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yKpoI14uk [10]
4 years ago
6

Problem 2 (20 points): Three prismatic bars, two of material A and one of material B, transmit a tensile load P. The two outer b

ars (material A) are identical. The cross-sectional area of the middle bar (material I} B) is 50% larger than the cross-sectional area of the outer bar. Also, the modulus of elasticity of material A is twice that ofmaterial B. Find: (a) What fraction ofthe load Pis transmitted by the middle bar? (b) What is the ratio of the stress in the middle bar to the stress in the outer bars? (c) What is the ratio of the strain in the middle bar to the strain in the outer bars
Engineering
1 answer:
abruzzese [7]4 years ago
3 0

Answer:

you said 20 points

Explanation:

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Which timeline shows the correct order of contributions made to the discovery of DNA?
Akimi4 [234]

Answer:

<em><u>THE ANSWER IS: B</u></em>

Explanation:

I took the Unit test and the answer is B

3 0
3 years ago
Read 2 more answers
A closed rigid tank contains water initially at 10,000 kPa and 520ºC and is cooled to a final temperature of 270° C. Determine t
jasenka [17]

Answer:

final pressure is 6847.41 kPa

Explanation:

given data:

P_1 =  10,000 kPa

T_1 =520\ degree\ celcius = 793 K

T_2  = 270 degree celcius = 543 K

as we can see all temperature are more than 100 degree, it mean this condition is refered to superheated stream

for ischoric process we know that

\frac{P_1}{T_1} =\frac{P_2}{T_2}

\frac{10*10^6}{793} = \frac{P_2}{543}

P_2 = 6.84741*10^6 Pa

final pressure is 6847.41 kPa

5 0
3 years ago
Oil with a density of 850 kg/m3 and kinematic viscosity of 0.00062 m2/s is being discharged by an 8-mm-diameter, 42-m-long horiz
sladkih [1.3K]

Answer:

The flow rate of oil through the pipe is 1.513E-7 m³/s.

Explanation:

Given

Density, ρ = 850 kg/m³

Kinematic viscosity, v = 0.00062 m²/s

Diameter, d = 8-mm = 0.008m

Length of horizontal pipe, L = 42-m

Height, h = 4-m.

We'll solve the flow rate of oil through the pipe by using Hagen-Poiseuille equation.

This is given as

∆P = (128μLQ)/πD⁴

Where ∆P = Rate of change of pressure

μ = Dynamic Viscosity

Q = Flow rate of oil through the pipe.

First, we need to determine the dynamic viscosity and the rate of change in pressure

Dynamic Viscosity, μ = Density (ρ) * Kinematic viscosity (v)

μ = 850 kg/m³ * 0.00062 m²/s

μ = 0.527kg/ms

Then, we calculate the rate of change of pressure.

Assuming that the velocity through the pipe is so small;

∆P = Pressure at the bottom of the tank

∆P = Density (ρ) * Acceleration of gravity (g) * Height (h)

Taking g = 9.8m/s²

∆P = 850kg/m³ x 9.8m/s² x 4m

∆P = 33320N/m²

Recall that Hagen-Poiseuille equation.

∆P = (128μLQ)/πD⁴ --- Make Q the subject of formula

Q = (πD⁴P)/(128μL)

By substituton;

Q = (π * 0.008⁴ * 33320)/(128 * 0.527 * 42)

Q = 0.00000015133693643099

Q = 1.513E-7 m³/s.

Hence, the flow rate of oil through the pipe is 1.513E-7 m³/s.

8 0
3 years ago
A homogeneous 800kg bar AB is supported at either end by a cable asshown in the figure
aleksandr82 [10.1K]

The smallest area of each cable if the stress is not to exceed 90MPa in bronze is 43.6 mm² and 120MPa in steel is 32.7 mm².

<h3>What is normal stress?</h3>

If the direction of deformation force is perpendicular to the cross-sectional area of ​​the body, the stress is called normal stress. Changes in wire length and body volume will be normal.

σ = P/A

Where, σ = Normal stress

P = Pressure

A = Area

1 Kg = 9.81 N

800 kg = 7848 N

Since the rod is half bronze and half steel

800 kg = 7848/2

= 3924 N

Pₙ = Fₙ = 3924 N                       [n = Bronze]

Pₓ =  3924 N                             [x = steel]

Given,

σₙ = 90MPa

σₓ = 120MPa

Aₙ = ?

Aₓ = ?

Aₙ = Pₙ/σₙ

Aₙ = 3924/90

Aₙ = 43.6 mm²

Aₓ = Pₓ/σₓ

Aₓ = 3924/120

Aₓ = 32.7 mm²

To know more about normal stress, visit:

brainly.com/question/28012990

#SPJ9

4 0
1 year ago
Drag each label to the correct location on the chart. Classify the organisms based on how they obtain food.
Arisa [49]

Autograph: Trees (Image 1), algae (Image 4), carrots (image 6)

Heterograph: Tiger (Image 2), deer (image 3), human (image 5)

Explanation:

The difference between autotroph and heterotroph organisms is the way they obtain energy. In the case of the first category, these include organisms that can produce energy by a process known as photosynthesis. This covers all plants and similar such as tress, algae, and carrots.

On the other hand, heterotrophic organisms cannot produce energy and need to feed on other organisms, which can include feeding on plants or animals. In this context, this category includes tigers because they feed on other animals, deer because they feed on plants, and humans because we feed on plants and animals.

3 0
3 years ago
Read 2 more answers
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