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3 years ago

How can you drop two eggs the fewest amount of times without them breaking?

Engineering

1 answer:

Sergio039 [100]3 years ago

4 0

Answer:

by catching it

Explanation:

if you catch it it will not break

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A 24-tooth gear has AGMA standard full-depth involute teeth with diametral pitch of 12. Calculate the pitch diameter, circular p

torisob [31]

Answer:

Explanation:

Given:

Tooth Number, N = 24

Diametral pitch pd = 12

pitch diameter, d = N/pd = 24/12 = 2in

circular pitch, pc = π/pd = 3.142/12 = 0.2618in

Addendum, a = 1/pd = 1/12 =0.08333in

Dedendum, b = 1.25/pd = 0.10417in

Tooth thickness, t = 0.5pc = 0,5 * 0.2618 = 0.1309in

Clearance, c = 0.25/pd = 0.25/12 = 0.02083in

5 0

3 years ago

Read 2 more answers

A 150 MVA, 24 kV, 123% three-phase synchronous generator supplies a large network. The network voltage is 27 kV. The phase angle

Aleks04 [339]

Answer:

the generator induced voltage is 60.59 kV

Explanation:

Given:

S = 150 MVA

Vline = 24 kV = 24000 V

$X_{s} =1.23(\frac{V_{line}^{2} }{s} )=1.23\frac{24000^{2} }{1500} =4723.2 ohms$

the network voltage phase is

$V_{phase} =\frac{V_{nline} }{\sqrt{3} } =\frac{27}{\sqrt{3} } =15.58kV$

the power transmitted is equal to:

$|E|=\frac{P*X_{s} }{3*|V_{phase}|sinO } ;if-O=60\\|E|=\frac{300*4.723}{3*15.58*sin60} =34.98kV$

the line induced voltage is

$|E_{line} |=\sqrt{3} *|E|=\sqrt{3} *34.98=60.59kV$

7 0

3 years ago

What type of engineer would be most likely to develop a design for cars? chemical civil materials mechanical

Andreyy89

I don’t know but good luck

4 0

3 years ago

Technologies that allow for instant worldwide communication include

hram777 [196]

Answer:

mobile phones and internet access

Explanation:

I got it right on my quiz

7 0

3 years ago

Read 2 more answers

A sand has a natural water content of 5% and bulk unit weight of 18.0 kN/m3. The void ratios corresponding to the densest and lo

Zinaida [17]

Answer:

Relative density = 0.545

Degree of saturation = 24.77%

Explanation:

Data provided in the question:

Water content, w = 5%

Bulk unit weight = 18.0 kN/m³

Void ratio in the densest state, $e_{min}$ = 0.51

Void ratio in the loosest state, $e_{max}$ = 0.87

Now,

Dry density, $\gamma_d=\frac{\gamma_t}{1+w}$

$=\frac{18}{1+0.05}$

= 17.14 kN/m³

Also,

$\gamma_d=\frac{G\gamma_w}{1+e}$

here, G = Specific gravity = 2.7 for sand

$17.14=\frac{2.7\times9.81}{1+e}$

e = 0.545

Relative density = $\frac{e_{max}-e}{e_{max}-e_{min}}$

= $\frac{0.87-0.545}{0.87-0.51}$

= 0.902

Also,

Se = wG

here,

S is the degree of saturation

therefore,

S(0.545) = (0.05)()2.7

S = 0.2477

S = 0.2477 × 100% = 24.77%

7 0

3 years ago