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
1 answer:
Answer:
Total bandwidth: 8 kHz
Explanation:
Data provided:
Transmitter frequency: 3.9 MHz
Modulation up to: 4 kHz
Solution:
For the upper side frequencies:
Upper side frequencies = 3.9 × + 4 × 10³
Upper side frequencies = 3.904 MHz
For the lower side frequencies:
Lower side frequencies = 3.9 × - 4 × 10³
Lower side frequencies = 3.896 MHz
Consequently, the total bandwidth is computed as:
Total bandwidth = upper side frequencies - lower side frequencies
Total bandwidth = 8 kHz
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Answer:
Refer to the attached document
1512 ft
Explanation:
Because the acceleration is constant or zero, the acceleration-time graph consists of horizontal segments. The values for t2 and a4 are derived as follows:
Acceleration - Time
0 < t < 6: Velocity change = area beneath the a–t graph
V_6 - 0 = (6 s)(4 ft/s²) = 24 ft/s
6 < t < t2: The velocity rises from 24 to 48 ft/s,
Velocity change = area beneath the a–t graph
48 - 24 = (t2 - 6) * 6
t2 = 10 s
t2 < t < 34: The velocity remains constant, meaning acceleration is zero.
34 < t < 40: Velocity change = area beneath the a–t graph
0 - 42 = 6*a4
a4 = - 8 ft / s²
A negative acceleration shows the area lies below the t-axis, indicating a decrease in speed.
Velocity - Time
Since acceleration remains constant or zero, the v−t graph is made up of linear segments connecting the calculated points.
Position change = area beneath the v−t graph
0 < t < 6: x6 - 0 = 0.5*6*24 = 72 ft
6 < t < 10: x10 - x6 = 0.5*4*(24 + 48) = 144 ft
10 < t < 34: x34 - x10 = 48*24 = 1152 ft
34 < t < 40: x40 - x34 = 0.5*6*48 = 144 ft
Summing these position changes yields the distance from A to B:
d = x40 - 0 = 1512 ft
Answer:
The duration is 17.43 minutes.
Explanation:
Based on the provided information, the initial diameter is 5 m
the velocity is 3 m/s
and the final diameter is 17 m.
To find the solution, we will use the volume change equation expressed as
ΔV = .............1
where ΔV represents the change in volume, rf is the final radius, and ri is the initial radius.
Calculating ΔV yields
ΔV =
ΔV = 2507 m³.
Thus,
Q = velocity × Area
Q = 3 × π ×(0.5)² = 2.356 m³/s.
Next, the change in time can be expressed as
Δt =
Δt =
Δt = 1046 seconds.
Therefore, the total change in time amounts to 17.43 minutes.