

As we can see, the effect depends not only on the circuit dimensions, but the frequency of the signal as well.

If L.E. is much smaller than λ, this means that we can neglect the effects of the delay, and we can use the circuit theory, namely KVL, KCL and Ohm's law to analyze the circuit, assuming that the current establishes instantaneously.

Otherwise, the current and voltage concepts are not valid anymore, as we need to think in terms of propagation of electromagnetic waves, like in transmission lines and antennas.

Roughly, if L.E. ≤ 20 λ, we say that we are still in the realm of the circuit theory.

3 0
3 years ago
4. A 25 km2 watershed has a time of concentration of 1.6 hr. Calculate the NRCS triangular UH for a 10-minute rainfall event and
Alika [10]

Answer:

The NRCS triangular UH for a 10-minute rainfall is Qp = 49.84 m³/s

Peak flow of the aggregated runoff hydrograph is 420.58 m³/s

The total volume of runoff is 2125000 m³/s

Explanation:

We have

A = 25 km²

tr = 10 min = 1/6 hr

tc = 1.6 hr

lag time = 0.6 tc = 0.96 hr

Tp = tr/2 + 0.6 tc = 1/12 + 0.96 = 1.043 hr

Qp = 2.08×25/1.043 = 49.84 m³/s

Tb = 8/3×Tp = 8/3×1.043 = 2.782 hr

 

Since the area is  

Time (min)           Runoff (cm)       Volume of runoff m³

0                   0                                     0

10                  4                                     1000000 m³

20                 2.5                                  625000 m³

30                 2                                      500000 m³

Total volume of runoff = 1000000 + 625000 + 500000 =  2125000 m³/s

For the 1st  10 minutes, we have

A = 25 km²

tr = 30 min = 1/2 hr

tc = 1.6 hr

lag time = 0.6 tc = 0.96 hr

Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr

Qp = 2.08×25×4/1.043 = 197.92 m³/s

Tb = 8/3×Tp = 8/3×1.21 = 3.227  hr

 

For the 2nd 10 minutes, we have

A = 25 km²

tr = 30 min = 1/2 hr

tc = 1.6 hr

lag time = 0.6 tc = 0.96 hr

Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr

Qp = 2.08×25×2.5/1.043 = 123.7 m³/s

Tb = 8/3×Tp = 8/3×1.21 = 3.227  hr

For the 3rd 10 minutes, we have

A = 25 km²

tr = 30 min = 1/2 hr

tc = 1.6 hr

lag time = 0.6 tc = 0.96 hr

Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr

Qp = 2.08×25×2.5/1.043 = 98.96 m³/s

Tb = 8/3×Tp = 8/3×1.21 = 3.227  hr

 

Peak flow of aggregate runoff is given by

Qp (total) = 98.96 + 123.7 +197.92 = 420.58 m³/s

Total volume of runoff is given by

Total volume of runoff = 1000000 + 625000 + 500000 =  2125000 m³/s

6 0
3 years ago
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