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2 years ago

Which solution causes cells to shrink

1 answer:

7 0

Answer: Hypertonic

Explain: a hypertonic solution has increased solute and a net movement of water outside causing the cell to shrink. A hypotonic has decreased solute concentration, and a net movement of water inside the cell, causing swelling or breakage.

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The aluminum rod AB (G 5 27 GPa) is bonded to the brass rod BD (G 5 39 GPa). Knowing that portion CD of the brass rod is hollow

Temka [501]

Answer:

Qcd=0.01507rad

QT= 0.10509rad

Explanation:

The full details of the procedure and answer is attached.

7 0

3 years ago

What do we need to do to get CO2 emissions all the way to zero?

Svetlanka [38]

Answer:

A key element is powering economies with clean energy, replacing polluting coal - and gas and oil-fired power stations - with renewable energy sources, such as wind or solar farms. This would dramatically reduce carbon emissions. Plus, renewable energy is now not only cleaner, but often cheaper than fossil fuels

Explanation:

here is your answer if you like my answer please follow

3 0

2 years ago

A 50 mm diameter shaft is subjected to a static axial load of 160 kN. If the yield stress of the material is 350 MPa, the ultima

zvonat [6]

In order to develop this problem it is necessary to take into account the concepts related to fatigue and compression effort and Goodman equation, i.e, an equation that can be used to quantify the interaction of mean and alternating stresses on the fatigue life of a materia.

With the given data we can proceed to calculate the compression stress:

$\sigma_c = \frac{P}{A}$

$\sigma_c = \frac{160*10^3}{\pi/4*0.05^2}$

$\sigma_c = 81.5MPa$

Through Goodman's equations the combined effort by fatigue and compression is expressed as:

$\frac{\sigma_a}{S_e}+\frac{\sigma_c}{\sigma_u}=\frac{1}{Fs}$

Where,

$\sigma_a=$ Fatigue limit for comined alternating and mean stress

$S_e =$ Fatigue Limit

$\sigma_c=$ Mean stress (due to static load)

$\sigma_u =$ Ultimate tensile stress

$Fs =$ Security Factor

We can replace the values and assume a security factor of 1, then

$\frac{\sigma_a}{320}+\frac{81.5}{400}=\frac{1}{1}$

Re-arrenge for $\sigma_a$

$\sigma_a = 254.8Mpa$

We know that the stress is representing as,

$\sigma_a = \frac{M_c}{I}$

Then,

Where $M_c$ =Max Moment

I= Intertia

The inertia for this object is

$I=\frac{\pi d^4}{64}$

Then replacing and re-arrenge for $M_c$

$M_c = \frac{\sigma_a*\pi*d^3}{32}$

$M_c = \frac{260.9*10^6*\pi*0.05^3}{32}$

$M_c = 3201.7N.m$

Thereforethe moment that can be applied to this shaft so that fatigue does not occur is 3.2kNm

5 0

3 years ago

There are 10 vehicles in a queue when an attendant opens a toll booth. Vehicles arrive at the booth at a rate of 4 per minute. T

dmitriy555 [2]

Answer:

as slated in your solution, if delay time is 2.30 mins, hence 9 vehicle will be on queue as the improved service commenced.

Explanation:

4 vehicle per min, in 2 mins of the delay time 8 vehicles while in 0.3 min average of 1 vehicle join the queue. making 9 vehicle maximum

3 0

2 years ago

A steel pipe of 400-mm outer diameter is fabricated from 10-mm-thick plate by welding along a helix that forms an angle of 20° w

Verdich [7]

Explanation:

Outer di ameter $d_{0}=400 \mathrm{mm}[tex] Thickness of the cylinder [tex]t=10 \mathrm{mm}$

$\therefore[tex] Inner diam eter [tex]d_{i}=d_{0}-2 t=400-2 \times 10$

$d_{1}=380 \mathrm{mm}$

Given loading on the cylinder $P=300 \mathrm{kN}$ Helix an gle of the weld form $\theta=20^{\circ}$

(i) Normal stress on the plane at angle $\theta=20^{\circ}$ is

$\sigma=\frac{P \cos ^{2} \theta}{A_{0}}$

$\text { Where } A_{0}=\frac{\pi}{4}\left(d_{0}^{2}-d_{1}^{2}\right)$

$\quad=\frac{\pi}{4}\left(400^{2}-380^{2}\right)$

$=12252.21 \mathrm{mm}^{2}$

$=12.25221 \times 10^{-9} \mathrm{m}^{2}$

$\sigma=\frac{-300 \times 10^{2} \times \cos ^{2} 20}{12.25221 \times 10^{-1}}$

$=-21.6 \mathrm{MPa}$

(ii) Shear stress along an angle of $\theta=20^{\circ}$ is $\tau=\frac{P}{A_{0}} \cos \theta \sin \theta$

$=\frac{-300 \times 10^{-1} \times \cos 20 \times \sin 20}{12.25221 \times 10^{-3}}$

$=-7.86 \mathrm{MPa}$

3 0

2 years ago