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 × + 4 × 10³
upper side frequencies = 3.904 MHz
and
now we get lower side frequencies that is
lower side frequencies = 3.9 × - 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
Answer:
18 kJ
Explanation:
Given:
Initial volume of air = 0.05 m³
Initial pressure = 60 kPa
Final volume = 0.2 m³
Final pressure = 180 kPa
Now,
the Work done by air will be calculated as:
Work Done = Average pressure × Change in volume
thus,
Average pressure = = 120 kPa
and,
Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³
Therefore,
the work done = 120 × 0.15 = 18 kJ
Explanation:
See attached file
Answer:
capacity = 0.555 mAh
capacity = 3600 mAh
Explanation:
given data
battery = 1800 mAh
OCV = 3.9 V
solution
we get here capacity when it is in series
so here Q = 2C
capacity = 2 × ampere × second ...............1
put here value and we get
and 1 Ah = 3600 C
capacity =
capacity = 0.555 mAh
and
when it is in parallel than capacity will be
capacity = Q1 +Q2 ...............2
capacity = 1800 + 1800
capacity = 3600 mAh