Electronic components are often mounted with good heat conduction paths to a finned aluminum base plate, which is exposed to a stream of cooling air from a fan. The sum of the mass times specific heat products for a base plate and components is 5000 J/K, and the effective heat transfer coefficient times surface area product is 10 W/K. The initial temperature of the plate and the cooling air temperature are

Answer:

Air speed in the wind-tunnel [tex]v_{2}[/tex] = 27.5 m/s

Explanation:

Given data :

Manometer reading ; p1 - p2 = 45 mm of water

Pressure at section ( I ) p1 = 100 kPa ( abs )

temperature ( T1 ) = 25°C

Pw ( density of water ) = 999 kg/m3

g = 9.81 m/s^2

next we apply Bernoulli equation at section 1 and section 2

p1 - p2 = [tex]\frac{PairV^{2} _{2} }{2}[/tex]     ----------  ( 1 )

considering  ideal gas equation

Pair ( density of air ) = [tex]\frac{P}{RT}[/tex] ------------------- ( 2 )

R ( constant ) = 287 NM/kg.k

T = 25 + 273.15 = 298.15 k

P1 = 100 kN/m^2 = 100 * 10^3 or N/m^2

substitute values into equation ( 2 )

= 100 * 10^3 / (287 * 298.15)

= 1.17 kg/m^3

Also note ; p1 - p2 = PwgΔh  ------- ( 3 )

finally calculate the Air speed in the wind-tunnel test section by equating equation ( 1 ) and ( 3 )

[tex]\frac{PairV^{2} _{2} }{2}[/tex]   =  PwgΔh  

[tex]V^{2} _{2}[/tex] = [tex]\frac{2*999* 9.81* 0.045}{1.17}[/tex]  =  753.86

[tex]v_{2}[/tex] = 27.5 m/s

Answer:

Sound waves are compression waves that cause energy transfer in air molecules

Sound waves have to have a medium to travel through

Loud sounds have high amplitude and vibrate with more energy than soft sounds

Explanation:

Sound waves is a form of energy composed of compression and rare factions. Sound waves are compression waves that cause energy transfer in air molecules.

Sound is an example of a mechanical wave hence it requires a material medium for propagation.

The amplitude of a sound wave determines its loudness or volume. A larger amplitude implies that we will have a louder sound, and a smaller amplitude means that we will have a softer sound.

Answer:

Without the indispensable advice of the business community, TVET Institutions will be unable to cover the gap in career knowledge required by the business community.  To develop workers who possess the knowledge and skills required by today's business entities, there is always the continual need for the educational institutions (gown) to regularly meet the business community (town).  This meeting provides the necessary ground for the institutions to develop programs that groom the workforce with skills that are needed in the current workplace.  Educational institutions that do not seek this important advice from the business community risk developing workers with outdated skills.

Explanation:

TVET Institutions mean Technical and Vocational Education and Training Institutions.  They play an important role in equipping young people to enter the world of work.  They also continue to develop programs that will improve the employability of workers throughout their careers.  They regularly respond to the changing labor market needs, adopt new training strategies and technologies, and expand the outreach of their training to current workers while grooming the young people for work.

Answer: Rc = 400 Ω and Rb = 57.2 kΩ

Explanation:

Given that;

VCE = 5V

VCC = 15 V

iC = 25 mA

β = 100

VD₀ = 0.7 V

taking a look at the image; at loop 1

-VCC + (i × Rc) + VCE = 0

we substitute

-15 + ( 25 × Rc) + 5 = 0

25Rc = 10

Rc = 10 / 25

Rc = 0.4 k

Rc = 0.4 × 1000

Rc = 400 Ω

iC = βib

25mA = 100(ib)

ib = 25 mA / 100

ib = 0.25 mA

ib = 0.25 × 1000

ib = 250 μAmp

Now at Loop 2

-Vcc + (ib×Rb) + VD₀ = 0

-15 (250 × Rb) + 0.7 = 0

250Rb = 15 - 0.7

250Rb = 14.3

Rb = 14.3 / 250

Rb = 0.0572 μ

Rb = 0.0572 × 1000

Rb = 57.2 kΩ

Therefore Rc = 400 Ω and Rb = 57.2 kΩ

Answer:

false

Explanation:

Answer:

The ability to read electrical schematics is a really useful skill to have. To start developing your schematic reading abilities, it's important to memorize the most common schematic symbols. ... You should also be able to get a rough idea of how the circuit works, just by looking at the schematic.

Explanation:

Answer:

B a cast part

Explanation:

Extrusion is defined as the process of shaping material, such as aluminum, by forcing it to flow through a shaped opening in a die. Extruded material emerges as an elongated piece with the same profile as the die opening.

Forcing a solid piece of heated aluminum through a die forms: a cast part. Hence, option B is correct.

A liquid element is more often filled with concrete that has a hollow chamber in the correct form during the casting manufacturing process, and the item is then let to harden.

A casting, which is the term for the solidified component, is ejected or broken out of the mould to complete the procedure.

Thus, option B is correct.

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Answer:a

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Answer:

When are resistors in series? Resistors are in series whenever the flow of charge, called the current, must flow through devices sequentially. For example, if current flows through a person holding a screwdriver and into the Earth, then  

R

1

 in Figure 1(a) could be the resistance of the screwdriver’s shaft,  

R

2

 the resistance of its handle,  

R

3

 the person’s body resistance, and  

R

4

 the resistance of her shoes.

Figure 2 shows resistors in series connected to a voltage source. It seems reasonable that the total resistance is the sum of the individual resistances, considering that the current has to pass through each resistor in sequence. (This fact would be an advantage to a person wishing to avoid an electrical shock, who could reduce the current by wearing high-resistance rubber-soled shoes. It could be a disadvantage if one of the resistances were a faulty high-resistance cord to an appliance that would reduce the operating current.)

Two electrical circuits are compared. The first one has three resistors, R sub one, R sub two, and R sub three, connected in series with a voltage source V to form a closed circuit. The first circuit is equivalent to the second circuit, which has a single resistor R sub s connected to a voltage source V. Both circuits carry a current I, which starts from the positive end of the voltage source and moves in a clockwise direction around the circuit.

Figure 2. Three resistors connected in series to a battery (left) and the equivalent single or series resistance (right).

To verify that resistances in series do indeed add, let us consider the loss of electrical power, called a voltage drop, in each resistor in Figure 2.

According to Ohm’s law, the voltage drop,  

V

, across a resistor when a current flows through it is calculated using the equation  

V

=

I

R

, where  

I

 equals the current in amps (A) and  

R

 is the resistance in ohms  

(

Ω

)

. Another way to think of this is that  

V

 is the voltage necessary to make a current  

I

 flow through a resistance  

R

.

So the voltage drop across  

R

1

 is  

V

1

=

I

R

1

, that across  

R

2

 is  

V

2

=

I

R

2

, and that across  

R

3

 is  

V

3

=

I

R

3

. The sum of these voltages equals the voltage output of the source; that is,

V

=

V

1

+

V

2

+

V

3

.

This equation is based on the conservation of energy and conservation of charge. Electrical potential energy can be described by the equation  

P

E

=

q

V

, where  

q

 is the electric charge and  

V

 is the voltage. Thus the energy supplied by the source is  

q

V

, while that dissipated by the resistors is

q

V

1

+

q

V

2

+

q

V

3

.

Explanation: