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3 years ago

15

The resistivity of mercury drops suddenly to zero at a critical temperature, maki mercury a superconductor below that temperatur

e. ( True , False )

1 answer:

seropon [69]3 years ago

4 0

Resistance zero meaning superconductor, so True.

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Select three types of lines that engineers use to help represent the shape of a design in a sketch.

Vikki [24]

Hidden lines

Used to describe the in shown lines (like diagonals inside cubes)

Extension lines:-

Used to explain the expansion of structures like building

Object lines

Used to describe the structure of objects and the lining to show borders

7 0

3 years ago

Consider a 400 mm × 400 mm window in an aircraft. For a temperature difference of 90°C from the inner to the outer surface of th

True [87]

Um I think the answer for this is 10l

5 0

3 years ago

A solid shaft and a hollow shaft of the same material have same length and outer radius R. The inner radius of the hollow shaft

alexandr402 [8]

Answer with Explanation:

By the equation or Torque we have

$\frac{T}{I_{p}}=\frac{\tau }{r}=\frac{G\theta }{L}$

where

T is the torque applied on the shaft

$I_{p}$ is the polar moment of inertia of the shaft

$\tau$ is the shear stress developed at a distance 'r' from the center of the shaft

$\theta$ is the angle of twist of the shaft

'G' is the modulus of rigidity of the shaft

We know that for solid shaft $I_{p}=\frac{\pi R^4}{2}$

For a hollow shaft $I_{p}=\frac{\pi (R_o^4-R_i^4)}{2}$

Since the two shafts are subjected to same torque from the relation of Torque we have

1) For solid shaft

$\frac{2T}{\pi R^4}\times r=\tau _{solid}$

2) For hollow shaft we have

$\tau _{hollow}=\frac{2T}{\pi (R^4-0.7R^4)}\times r=\frac{2T}{\pi 0.76R^4}$

Comparing the above 2 relations we see

$\frac{\tau _{solid}}{\tau _{hollow}}=0.76$

Similarly for angle of twist we can see

$\frac{\theta _{solid}}{\theta _{hollow}}=\frac{\frac{LT}{I_{solid}}}{\frac{LT}{I_{hollow}}}=\frac{I_{hollow}}{I_{solid}}=1.316$

Part b)

Strength of solid shaft = $\tau _{max}=\frac{T\times R}{I_{solid}}$

Weight of solid shaft = $\rho \times \pi R^2\times L$

Strength per unit weight of solid shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{solid}}\times \frac{1}{\rho \times \pi R^2\times L}=\frac{2T}{\rho \pi ^2R^5L}$

Strength of hollow shaft = $\tau '_{max}=\frac{T\times R}{I_{hollow}}$

Weight of hollow shaft = $\rho \times \pi (R^2-0.7R^2)\times L$

Strength per unit weight of hollow shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{hollow}}\times \frac{1}{\rho \times \pi (R^2-0.7^2)\times L}=\frac{5.16T}{\rho \pi ^2R^5L}$

Thus $\frac{Strength/Weight _{hollow}}{Strength/Weight _{Solid}}=5.16$

3 0

4 years ago

A. Name the major strengthening mechanisms in metals and explain the working principle under each mechanism.Give the relevant eq

Sever21 [200]

Answer:

a) Solid solution strengthening and alloying, Precipitation hardening, work hardening

b) Absence of enough crystallographic misalignment in the grain boundary region for a small-angle

Explanation:

<u>A) strengthening mechanism</u>

i) Solid solution strengthening and alloying:

In solid solution strengthening and alloying mechanism there is an addition of one atom of solute to another during this process, there might be substitution of interstitial point defect in crystal

also the shear stress required can be represented as: Δz = Gb√Ce^3/2

where : C = solute concentration , e = strain on material

ii) Precipitation hardening:

During precipitation hardening the alloying above the concentrate will lead to the formation of a second phase also under precipitation hardening a second phase can also be created via thermal treatments

particle bowing cab be written as : Δz = Gb / L-2x

iii) work hardening :

Dislocation caused by stress fields been generated hardens metals under the work hardening mechanism

dislocation can be represented as ; Gb √ p

where : G = shear modulus , b = Burgess vector, p = dislocation density

B) The small angle grain boundaries are not effective enough because there is less crystallographic misalignment in the grain boundary region for a small-angle

3 0

3 years ago

How much extra water does a 21.5 ft, 175-lb concrete canoe displace compared to an ultra-lightweight 38-lb Kevlar canoe of the s

pantera1 [17]

Answer:

The volume of the extra water is $2.195 ft^{3}$

Solution:

As per the question:

Mass of the canoe, $m_{c} = 175 lb + w$

Height of the canoe, h = 21.5 ft

Mass of the kevlar canoe, $m_{Kc} = 38 lb + w$

Now, we know that, bouyant force equals the weight of the fluid displaced:

Now,

$V\rho g = mg$

$V = \frac{m}{\rho}$ (1)

where

V = volume

$\rho = 62.41 lb/ft^{3}$ = density

m = mass

Now, for the canoe,

Using eqn (1):

$V_{c} = \frac{m_{c} + w}{\rho}$

$V_{c} = \frac{175 + w}{62.41}$

Similarly, for Kevlar canoe:

$V_{Kc} = \frac{38 + w}{62.41}$

Now, for the excess volume:

V = $V_{c} - V_{Kc}$

V = $\frac{175 + w}{62.41} - \frac{38 + w}{62.41} = 2.195 ft^{3}$

8 0

3 years ago