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katrin2010 [14]
3 years ago
12

An AM radio transmitter operating at 3.9 MHz is modulated by frequencies up to 4 kHz. What are the maximum upper and lower side

frequencies. What is the total bandwidth of the AM signal
Engineering
1 answer:
Aliun [14]3 years ago
6 0

Answer:

total width bandwidth = 8kHz

Explanation:

given data

transmitter operating = 3.9 MHz

frequencies up to =  4 kHz

solution

we get here upper side frequencies that is

upper side frequencies = 3.9 × 10^{6} + 4  × 10³

upper side frequencies = 3.904 MHz

and

now we get lower side frequencies that is

lower side frequencies = 3.9 × 10^{6} - 4  × 10³

lower side frequencies =  3.896 MHz

and now we get total width bandwidth

total width bandwidth = upper side frequencies  - lower side frequencies

total width bandwidth = 8kHz

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Every two years or at recommendation by manufacturer.
9966 [12]

Answer:

Manufacturer’s Recommendations means the instructions, procedures, and recommendations which are issued by the manufacturer of any equipment used at the Facility relating to the operation, maintenance, or repair of such equipment, and any revisions or updates thereto from time to time issued by the manufacturer.

Manufacturer’s Recommendations means the instructions, procedures and recommendations which are issued by any manufacturer of the Equipment relating to the operation, maintenance and repair of the Equipment and any revisions to such instructions, procedures and recommendations agreed to by any manufacturer of the Equipment and which are valid at the time such operation, repair and maintenance is being carried out.

Manufacturer’s Recommendations means the written instructions, procedures and recommendations which are issued by the original equipment manufacturer of any plant or equipment used at the Utility Plant relating to the operation, maintenance and repair of such plant or equipment and any revisions thereto issued by the manufacturer, which are valid and applicable at the time such operation, maintenance or repair is undertaken. Notwithstanding the above, Manufacturer’s Recommendations shall not include any instructions, procedures or recommendations of a manufacturer of any plant or equipment that the Owner and the Operator have agreed in writing to exclude from this definition or have agreed in writing should not be followed.

Explanation:

4 0
2 years ago
Determine the hydraulic radius for the following rectangular open channel width =23m water depth =3m
Romashka-Z-Leto [24]

Answer:

2.379m

Explanation:

The width = 23m

The depth = 3m

The radius is denoted as R

The wetted area is = A

The perimeter perimeter = P

Hydraulic radius

R = A/P

The area of a rectangular channel

= Width multiplied by Depth

A = 23x3

A = 69m²

Perimeter = (2x3)+23

P = 6+23

P= 29

Hydraulic radius R = 69/29

= 2.379m

This answers the question

Thank you!

8 0
3 years ago
Assume the work done compressing the He gas is -63 kJ and the internal energy change of the gas is 79 kJ. What is the heat loss
klemol [59]

Answer:

Heat gain of 142 kJ

Explanation:

We can see that job done by compressing the He gas is negative, it means that the sign convention we are going to use is negative for all the work done by the gas and positive for all the job done to the gas. With that being said, the first law of thermodynamics equation will help us to solve this problem.

ΔU = Q + W ⇒ Q = ΔU -W

Q = 79 - (-63) = 142 kJ

Therefore, the gas gained heat by an amount of 142 kJ.

3 0
3 years ago
A sheet of steel 1.5 mm thick has nitrogen (N2) atmospheres on both sides at 1200°C and is permitted to achieve steady-state dif
Gelneren [198K]

Answer:

do the wam wam

Explanation:

3 0
3 years ago
I have a molten Ni-Cu alloy with 60 wt%Ni. (ii) I cool it down to a temperature where I have both solid and liquid phases. At th
SIZIF [17.4K]

Answer:

The percentage of the remaining alloy would become solid is 20%

Explanation:

Melting point of Cu = 1085°C

Melting point of Ni = 1455°C

At 1200°C, there is a 30% liquid and 70% solid, the weight percentage of Ni in alloy is the same that percentage of solid, then, that weight percentage is 70%.

The Ni-Cu alloy with 60% Ni and 40% Cu, and if we have the temperature of alloy > temperature of Ni > temperature of Cu, we have the follow:

60% Ni (liquid) and 40% Cu (liquid) at temperature of alloy

At solid phase with a temperature of alloy and 50% solid Cu and 50% liquid Ni, we have the follow:

40% Cu + 10% Ni in liquid phase and 50% of Ni is in solid phase.

The percentage of remaining alloy in solid is equal to

Solid = (10/50) * 100 = 20%

4 0
3 years ago
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