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2 years ago

Convert A'B'C'D' + A'B'C'D + A'B'CD' + A'BC'D + AB'C'D' + AB'C'D+ AB'CD' to SOP form

Engineering

2 answers:

bazaltina [42]2 years ago

6 0

Answer:

thats really hard how could you answerthis hhhhhhh

Viefleur [7K]2 years ago

5 0

Just put on dofferent order take off the sighn

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What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 2.5×10-4

krok68 [10]

Answer:

1788.9 MPa

Explanation:

The magnitude of the maximum stress (σ) can be calculated usign the following equation:

$\sigma = 2\sigma_{0} \sqrt{\frac{a}{\rho}}$

Where:

ρ: is the radius of curvature = 2.5x10⁻⁴ mm (0.9843x10⁻⁵ in)

σ₀: is the tensile stress = 100x10⁶ Pa (14500 psi)

2a: is the crack length = 4x10⁻² mm (1.575x10⁻³ in)

Hence, the maximum stress (σ) is:

$\sigma = 2*100\cdot 10^{6} Pa \sqrt{\frac{(4 \cdot 10^{-2} mm)/2}{2.5 \cdot 10^{-4} mm}} = 1.79 \cdot 10^{6} Pa = 1788.9 MPa$

Therefore, the magnitude of the maximum stress is 1788.9 MPa.

I hope it helps you!

5 0

3 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

2 years ago

Plssssssssssssss Alexi is writing a program which prompts users to enter their age. Which function should she use?

aleksandr82 [10.1K]

Answer:

int()

Explanation:

float() is using decimals, so that can't be it, like float(input( "how much does this cost?"))

print() is used to print something, not a user asking, like print("hello")

string() means like a whole, like string( I am good)

By elimination, int() is correct.

Hope this helps!

7 0

2 years ago

Socket Programming: (30 points) Use Python TCP socket to implement an application with client-server architecture. In this appli

ololo11 [35]

Answer:

Explanation:

Run the code given in text file following instructions.

Download txt

8 0

2 years ago

.The war of the currents in the 1880's involved Thomas Edison and Nikola Tesla on a reality TV show stranded on an island. Each

natali 33 [55]

Answer:

True

Explanation:

Nikola Tesla defeated Thomas Edison in the AC/DC battle of electric current.

7 0

2 years ago