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saveliy_v [14]
3 years ago
12

Why did my dad leave me yesterday like whatd i do?

Engineering
1 answer:
Vsevolod [243]3 years ago
4 0

Answer:

hmm i would try calling him. ask your mom or other adults where he is!

Explanation:

hope you get help soon!

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The melting point of Pb (lead) is 327°C, is the processing at 20°C hot working or cold working?
bonufazy [111]

Answer:

Explained

Explanation:

Cold working: It is plastic deformation of material at temperature below   recrystallization temperature. whereas hot working is deforming material above the recrystallization temperature.

Given melting point temp of lead is 327° C and lead recrystallizes at about

0.3 to 0.5 times melting temperature which will be higher that 20°C. Hence we can conclude that at 20°C lead will under go cold working only.

6 0
3 years ago
A 4140 steel shaft, heat-treated to a minimum yield strength of 100 ksi, has a diameter of 1 7/16 in. The shaft rotates at 600 r
velikii [3]
Answer:










Explanation:



4140-40 I’d pick wood




I hope this helps! :)
4 0
3 years ago
Read 2 more answers
An ac source of period T and maximum voltage V is connected to a single unknown ideal element that is either a resistor, and ind
Misha Larkins [42]

Answer:

The unknown element is Capacitor.

Explanation:

The sinusoidal voltage is given as:

v(t) = V Sin (ωt + Ф)

Where:

V = Amplitude of Voltage

ω = 2π / T = > Time period (T)

Ф = phase shift

Considering no horizontal phase shift in the wave form, the equation can be written as:

v(t) = V Sin (ωt)--------(1)

Since, current in the capacitor can be given as:

i(t) = C dv(t)/dt = ωCV Cos (ωt)--------(2)

Now, checking all conditions:

At t=0 :

Equation (1) implies:

v(t) = V Sin [(2π/T)(0)] = V Sin (0)

v(t) = 0

The above finding satisfies the condition in the question. Now checking other conditions.

At t = T/4:

Equation (2) implies:

i(t) = ωCV Cos [(2π/T)(T/4)] = ωCV Cos [(π/2)] = ωCV (0)

i(t) = 0

At t = T/2 :

Equation (2) implies:

i(t) = ωCV Cos [(2π/T)(T/2)] = ωCV Cos [(π)] = ωCV (-1)

i(t) = - ωCV = max amplitude of current in negative direction

All three conditions of voltage and currents of question are satisfied with equations of capacitor hence, the unknown element is capacitor.

4 0
3 years ago
A vehicle is considered to be legally parked if it is parked _____ or more from a pedestrian crosswalk or a marked or unmarked i
Yuki888 [10]

Answer:

A vehicle is considered to be legally parked if it is parked 20 feet (6 m) or more from a pedestrian crosswalk or a marked or unmarked intersection.

Explanation:

Hello!

I obtained the provided data from the New York State Driver's Manual. I wish it was useful to help you.

Success in your homework!

7 0
3 years ago
A rigid tank with a total volume of 0.05 m3 initially contains a two-phase liquid-vapor mixture of water at a pressure of 15 bar
Westkost [7]

Answer:

a) m_{2} = 0.753\,kg, b) Q_{in} = 2122.963\,kJ

Explanation:

A rigid tank means a storage whose volume is constant. Process is entirely isobaric. Initial and final properties of water are included below:

State 1 - Gas-Vapor Mixture

P = 1500\,kPa

T = 198.29^{\textdegree}C

\nu = 0.02726\,\frac{m^{3}}{kg}

u = 1192.94\,\frac{kJ}{kg}

h_{g} = 2791.0\,\frac{kJ}{kg}

x = 0.2

State 2 - Gas-Vapor Mixture

P = 1500\,kPa

T = 198.29^{\textdegree}C

\nu = 0.06643\,\frac{m^{3}}{kg}

u = 1718.12\,\frac{kJ}{kg}

h_{g} = 2791.0\,\frac{kJ}{kg}

x = 0.5

The model for the rigid tank is created by using the First Law of Thermodynamics:

Q_{in} - (m_{1}-m_{2})\cdot h_{g} = m_{2}\cdot u_{2}-m_{1}\cdot u_{1}

Initial and final masses are:

m_{1} = \frac{V_{1}}{\nu_{1}}

m_{1} = \frac{0.05\,m^{3}}{0.02726\,\frac{m^{3}}{kg} }

m_{1} = 1.834\,kg

m_{2} = \frac{V_{2}}{\nu_{2}}

m_{2} = \frac{0.05\,m^{3}}{0.06643\,\frac{m^{3}}{kg} }

m_{2} = 0.753\,kg

a) The final mass within the tank is:

m_{2} = 0.753\,kg

b) The total amount of heat transfer is:

Q_{in} = m_{2}\cdot u_{2}-m_{1}\cdot u_{1}+ (m_{1}-m_{2})\cdot h_{g}

Q_{in} = (0.753\,kg)\cdot (1718.12\,\frac{kJ}{kg} )- (1.834\,kg)\cdot (1192.94\,\frac{kJ}{kg} ) + (1.081\,kg)\cdot (2791.0\,\frac{kJ}{kg} )

Q_{in} = 2122.963\,kJ

5 0
3 years ago
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