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3 years ago

Một doanh nghiệp có tư bản đầu tư là 600,000 usd, cấu tạo hữu cơ tư bản 3/1. Xác định tiền công trả cho người lao động

Engineering

1 answer:

Maru [420]3 years ago

3 0

Answer:

450.000 USD

Explanation:

Đây,

Cấu trúc hữu cơ 3/1 có nghĩa là ¾ một phần vốn được chuyển vào chi phí cố định và ¼ một phần đi vào chi phí biến đổi.

Tiền lương của người lao động là chi phí cố định và do đó, mức lương

= ¾ * 600.000 USD

= 450.000 USD

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state & prove parallelogram law of vector addition &Also determine magnitude &direction of resultant vector.

ludmilkaskok [199]

Answer:

Parallelogram law of vector addition states that if two vectors are considered to be the adjacent sides of a parallelogram, then the resultant of the two vectors is given by the vector that is diagonal passing through the point of contact of two vectors.

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3 years ago

Steam enters a turbine steadily at 7 MPa and 600°C with a velocity of 60 m/s and leaves at 25 kPa with a quality of 95 percent.

Rufina [12.5K]

Answer:

a) $\dot m = 16.168\,\frac{kg}{s}$ , b) $v_{out} = 680.590\,\frac{m}{s}$ , c) $\dot W_{out} = 18276.307\,kW$

Explanation:

A turbine is a steady-state devices which transforms fluid energy into mechanical energy and is modelled after the Principle of Mass Conservation and First Law of Thermodynamics, whose expressions are described hereafter:

Mass Balance

$\frac{v_{in}\cdot A_{in}}{\nu_{in}} - \frac{v_{out}\cdot A_{out}}{\nu_{out}} = 0$

Energy Balance

$-q_{loss} - w_{out} + h_{in} - h_{out} = 0$

Specific volumes and enthalpies are obtained from property tables for steam:

Inlet (Superheated Steam)

$\nu_{in} = 0.055665\,\frac{m^{3}}{kg}$

$h_{in} = 3650.6\,\frac{kJ}{kg}$

Outlet (Liquid-Vapor Mix)

$\nu_{out} = 5.89328\,\frac{m^{3}}{kg}$

$h_{out} = 2500.2\,\frac{kJ}{kg}$

a) The mass flow rate of the steam is:

$\dot m = \frac{v_{in}\cdot A_{in}}{\nu_{in}}$

$\dot m = \frac{\left(60\,\frac{m}{s} \right)\cdot (0.015\,m^{2})}{0.055665\,\frac{m^{3}}{kg} }$

$\dot m = 16.168\,\frac{kg}{s}$

b) The exit velocity of steam is:

$\dot m = \frac{v_{out}\cdot A_{out}}{\nu_{out}}$

$v_{out} = \frac{\dot m \cdot \nu_{out}}{A_{out}}$

$v_{out} = \frac{\left(16.168\,\frac{kg}{s} \right)\cdot \left(5.89328\,\frac{m^{3}}{kg} \right)}{0.14\,m^{2}}$

$v_{out} = 680.590\,\frac{m}{s}$

c) The power output of the steam turbine is:

$\dot W_{out} = \dot m \cdot (-q_{loss} + h_{in}-h_{out})$

$\dot W_{out} = \left(16.168\,\frac{kg}{s} \right)\cdot \left(-20\,\frac{kJ}{kg} + 3650.6\,\frac{kJ}{kg} - 2500.2\,\frac{kJ}{kg}\right)$

$\dot W_{out} = 18276.307\,kW$

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3 years ago

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.

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3 years ago

Most deformation occurs along plate boundaries because ________.

Svetlanka [38]

B) the plates are in constant motion and as a result the boundaries are where they interact

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3 years ago

What are the de Broglie frequencies and wavelengths of (a) an electron accelerated to 50 eV (b) a proton accelerated to 100 eV

DaniilM [7]

Answer:

(a) De-Brogie wavelength is 0.173 nm and frequency is 2.42 x 10^16 Hz

(b) De-Brogie wavelength is 2.875 pm and frequency is 4.8 x 10^16 Hz

Explanation:

(a)

First, we need to find velocity of electron. Since, it is accelerated by electric potential. Therefore,

K.E of electron = (1/2)mv² = (50 eV)(1.6 x 10^-19 J/1 eV)

(1/2)mv² = 8 x 10^(-18) J

Mass of electron = m = 9.1 x 10^(-31) kg

Therefore,

v² = [8 x 10^(-18) J](2)/(9.1 x 10^(-31) kg)

v = √1.75 x 10^13

v = 4.2 x 10^6 m/s

Now, the de Broglie's wavelength is given as:

λ = h/mv

where,

h = Plank's Constant = 6.626 x 10^(-34) kg.m²/s

Therefore,

λ = (6.626 x 10^(-34) kg.m²/s)/(9.1 x 10^(-31) kg)(4.2 x 10^6 m/s)

λ = 0.173 x 10^(-9) m = 0.173 nm

The frequency is given as:

Frequency = f = v/λ

f = (4.2 x 10^6 m/s)/(0.173 x 10^(-9) m)

f = 2.42 x 10^16 Hz

(b)

First, we need to find velocity of proton. Since, it is accelerated by electric potential. Therefore,

K.E of proton = (1/2)mv² = (100 eV)(1.6 x 10^-19 J/1 eV)

(1/2)mv² = 1.6 x 10^(-17) J

Mass of proton = m = 1.67 x 10^(-27) kg

Therefore,

v² = [1.6 x 10^(-17) J](2)/(1.67 x 10^(-27) kg)

v = √1.916 x 10^10

v = 1.38 x 10^5 m/s

Now, the de Broglie's wavelength is given as:

λ = h/mv

where,

h = Plank's Constant = 6.626 x 10^(-34) kg.m²/s

Therefore,

λ = (6.626 x 10^(-34) kg.m²/s)/(1.67 x 10^(-27) kg)(1.38 x 10^5 m/s)

λ = 2.875 x 10^(-12) m = 2.875 pm

The frequency is given as:

Frequency = f = v/λ

f = (1.38 x 10^5 m/s)/(2.875 x 10^(-12) m)

f = 4.8 x 10^16 Hz

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3 years ago