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Naddik [55]
3 years ago
11

Select the creative imaging fields that require knowledge of programming.

Engineering
1 answer:
denis23 [38]3 years ago
8 0
Video game designer for sure
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An amplifier which needs a high input resistance and a high output resistance is : Select one: a. A voltage amplifier b. None of
True [87]

Answer:

None of these

Explanation:

There are different types of amplifiers, and each has different characteristics.

  • Voltage amplifier needs high input and low output  resistance.
  • Current amplifier needs Low Input and High Output  resistance.
  • Trans-conductance amplifier Low Input and High Output resistance.
  • Trans-Resistance amplifier requires High Input and Low output  resistance.

Therefore, the correct answer is "None of these "

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A. A 3-kg plastic tank that has a volume of 0.2 m^3 is lled with liquid water. Assuming the density of water is 1000 kg=m^3, det
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Answer:

The answer is below

Explanation:

a) The weight of the combined system is the sum of the weight of the water and the weight of the tank

m_{water}=V_{tank}.\rho_{wtaer}\\\\m_{water}=0.2m^3*1000kg/m^3\\\\m_{water}=200 \ kg\\\\m_{total} = m_{water}+m_{tank}\\\\But\ m_{tank}=3kg,therefore:\\\\m_{total} =200kg+3kg\\\\m_{total} =203\ kg\\\\weight_{total}=m_{total}g\\\\weight_{total}=203kg*9.81m/s^2\\\\weight_{total}=1991.43\ N

b) Since the weight of a system can be divided into smaller portions, hence weight is an extensive property.

c) When analyzing the acceleration of gases as they flow through a nozzle, the geometry of the nozzle which is an open system can be chosen as our system.

d) Given that:

\rho_{water}=1000kg/m^3\\\\1kg/m^3=0.062428lb/ft^3\\\\1000kg/m^3=1000kg/m^3*\frac{0.062428lb/ft^3}{kg/m^3}=62.43lb/ft^3\\ \\\rho=SG*\rho_{water}=1.03*62.43=64.272lb/ft^3\\\\P=P_{atm}+\rho g H\\\\P=14.7\ psia+64.272\ lb/ft^3*32.2\ ft/s^2*175\ ft*\frac{1\ ft^2}{12^2\ in^2}*\frac{1\ lbf}{32.2\ lbm.ft/s^2}  \\\\P=92.8\ psia

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3 years ago
The human eye, as well as the light-sensitive chemicals on color photographic film, respond differently to light sources with di
jeka57 [31]

Answer:

a) at T = 5800 k  

  band emission = 0.2261

at T = 2900 k

  band emission = 0.0442

b) daylight (d) = 0.50 μm

    Incandescent ( i ) =  1 μm

Explanation:

To Calculate the band emission fractions we will apply the Wien's displacement Law

The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as

F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )

<em>Values are gotten from the table named: blackbody radiati</em>on functions

<u>a) Calculate the band emission fractions for the visible region</u>

at T = 5800 k  

  band emission = 0.2261

at T = 2900 k

  band emission = 0.0442

attached below is a detailed solution to the problem

<u>b)calculate wavelength corresponding to the maximum spectral intensity</u>

For daylight ( d ) = 2898 μm *k / 5800 k  = 0.50 μm

For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm

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Need help solving math problem using integration
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Ummm did you try to add or subtract and multiply or divide that can get your answer
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A design team is working on creating a new locker organizer. They have
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A. Present a Solution
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