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larisa86 [58]
3 years ago
15

Using your knowledge of how an ATM is used, develop a set of use cases that could serve as a basis for understanding the require

ments for an ATM system.
Engineering
1 answer:
kenny6666 [7]3 years ago
4 0

Answer:

Explanation:

bces

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9.For a single-frequency sine wave modulating signal of 3 kHz with a carrier frequency of 36 MHz, what is the spacing between si
nikdorinn [45]

The spacing between sidebands is equal to 6 kHz.

<u>Given the following data:</u>

  • Modulating signal = 3 kHz.
  • Carrier frequency = 36 MHz.

<h3>What is a sideband?</h3>

A sideband can be defined as a band of frequencies that are lower or higher than the carrier frequency due to the modulation process. Thus, it will either be lower than or higher than the carrier frequency.

Generally, the frequency of the modulating signal is equal to the spacing between the sidebands. Therefore, a modulating signal of 3 kHz simply means that the lower sideband is <u>3 kHz</u> higher while the upper sideband is <u>3 kHz</u> lower.

Spacing = 3 kHz + 3 kHz = 6 kHz.

Read more on frequency here: brainly.com/question/3841958

8 0
2 years ago
You wish to design a cantilever beam with a square cross section and length L that can support an end load of F without yielding
koban [17]

Answer:

with a square cross section and length L that can support an end load of F without yielding. You also wish to minimize the amount the beam deflects under load. What is the free variable(s) (other than the material) for this design problem?

a. End load, F.

b. Length, L.

c. Beam thickness, b

d. Deflection, δ

e. Answers b and c.

f. All of the above.

8 0
3 years ago
A stainless steel ball (rho = 8055 kg/m3, cp = 480 J/kg·K) of diameter D = 0.21 m is removed from the oven at a uniform temperat
Nataliya [291]

Answer:

Explanation:

The complete detailed  explanation which answer the question efficiently is shown in the attached files below.

I hope it helps a lot !

5 0
3 years ago
Read 2 more answers
Beryllium (Be) has an HCP unit cell for which the ratio of the lattice parameters c/a is 1.568. If the radius of the Be atom is
sweet-ann [11.9K]

Answer:

Unit cell volume will be

4.866*10^{-2}nm^{3}

8 0
3 years ago
An adiabatic air compressor compresses 10 L/s of air at 120 kPa and 20 degree C to 1000 kPa and 300 degree C.
Oksana_A [137]

Answer:

work=281.4KJ/kg

Power=4Kw

Explanation:

Hi!

To solve follow the steps below!

1. Find the density of the air at the entrance using the equation for ideal gases

density=\frac{P}{RT}

where

P=pressure=120kPa

T=20C=293k

R= 0.287 kJ/(kg*K)= gas constant ideal for air

density=\frac{120}{(0.287)(293)}=1.43kg/m^3

2.find the mass flow by finding the product between the flow rate and the density

m=(density)(flow rate)

flow rate=10L/s=0.01m^3/s

m=(1.43kg/m^3)(0.01m^3/s)=0.0143kg/s

3. Please use the equation the first law of thermodynamics that states that the energy that enters is the same as the one that must come out, we infer the following equation, note = remember that power is the product of work and mass flow

Work

w=Cp(T1-T2)

Where

Cp= specific heat for air=1.005KJ/kgK

w=work

T1=inlet temperature=20C

T2=outlet temperature=300C

w=1.005(300-20)=281.4KJ/kg

Power

W=mw

W=(0.0143)(281.4KJ/kg)=4Kw

5 0
3 years ago
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