Answer:
Specific heat at constant pressure is = 1.005 kJ/kg.K
Specific heat at constant volume is = 0.718 kJ/kg.K
Explanation:
given data
temperature T1 = 50°C
temperature T2 = 80°C
solution
we know energy require to heat the air is express as
for constant pressure and volume
Q = m × c × ΔT ........................1
here m is mass of the gas and c is specific heat of the gas and Δ
T is change in temperature of the gas
here both Mass and temperature difference is equal and energy required is dependent on specific heat of air.
and here at constant pressure Specific heat is greater than the specific heat at constant volume,
so the amount of heat required to raise the temperature of one unit mass by one degree at constant pressure is
Specific heat at constant pressure is = 1.005 kJ/kg.K
and
Specific heat at constant volume is = 0.718 kJ/kg.K
Answer:
More force
Explanation:
Pressure and force are related by the equation:
where
p is the pressure
F is the force
A is the area
We can re-arrange the equation as
In this problem, the pressure is kept the same (p' = p) while the area is increased. As we can see from the previous equation, the force applied is directly proportional to the area: therefore, a greater area means also a greater force.
Answer:
Explanation:
m = ρV = 1.03( 1000 kg/m³)(π(2² m²)(3.0 m)) = 12360π kg
m ≈ 38,830 kg
Answer:
The average power developed by a motor is 4905 Watts.
Explanation:
The average power is given by:
Where:
W: is the work
t: is the time = 8.0 s
First, let's find the work:
Where:
F: is the force
d: is the displacement = 10.0 m
The force is in the vertical motion, and since the movement of the mass is at constant speed the force is:
Hence, the average power is:
Therefore, the average power developed by a motor is 4905 Watts.
I hope it helps you!
@AL2006 had answered this before: Well, first of all, wherever you got this question from has done
a really poor job of question-writing. There are a few assorted
blunders in the question, both major and minor ones:
-- 22,500 is the altitude of a geosynchronous orbit in miles, not km.
-- That figure of 22,500 miles is its altitude above the surface,
not its radius from the center of the Earth.
-- The orbital period of a synchronous satellite has to match
the period of the Earth's rotation, and that's NOT 24 hours.
It's about 3 minutes 56 seconds less ... about 86,164 seconds.
Here's my solution to the question, using some of the wreckage
as it's given, and correcting some of it. If you turn in these answers
as homework, they'll be marked wrong, and you'll need to explain
where they came from. If that happens, well, serves ya right for
turning in somebody else's answers for homework.
The satellite is traveling a circle. The circle's radius is 26,200 miles
(not kilometers) from the center of the Earth, so its circumference
is (2 pi) x (26,200 miles) = about 164,619 miles.
Average speed = (distance covered) / (time to cover the distance)
= (164,619 miles) / day
(264,929 km)
= 6,859 miles per hour
(11,039 km)
= 1.91 miles per second
(3.07 km)
