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lana [24]
4 years ago
8

Electron microscopes can form images of extremely small objects because the electrons used in them typically have wavelengths ab

out 100,000 times smaller than visible light (on the order of 10-12 m). Show that the electron in
(a) would be useful in an electron microscope by finding its de Broglie wavelength.
Engineering
1 answer:
SCORPION-xisa [38]4 years ago
7 0

Answer:

The Broglie wavelength of the electron is 6.171x10⁻¹² m

Explanation:

the solution is in the attached Word document

Download docx
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A stationary gas-turbine power plant operates on a simple ideal Brayton cycle with air as the working fluid. The air enters the
ololo11 [35]

Answer:

A) W' = 15680 KW

B) W' = 17113.87 KW

Explanation:

We are given;

Temperature at state 1; T1 = 290 K

Temperature at state 3; T3 = 1100 K

Rate of heat transfer; Q_in = 35000 kJ/s = 35000 Kw

Pressure of air into compressor; P_c = 95 kPa

Pressure of air into turbine; P_t = 760 kPa

A) The power assuming constant specific heats at room temperature is gotten from;

W' = [1 - ((T4 - T1)/(T3 - T2))] × Q_in

Now, we don't have T4 and T2 but they can be gotten from;

T4 = [T3 × (r_p)^((1 - k)/k)]

T2 = [T1 × (r_p)^((k - 1)/k)]

r_p = P_t/P_c

r_p = 760/95

r_p = 8

Also,k which is specific heat capacity of air has a constant value of 1.4

Thus;

Plugging in the relevant values, we have;

T4 = [(1100 × (8^((1 - 1.4)/1.4)]

T4 = 607.25 K

T2 = [290 × (8^((1.4 - 1)/1.4)]

T2 = 525.32 K

Thus;

W' = [1 - ((607.25 - 290)/(1100 - 525.32))] × 35000

W' = 0.448 × 35000

W' = 15680 KW

B) The power accounting for the variation of specific heats with temperature is given by;

W' = [1 - ((h4 - h1)/(h3 - h2))] × Q_in

From the table attached, we have the following;

At temperature of 607.25 K and by interpolation; h4 = 614.64 KJ/K

At T3 = 1100 K, h3 = 1161.07 KJ/K

At T1 = 290 K, h1 = 290.16 KJ/K

At T2 = 525.32 K, and by interpolation, h2 = 526.12 KJ/K

Thus;

W' = [1 - ((614.64 - 290.16)/(1161.07 - 526.12))] × 35000

W' = 17113.87 KW

4 0
3 years ago
Air at T1 = 32°C, p1 = 1 bar, 50% relative humidity enters an insulated chamber operating at steady state with a mass flow rate
RUDIKE [14]

Answer:

4.5kg/min

Explanation:

Given parameters

T_1 = 32^0 C,  m_1 = 3 kg/min, T_2 = 7^0 C ,T_3 = 17^0

if we take  

The mass flow rate of the second stream = m_2(kg/min)

The mass flow rate of mixed exit stream = m_3 (kg/min)

Now from mass conservation

m_3 = m_2 + m_1

m_3 = m_2 + 3 (kg/min)

The temperature of the mixed exit stream given as

T_3m_3 = T_2m_2 +T_1m_1\\\\17 ( 3 + m_2) = 7 \times m_2 + 32 \times 3\\\\51 + 17 m_2 = 7 m_2 + 96\\\\10 m_2 = 96 - 51\\\\m_2 = 4.5 kg/min\\\\\\\\

Therefore the mass flow rate of second stream will be 4.5 kg/min.

7 0
3 years ago
A storage tank, used in a fermentation process, is to be rotationally molded from polyethylene plastic. This tank will have a co
NNADVOKAT [17]

Answer:

The volume up to cylindrical portion is approx  32355 liters.

Explanation:

The tank is shown in the attached figure below

The volume of the whole tank is is sum of the following volumes

1) Hemisphere top

Volume of hemispherical top of radius 'r' is

V_{hem}=\frac{2}{3}\pi r^3

2) Cylindrical Middle section

Volume of cylindrical middle portion of radius 'r' and height 'h'

V_{cyl}=\pi r^2\cdot h

3) Conical bottom

Volume of conical bottom of radius'r' and angle \theta is

V_{cone}=\frac{1}{3}\pi r^3\times \frac{1}{tan(\frac{\theta }{2})}

Applying the given values we obtain the volume of the container up to cylinder is

V=\pi 1.5^2\times 4.0+\frac{1}{3}\times \frac{\pi 1.5^{3}}{tan30}=32.355m^{3}

Hence the capacity in liters is V=32.355\times 1000=32355Liters

3 0
4 years ago
Technician A says it will save you money if you get a high interest rate from a bank on the loan. Technician B says it is always
zubka84 [21]
Is there a other option
5 0
3 years ago
There are four distinct events that occur in the Carnot cycle. Name the events and describe each one.
OLEGan [10]

Answer and Explanation:

There four events or stages of reversible nature in a Carnot cycle are:

1. Isothermal expansion of gas

2. Adiabatic expansion

3. Isothermal compression of gas

4.Adiabatic compression of gas

All these 4 processes are reversible processes.

1. A reversible Isothermal expansion of gas takes place in Carnot cycle where an ideal gas  absorbs or intake certain quantity of heat from a heat reservoir or source at an elevated temperature which results in the expansion of gas and work is done on the  surroundings.

2. A reversible expansion of gas adiabatically takes place in Carnot cycle in an environment with thermal insulation where the gas expand and work is done on the surrounding which results in the lowering of the temperature of the system.

3. A reversible compression of gas  isothermally takes place such that the work is done on the system by the surrounding resulting in heat loss.

4. A reversible compression of gas adiabatically, takes place in an environment with thermal insulation and the work is done continuously on the system by the surroundings resulting in the rise in temperature.

8 0
4 years ago
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