How does load transfer of space needle
1 answer:
You might be interested in
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
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:
Equilibrium Temperature is 382.71 K
Total entropy is 0.228 kJ/K
Solution:
As per the question:
Mass of the Aluminium block, M = 28 kg
Initial temperature of aluminium, = 273 + 140 = 413 K
Mass of Iron block, m = 36 kg
Temperature for iron block, = 273 + 60 = 333 K
At 400 k
Specific heat of Aluminium,
At room temperature
Specific heat of iron,
Now,
To calculate the final equilibrium temperature:
Amount of heat loss by Aluminium = Amount of heat gain by Iron
Thus
= 273 + 109.71 = 382.71 K
where
= Equilibrium temperature
Now,
To calculate the changer in entropy:
Now,
For Aluminium:
For Iron:
Thus