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3 years ago

Which two is right about febuary 14

Engineering

2 answers:

NemiM [27]3 years ago

6 0

Answer:

Valentine's Day and your birthday :Þ

My new best friend <3

~ Nicki

igor_vitrenko [27]3 years ago

3 0

Answer:A and B

Explanation:

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A cylindrical specimen of a cold-worked brass has a ductility (%EL) of 27%. If its cold-worked radius is 14 mm, what was its rad

blagie [28]

Answer:

Initial radius of cylindrical specimen

ri = 15.46mm

Explanation:

Cw = [(A1 - A2)/A1] * 100% ...equation 1

Where CW = cold worked percentage gain for( 27% EL) ductile

A2 = final area of cylindrical specimen = pai * r²

Where r = final radius of cylindrical specimen

So therefore expanding equation 1 will give

A1 * Cw = (100 * A1) - (100 * A2)

Cw = 18% from attached graph

A2 = area of cylindrical specimen after cold work = pai * 14² = 615.75mm²

So therefore

18A1 = 100A1 - (100 * 615.75)

-82A1 = -61575.22

A1 = 750.92mm²

So solving for initial radius from initial area of specimen

r1 = (A1/pai)^½

r1 = (750.92/pai)^½

r1= 15.46mm

5 0

3 years ago

In alternating current, how often does the current alternate between negative and positive?

Novay_Z [31]

Answer:

AC voltages alternate/cycle, at a rate of 60 times each second.

7 0

2 years ago

Read 2 more answers

Air is drawn from the atmosphere into a turbomachine. At the exit, conditions are 500 kPa (gage) and 130oC. The exit speed is 10

finlep [7]

Answer:

P=- 88.41 KW

Negative sign indicates that power is given to the system.

Explanation:

Given that

P₂=500 KPa

T₂=130°C

V₂=100 m/s

mass flow rate ,m= 0.8 kg/s

Lets take inlet condition for air

T₁=25°C

P₁=100 KPa

V₁=0 m/s

We know that

Heat capacity for air Cp=1.005 KJ/kg.k

We know that for air change in enthalpy only depends only on temperature

Now from first law for open system

$h_1+\dfrac{V_1^2}{2000}=h_2+\dfrac{V_2^2}{2000}+W$

$1.005\times 298+\dfrac{0^2}{2000}=1.005\times 403+\dfrac{100^2}{2000}+W$

$W=1.005\times 298-1.005\times 403-\dfrac{100^2}{2000}$

W=-110.52 KJ/kg

Shaft power P = m .W

P = -110.52 x 0.8

P=- 88.41 KW

Negative sign indicates that power is given to the system.

3 0

3 years ago

What’s the answer to this I don’t understand

kari74 [83]

Answer:

Electrons are the negatively charged particles of atom. Electrons are extremely small compared to all of the other parts of the atom. So, the negative green particles of this atom are the electrons.

Explanation:

Hope it helps:)

3 0

2 years ago

The disk of radius 0.4 m is originally rotating at ωo=4 rad/sec. If it is subjected to a constant angular acceleration of α=5 ra

Lina20 [59]

Answer:32.4m/ $s^2$

Explanation:

Given data

$radius\left ( r\right )$ =0.4m

Intial angular velocity $\left ( \omega_0\right )$ =4rad/s

angular acceleration $\left ( \alpha\right )$ =5rad/ $s^2$

angular velocity after 1 sec

$\omega$ = $\omega_0$ + $\alpha\times\t$

$\omega$ =4+5 $\left ( 1\right )$

$\omega$ =9rad/s

Velocity of point on the outer surface of disc $\left ( v\right )$ = $\omega_0\timesr$

v= $9\times0.4$ m/s=3.6m/s

Normal component of acceleration $\left ( a_c\right )$ = $\frac{v^2}{r}$

$a_c$ = $\frac{3.6\times3.6}{0.4}$ =32.4m/ $s^2$

3 0

4 years ago