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Leviafan [203]
3 years ago
13

Explain the following statement: In a windowless proportional counter, the output pulse height from an alpha particle source wil

l increase with increasing alpha energy whereas for beta particles the opposite is true.
Engineering
1 answer:
-BARSIC- [3]3 years ago
5 0

Answer / Explanation:

Regarding α and β Particles in Windowless Counter, the range of α particles is lower than β particles. Alpha particles typically have range less than the dimensions of the gas chamber so that proportional  counters are able to easily record. Hence,  with almost 100% efficiency, each particle which enters the so called active volume.

Then, since the pulse height spectra depends on the number of ion pairs which have formed,  an aplha particle with higher energy creates more ion pairs in the chamber.  However, beta particle range usually exceeds the dimensions of the chamber and therefore  most of the betas hit the walls where they deposit energy. Then, fewer ion pairs are formed  because very few β’s give their energy to the bulk gas.

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1. A turbine in a steam power plant operates isentropically with an inlet pressure (P3) of 3.5 MPa and inlet temperature (T3) of
sergey [27]

Answer:

See attached picture.

Explanation:

See attached picture for explanation.

4 0
3 years ago
1). Mention any four operations that requires airlines. 2). Explain how airflow is applicable to the above mentioned operations.
vladimir2022 [97]

Answer:

Following are the answer to this question:

Explanation:

1)

Following the four operations in the airlines:

Landside operations:

In Airlines, the airports are divided into areas on the countryside and on the airside, in which landside region is available to the public, although strictly controlled access to the airside zone. Its area covers all areas of the airport across the aircraft, including parts of the buildings which can only be reached by customers and employees.

Airside Operations:

It's also committed to ensuring which air operations military exercises Ballarat airfields are safe and secure. It includes the provision of parking and flight escort services to itinerant and automates. Organizing operational response to incidents, accidents, or emergencies at the airport.

Billing and invoicing Operations:

This requires several steps, each of which must be performed with absolute accuracy to ensure that perhaps the airport operator is adequately paid for supplying passengers with all the services and infrastructure. After this, the receipts want to be produced and sent to customers on the airline.

Information management:

In this, it collects all the data about the customers and employees and it also helps in finding new routes and after collecting the data it processes on them.  

2)

It involves many activities in the airline, including dispatch, flight preparation, flight watch, weather information source, activities control, ground-to-air communications, and staff coordination, scheduling, and maintenance planning. Computing and expert programs are constantly being used to handle unpredictable activities.

4 0
3 years ago
The yellow rectangle area is 25% (or 1/4) the area of the blue rhombus. The height (H) of the yellow rectangle is twice as long
Kitty [74]

Answer:

I don't know sry

Explanation:

6 0
3 years ago
A 12-ft circular steel rod with a diameter of 1.5-in is in tension due to a pulling force of 70-lb. Calculate the stress in the
padilas [110]

Answer:

The stress in the rod is 39.11 psi.

Explanation:

The stress due to a pulling force is obtained dividing the pulling force by the the area of the cross section of the rod. The respective area for a cylinder is:

A=\pi*D^2/4

Replacing the diameter the area results:

A= 17.76 in^2

Therefore the the stress results:

σ = 70/17.76 lb/in^2 = 39.11 lb/in^2= 39.11 psi

5 0
3 years ago
. Air at 200 C blows over a 50 cm x 75 cm plain carbon steel (AISI 1010) hot plate with a constant surface temperature of 2500 C
MrRissso [65]

Answer:

The inside temperature, T_{in} is approximately 248 °C.

Explanation:

The parameters given are;

Temperature of the air = 20°C

Carbon steel surface temperature 250°C

Area of surface = 50 cm × 75 cm = 0.5 × 0.75 = 0.375 m²

Convection heat transfer coefficient = 25 W/(m²·K)

Heat lost by radiation = 300 W

Assumption,

Air temperature = 20 °C

Hot plate temperature = 250 °C

Thermal conductivity K = 65.2 W/(m·K)

Steady state heat transfer process

One dimensional heat conduction

We have;

Newton's law of cooling;

q = h×A×(T_s - T_{\infty) + Heat loss by radiation

= 25×0.325×(250 - 20) + 300

= 2456.25 W

The rate of energy transfer per second is given by the following relation;

P = \dfrac{K \times A \times \Delta T}{L}

Thermal conductivity K = 65.2 W/(m·K)

Therefore;

2456.25  = \dfrac{65.2 \times 0.375 \times (250 - T_{in})}{0.02}

T_{in} = 250 - \dfrac{2456.25  \times 0.02}{65.2 \times 0.375} = 247.99 ^{\circ}C

The inside temperature, T_{in} = 247.99 °C  ≈ 248 °C.

3 0
4 years ago
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