A transistor with a height of 0.4 cm and a diameter of 0.6 cm is mounted on a circuit board. The transistor is cooled by air flowing over it with an average heat transfer coefficient of 30 W/m2°C air = 30°C If the air temperature is 55°C and the transistor case temperature is not to exceed 70°C, determine the amount of power this transistor can dissipate safely. Disregard any heat transfer from the transistor base.

The thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.

The thickness of the aluminum can be determined using from distance of closest approach of the particle.

[tex]K.E = \frac{2KZe^2}{r}[/tex]

where;

[tex]r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (1.6\times 10^{-19})^2}{2.5 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m[/tex]

Since the magnitude of charge of electron and proton is the same, at equal kinetic energy, the thickness will be same. r = 1.5 x 10⁻¹⁴ m.

Charge of alpah particle = 2e

[tex]r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (2 \times 1.6\times 10^{-19})^2}{10 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m[/tex]

Thus, the thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.

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A: Copper has a higher elastic modulus than aluminum.

B: The density of copper is closer to that of aluminum than it is to iron.

C: Bronze is an alloy of copper and zinc.

D: Copper and its alloys form a green tarnish over time, consisting of sulfides and carbonates.

E: Copper is relatively resistant to corrosion by neutral and even mildly basic water, making it useful for freshwater plumbing applications.

scale is used to make a scale drawing. every scale drawing has it's sclae written for proper interpreatation

A Scale is the ratio of a sketched model and the actual measuremnts while a scale drawing is a sketched model that was carefully made such that every measurement in the sketch could be represented to the site or actual mearuement through the scale with very little or no error

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To make a scale drawing, scale is used. For correct interpretation, every scale drawing has its own scale inscribed on it.

A scale drawing is a sketched model that has been meticulously prepared so that every measurement in the sketch may be conveyed to the site or actual measurement through the scale with very little or no mistake.

A weighing apparatus or equipment with two pans of equal weight hung from its ends —usually used in the plural, either pan or tray of a balance scale.

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Answer:

gas

Explanation:

As we speak gas is to high and is making it unconfortable for people to live normaly.

Due to the fact that there is a 14.3inch deflection in the manometer, water is flowing towards the regions that have low pressure.

The fluid that is contained in the manometer is moving towards areas of low pressure. This means they are going to areas with high velocity.

The areas that have low cross sections would flow with a constant rate. This venturi would have low pressure because it has low cross sectional area.

The fluid in the device would then have to move from bottom to top towards the neck of the mercury.

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Since this traffic flow has a jam density of 122 veh/km, the maximum flow is equal to 3,599 veh/hr.

Given the following data:

According to Greenshield Model, maximum flow is given by this formula:

[tex]q_{max}=\frac{V_f \times K_i}{4}[/tex]

Where:

In order to calculate the free flow speed, we would use this formula:

[tex]V_f =2 V\\\\V_f =2\times 59\\\\V_f=118\;km/hr[/tex]

Substituting the parameters into the model, we have:

[tex]q_{max}=\frac{118 \times 122}{4}\\\\q_{max}=\frac{14396}{4}[/tex]

Max flow = 3,599 veh/hr.

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Answer:

You may not operate a vehicle unless all child passengers under the age of 16 years are either wearing a safety belt assembly or are securely fastened into an approved child restraint device.

Explanation:

Answer:

Implements a reordering of the letters of the alphabet.

Explanation: GIVE ME 5 STARS AND a HEART!!!                                      Those contacts are wired across the rotor so that each contact on the left connects to the contact on the right in some scrambled arrangement. Each rotor, therefore, implements a reordering of the letters of the alphabet, which mathematicians call a permutation.

The disadvantage of using a linear model is that; It is sensitive to outliers and prone to multicollinearity.

Linear regression is defined as a regression technique whereby the independent variable has a linear relationship with the dependent variable.

Now, there are some advantages of this model in statistics such as;

However, there are some disadvantages such as;

It is very sensitive to outliers.

It is prone to having multicollinearity.

It assumes linearity between dependent and independent variables.

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U lungul 200 m. If A = 100101 and B = 110110 are applied to X-NOR gate, then the output will be :- plo (1) 101100 (2)011011 (31110111 (4) 010011 As shown in ..

Answer:

With the ongoing pandemic, businesses have shifted their operations to online platforms while people have been forced to resort to work-from-home lifestyles to prevent the further prevent the spread of the virus.

Electromagnetic waves can transfer energy without a material medium.

3 examples of elctromagnetic energy include: Radio wave, X-rays and gamma rays

A wave is a disturbance that transfers energy from one point to another.

Electromagnetic waves are waves that does not require a material medium for the tranfer of energy

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This question is about Circle Geometry. it evaluates connected and broken lines with respect to circles.

This refers to the body of knowledge in mathematics that has to do with the various problems associated with the Circle.

In real-world scenarios, circle geometry is used in technologies involving:

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Answer:

A direct-current (DC) generator is a rotating machine that supplies an electrical output with unidirectional voltage and current.

Explanation:

Hope this helps

Engineers take periodic site measurements to verify the scale conveys true information

measureants is the amount or extent of something ascertained through use of calibrated instrument called measuring instrument. When measurements are done on site it is called site measurements.

Engineers use the site measurements to scale their sketches hence to ensure the scale is always correct periodic measurements should be taken.

On-site measurements value and sketch measurement value gives the scale which  is the ratio, should be maintained always

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The aliased frequency that is due to small sampling frequency is 675 Hz.

The proposed sampling frequency is 2f = 120

The signal frequency that is given in the question is 60Hz

The sampling frequency is 90Hz

a. Using Nyquist frequency,

fn = 0.5*90Hz = 45Hz

The aliased frequency =  |n*fn - fs|

aliased frequency = |n*fn - fs|

We have  n >= 1/Fs = 17ms

Then the aliased frequency is = 17 x 45 - 90

= 675 Hz

In order to avoid aliasing the sampling frequency, this has to be at least two times the highest frequency.

Therefore Fs =

2xF = 120Hz

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A geostationary orbit is in same direction as the spinning of the earth while polar orbit is along the poles

Orbit describes the pathway a satellite moves. the pathway is usually circular.

Polar orbit refers to orbits that traces along the poles that is North pole and South pole. owing to the spherical nature of the earth polar orbits are shorter. This allows for a shorter time to complete a revolution.

Geostationary orbit as against the polar orbit , geostationary orbits traces same path as the Earth's spinning. This path os longer and will require more time to complete. Satellites tracing this orbits are launched further away about 36000 km above the earths surface while satellites on polar orbits is shorter about 200 km - 2000 km

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Explanation:

mathematics education grade 7

Answer:

if you encounter fishing or hunting boats while you are paddling, be courteous.

Give them a wide berth.

Never run overfishing lines. Be aware that anglers may have lines out to the sides of their boat or trolling behind them.

Explanation:

The photon has a wavelength of 588 nm

Wavelength is the distance between two points along the path of the wave.

it is usually represented by λ

Energy of a photon, E = ( h c ) /  λ

where:

h is Planck's constant = 6.626 × 10 -34 Js

c is speed of light = 3.0 * 10^8 m/s

λ = ( h c ) / E

λ = ( ( 6.626 × 10 -34 ) * ( 3.0 * 10^8 ) ) / 3.38 * 10^-19

λ = 5.88 * 10^-7 m

λ = 588 * 10^-9 m = 588 nm

λ = 588 nm = 588 nanometers

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The mean kinetic energy of the N₂ molecules at 25 °C is 3593.519 joules per mole.

Let suppose that the gas sample behaves ideally. The Graham's law establishes a connection between the ideal gas model and the kinetic theory of gases.

In this question we need to use this law to estimate the average kinetic energy (Kₐ), in joules per mole, of a diatomic gas (N₂), which is defined by the following expression:

[tex]K_{a} = \frac{3}{2}\cdot R_{u}\cdot T[/tex]   (1)

Where:

If we know that [tex]R_{u} = 8.314\,\frac{J}{mol\cdot K}[/tex] and [tex]T = 288.15\,K[/tex], then the average kinetic energy is:

[tex]K_{a} = \frac{3}{2}\cdot \left(8.314\,\frac{J}{mol\cdot K} \right) \cdot (288.15\,K)[/tex]

[tex]K_{a} = 3593.519\,\frac{J}{mol}[/tex]

The mean kinetic energy of the N₂ molecules at 25 °C is 3593.519 joules per mole. [tex]\blacksquare[/tex]

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The Given Data were properly outline and the relation for the quantity of heated at equilibrium was used to compute the temperature as 18.84°C

Given Data

At equilibrium, the final temperature of the system T2 will be same for both copper and water

McΔT(copper) = McΔT(water)

0.3*490*(T2 - 150) = 4*4180*(T2-20)

147*(T2 - 150) =16720*(T2-20)

147T2 - 22050 = 16720T2 - 334400

collect like terms

147T2 -16720T2  = - 334400+22050

147T2 -16720T2  = - 334400+22050

-16573T2 = -312350

16573T2 = 312350

Divide both sides by 16573

T2 = 312350/16573

T2 = 18.84°C

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With a high temperature of 1000k and a cold temperature of 308K, The rate of gain for the high reservoir would be 780 kj/s.

A. η = 35%

[tex]\frac{w}{Q1} = \frac{35}{100}[/tex]

W = [tex]1.2*\frac{35}{100}*1000kj/s[/tex]

W = 420 kj/s

Q2 = Q1-W

= 1200-420

= 780 kJ/S

B. W = 420 kj/s

= 420x1000 w

= 4.2x10⁵W

The work done is 4.2x10⁵W

c. 780/308 - 1200/1000

= 2.532 - 1.2

= 1.332kj

The total enthropy gain is 1.332kj

D. Q1 = 1200

T1 = 1000

[tex]\frac{1200}{1000} =\frac{Q2}{308} \\\\Q2 = 369.6 KJ[/tex]

= 1200 - 369.6/1200

= 69.2 percent

change in s is zero for the reversible heat engine.

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Most open-drive compressors use bolt-on service valves, which cannot be replaced is a false statement.

Open Drive Compressor is a term that is often called a coupler drive. This is known to be a fully Serviceable motor that is said to drives the compressor through the use of a Coupling or Belt.

Most open-drive compressors use bolt-on service valves, which cannot be replaced is a false because open drive compressors are said to be  driven by a motor.

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The way in which data mining helps interactive marketing for a business firm is: D. By predicting what each individual accessing a website is most likely interested in seeing.

Data mining can be defined as a strategic process that typically involves extracting and analyzing data that are contained in large databases, especially through the use of specialized software to generate new information.

In this context, a business firm can adopt the use of data mining to enhance and facilitate interactive marketing by creating a process that can effectively and efficiently predict what each individual (customer) accessing their website is most likely interested in seeing.

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Answer:

tama po ba

Explanation:

correct me if im wrong

Answer:

When the AC power is larger than the DC load's needs, a big filter capacitor is typically employed to absorb and store energy, and to give energy to the load when the AC power is lower.

Explanation:

By using the boundary layer equation, the average heat transfer coefficient for the plate is equal to 4.87 W/m²k.

Given the following data:

Surface temperature = 15°C

Bulk temperature = 75°C

Side length of plate = 150 mm to m = 0.15 meter.

Since we have a quiescent room air and a uniform pole surface temperature, the film temperature is given by:

[tex]T_f=\frac{T_{s} + T_{\infty} }{2} \\\\T_f=\frac{15 + 75 }{2} \\\\T_f = 45[/tex]

Film temperature = 45°C to K = 273 + 45 = 318 K.

For the coefficient of thermal expansion, we have:

[tex]\beta =\frac{1}{T_f} \\\\\beta =\frac{1}{318}[/tex]

From table A-9, the properties of air at a pressure of 1 atm and temperature of 45°C are:

Next, we would solve for the Rayleigh number to enable us determine the heat transfer coefficient by using the boundary layer equations:

[tex]R_{aL}=\frac{g\beta \Delta T l^3}{v\alpha } \\\\R_{aL}=\frac{9.8 \;\times \;\frac{1}{318} \;\times \;(75-15) \;\times \;0.15^3 }{1.750 \times 10^{-5}\; \times \;2.416 \times 10^{-5} } \\\\R_{aL}=\frac{9.8\; \times 0.00315 \;\times \;60\; \times\; 0.003375 }{4.228 \times 10^{-10} }\\\\R_{aL}=1.48 \times 10^{7}[/tex]

Also take note, g(Pr) is given by this equation:

[tex]g(P_r)=\frac{0.75P_r}{[0.609 \;+\;1.221\sqrt{P_r}\; +\;1.238P_r]^\frac{1}{4} } \\\\g(P_r)=\frac{0.75(0.7241)}{[0.609 \;+\;1.221\sqrt{0.7241}\; +\;1.238(0.7241)]^\frac{1}{4} }\\\\g(P_r)=\frac{0.543075}{[0.609 \;+\;1.221\sqrt{0.7241}\; +\;1.238(0.7241)]^\frac{1}{4} }\\\\g(P_r)=\frac{0.543075}{[2.5444]^\frac{1}{4} }\\\\g(P_r)=\frac{0.543075}{1.2630 }[/tex]

g(Pr) = 0.430

For GrL, we have:

[tex]G_{rL}=\frac{R_{aL}}{P_r} \\\\G_{rL}=\frac{1.48 \times 10^7}{0.7241} \\\\G_{rL}=1.99 \times 10^7[/tex]

Since the Rayleigh number is less than 10⁹, the flow is laminar and the condition is given by:

[tex]N_{uL}=\frac{h_{L}L}{k} = \frac{4}{3} (\frac{G_{rL}}{4} )^\frac{1}{4} g(P_r)\\\\h_{L}=\frac{0.02699}{0.15} \times [\frac{4}{3} \times (\frac{1.99 \times 10^7}{4} )^\frac{1}{4} ]\times 0.430\\\\h_{L}= 0.1799 \times 62.9705 \times 0.430\\\\h_{L}=4.87\;W/m^2k[/tex]

Based on empirical correlation method, the average heat transfer coefficient for the plate is given by this equation:

[tex]N_{uL}=\frac{h_{L}L}{k} =0.68 + \frac{0.670 R_{aL}^\frac{1}{4}}{[1+(\frac{0.492}{P_r})^\frac{9}{16}]^\frac{4}{19} } \\\\h_{L}=\frac{0.02699}{0.15} \times ( 0.68 + \frac{0.670 (1.48 \times 10^7)^\frac{1}{4}}{[1+(\frac{0.492}{0.7241})^\frac{9}{16}]^\frac{4}{19} })\\\\h_{L}=4.87\;W/m^2k[/tex]

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