1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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

You might be interested in
Explain what a margin of safety is in driving as well as how it can help minimize risk.
Yakvenalex [24]

Answer:

A safety margin is the space left between your vehicle and the next to provide room, time and visibility at every instant

Explanation:

A safety margin is defined as an allowance given between your vehicle and the next vehicle in front to provide enough room, visibility and time to move in a safe manner to prevent the occurrence of an accident at anytime the frontal vehicle suddenly stops or slows down

Safety margins help minimize risks in the following way

1) A common knowledge of safety margins, improves predictability among road users, thereby minimizing the risk traffic accidents caused due to late communication

2) The use of safety margins helps minimize the risk due to a change in driving conditions such as when the road becomes more slippery from being covered with fluid that is being wetted

3) Safety margin can help prevent the occurrence of an accident between vehicles due to failure of a car system, such as a punctured tire or failed breaking system

4) Safety margin helps to protect road users from the introduction of obstacles on the main roads such as ongoing road construction, broken down vehicles, road blockage by vehicles involved in an accident etc

5) Safety margin help protect road users from being involved in an accident due to the loss of driving focus of the driver of the frontal vehicle

6 0
2 years ago
(35-39) A student travels on a school bus in the middle of winter from home to school. The school bus temperature is 68.0° F. Th
arlik [135]

Answer:

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

Explanation:

From Heat Transfer we determine that heat transfer rate due to electromagnetic radiation (\dot Q), measured in BTU per hour, is represented by this formula:

\dot Q = \epsilon\cdot A\cdot \sigma \cdot (T_{s}^{4}-T_{b}^{4}) (1)

Where:

\epsilon - Emissivity, dimensionless.

A - Surface area of the student, measured in square feet.

\sigma - Stefan-Boltzmann constant, measured in BTU per hour-square feet-quartic Rankine.

T_{s} - Temperature of the student, measured in Rankine.

T_{b} - Temperature of the bus, measured in Rankine.

If we know that \epsilon = 0.90, A = 16.188\,ft^{2}, \sigma = 1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}}, T_{s} = 554.07\,R and T_{b} = 527.67\,R, then the heat transfer rate due to electromagnetic radiation is:

\dot Q = (0.90)\cdot (16.188\,ft^{2})\cdot \left(1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}} \right)\cdot [(554.07\,R)^{4}-(527.67\,R)^{4}]

\dot Q = 417.492\,\frac{BTU}{h}

Under the consideration of steady heat transfer we find that the net energy transfer from the student's body during the 20 min-ride to school is:

Q = \dot Q \cdot \Delta t (2)

Where \Delta t is the heat transfer time, measured in hours.

If we know that \dot Q = 417.492\,\frac{BTU}{h} and \Delta t = \frac{1}{3}\,h, then the net energy transfer is:

Q = \left(417.492\,\frac{BTU}{h} \right)\cdot \left(\frac{1}{3}\,h \right)

Q = 139.164\,BTU

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

7 0
2 years ago
In the lab, a container of saturated soil had a mass of 113.27 g before it was placed in the oven and100.06 g after the soil had
kipiarov [429]

Answer:

  • Moisture/ water content w = 26%
  • Void ratio , e =  0.73

Explanation:

  • Initial mass of saturated soil w1 = mass of soil - weight of container

                                                 = 113.27 g - 49.31 g = 63.96 g

  • Final mass of soil after oven w2 = mass of soil - weight of container

                                                  = 100.06 g - 49.31 g = 50.75

Moisture /water content, w =   \frac{w1-w2}{w2} = \frac{63.96-50.75}{50.75} = 0.26 = 26%

Void ratio =  water content X specific gravity of solid

                  = 0.26 X 2.80 =0.728

5 0
3 years ago
A 100 ft long steel wire has a cross-sectional area of 0.0144 in.2. When a force of 270 lb is applied to the wire, its length in
blondinia [14]

Answer:

(a) The stress on the steel wire is 19,000 Psi

(b) The strain on the steel wire is 0.00063

(c) The modulus of elasticity of the steel is 30,000,000 Psi

Explanation:

Given;

length of steel wire, L = 100 ft

cross-sectional area, A = 0.0144 in²

applied force, F = 270 lb

extension of the wire, e = 0.75 in

<u>Part (A)</u> The stress on the steel wire;

δ = F/A

   = 270 / 0.0144

δ  = 18750 lb/in² = 19,000 Psi

<u>Part (B)</u> The strain on the steel wire;

σ = e/ L

L = 100 ft = 1200 in

σ = 0.75 / 1200

σ = 0.00063

<u>Part (C)</u> The modulus of elasticity of the steel

E = δ/σ

   = 19,000 / 0.00063

E = 30,000,000 Psi

4 0
3 years ago
A seamless pipe 800mm diameter contains a fluid under a pressure of 2N/mm2. If the permissible tensile stress is 100N/mm2, find
Bad White [126]

Answer:

8 mm

Explanation:

Given:

Diameter, D = 800 mm

Pressure, P = 2 N/mm²

Permissible tensile stress, σ = 100 N/mm²

Now,

for the pipes, we have the relation as:

\sigma=\frac{\textup{PD}}{\textup{2t}}

where, t is the thickness

on substituting the respective values, we get

100=\frac{\textup{2\times800}}{\textup{2t}}

or

t = 8 mm

Hence, the minimum thickness of pipe is 8 mm

3 0
3 years ago
Other questions:
  • Which phrases describe an irregular galaxy?
    8·1 answer
  • What is the first step in the problem-solving process, as well as in the engineering design process?
    7·1 answer
  • Explain the four criteria for proving the correctness of a logical pretest loop construct of the form "while B do S end". And pr
    12·1 answer
  • An air conditioner using refrigerant-134a as the working fluid and operating on the ideal vapor-compression refrigeration cycle
    7·2 answers
  • 4. Which of the following is the first thing you should do when attempting
    13·2 answers
  • These are the most widely used tools and most often abuse tool​
    15·2 answers
  • Which component found in fertilizer is a known cancer-causing agent?
    11·2 answers
  • I know this answer i just want too see if people know it too
    9·2 answers
  • Which of the following maintenance items helps to ensure the vehicles engine lasts as long as possible?
    6·1 answer
  • When you approach an uncontrolled intersection, you should treat it as though which sign is present?
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!