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3 years ago

Is 4/16 equal in measurement to 1/4

Engineering

2 answers:

Alja [10]3 years ago

5 0

Answer:yes

Explanation:

rodikova [14]3 years ago

4 0

Yes It is. The first fraction can be reduced to equal 1/4

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Which option identifies the type of engineering technician most likely to be involved in the following scenario?

aleksandrvk [35]

Answer:

I believe the answer to your question would be A.

Explanation:

Electrical engineering technician

6 0

2 years ago

An engineer is designing a total hip implant. She intends to make the femoral stem out of titanium because it forms a good inter

makvit [3.9K]

Answer:

1. Yes.

2. Localized corrosion

Explanation:

Should she be worried about corrosion?

Yes, the engineer needs to be worried about corrosion as stainless steel has a lower resistance to corrosion, in other words, stainless steel corrodes faster than Titanium.

If so, what types of corrosion could take place?

The type of corrosion that takes place is called Localized corrosion. Localized corrosion occurs when a small part of a component experiences corrosion. In this case, the ball component of the femoral stem is made of stainless steel which will corrode faster than the other parts of the femoral stem which is made of Titanium.

3 0

3 years ago

The toggle (t) flip-flop has one input, clk, and one output, q. on each rising edge of clk, q toggles to the complement of its p

jeyben [28]

Answer:

See attached

Explanation:

The next state of a toggle flip-flop is the inverse of the present state. This behavior can be produced using a D flip-flop that has its input connected to the inverse of its output.

A schematic is attached.

7 0

2 years ago

A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h

hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

$D_{1}$ = 150 mm

$A_{1}= \frac{\pi }{4}\times 150^{2}$

= 17671.45 $mm^{2}$

$D_{2}$ = 250 mm

$A_{2}= \frac{\pi }{4}\times 250^{2}$

= 49087.78 $mm^{2}$

The centrifugal force acting on the flywheel is fiven by

F = M ( $R_{2}$ - $R_{1}$ ) x $w^{2}$ ------------(1)

Here F = ( -UTS x $A_{1}$ + UCS x $A_{2}$ )

Since density, $\rho = \frac{M}{V}$

$\rho = \frac{M}{A\times t}$

$M = \rho \times A\times t$ $M = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t$

$M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37$

$M = 8252963901$

∴ $R_{2}$ - $R_{1}$ = 50 mm

∴ F = $763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}$

F = 33618968.38 N --------(2)

Now comparing (1) and (2)

$33618968.38 = 8252963901\times 50\times \omega ^{2}$

∴ ω = 4036.61

We know

$\omega = \frac{2\pi N}{60}$

$4036.61 = \frac{2\pi N}{60}$

∴ N = 38546.82 rpm

7 0

3 years ago

The aluminum rod (E1 = 68 GPa) is reinforced with the firmly bonded steel tube (E2 = 201 GPa). The diameter of the aluminum rod

Vsevolod [243]

Answer:

Explanation:

From the information given:

$E_1 = 68 \ GPa \\ \\ E_2 = 201 \ GPa \\ \\ d = 25 \ mm \ \\ \\ D = 45 \ mm \ \\ \\ L = 761 \ mm \\ \\ P = -88 kN$

The total load is distributed across both the rod and tube:

$P = P_1+P_2 --- (1)$

Since this is a composite column; the elongation of both aluminum rod & steel tube is equal.

$\delta_1=\delta_2$

$\dfrac{P_1L}{A_1E_1}= \dfrac{P_2L}{A_2E_2}$

$\dfrac{P_1 \times 0.761}{(\dfrac{\pi}{4}\times .0025^2 ) \times 68\times 10^4}= \dfrac{P_2\times 0.761}{(\dfrac{\pi}{4}\times (0.045^2-0.025^2))\times 201 \times 10^9}$