A piston–cylinder device contains 0.15 kg of air initially at 2 MPa and 350°C. The air is first expanded isothermally to 500 kPa
1 answer:
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Answer:
Explanation:
The distance between successive wave crests at sea is 20 m. The boat descends from the crest of the wave, rises and reaches the highest position again within 5 s. Calculate the wave propagation speed.
Given that,
The distance between two successive crest is 20m
Wavelength is the distance between two successive crest or trough
Then, it's wavelength is λ = 20m
The time to reached the maximum height is 5seconds, then it will take (5×4) to complete one period
Then,
Period T = 20seconds
From wave equation
v = fλ
Where
v is speed
f is frequency and
λ is wavelength
The frequency is related to the period
f = 1 / T
Then,
v = λ / T
So, v = 20 / 20
v = 1 m/s
The speed of propagation of the wave is 1m/s
To Polish
Jeśli się uwzględni,
Odległość między dwoma kolejnymi grzebieniami wynosi 20 m
Długość fali to odległość między dwoma kolejnymi grzebieniami lub dolinami
Zatem jego długość fali wynosi λ = 20 m
Czas do osiągnięcia maksymalnej wysokości wynosi 5 sekund, a następnie ukończenie jednego okresu zajmie (5 × 4)
Następnie,
Okres T = 20 sekund
Z równania falowego
v = fλ
Gdzie
v to prędkość
f oznacza częstotliwość, a
λ jest długością fali
Częstotliwość jest związana z okresem
f = 1 / T
Następnie,
v = λ / T
Zatem v = 20/20
v = 1 m / s
Prędkość propagacji fali wynosi 1m/s
The question is missing some parts. Here is the complete question.
An ideal gas is contained in a vessel at 300K. The temperature of the gas is then increased to 900K.
(i) By what factor does the average kinetic energy of the molecules change, (a) a factor of 9, (b) a factor of 3, (c) a factor of , (d) a factor of 1, or (e) a factor of
?
Using the same choices in part (i), by what factor does each of the following change: (ii) the rms molecular speed of the molecules, (iii) the average momentum change that one molecule undergoes in a colision with one particular wall, (iv) the rate of collisions of molecules with walls, and (v) the pressure of the gas.
Answer: (i) (b) a factor of 3;
(ii) (c) a factor of ;
(iii) (c) a factor of ;
(iv) (c) a factor of ;
(v) (e) a factor of 3;
Explanation: (i) Kinetic energy for ideal gas is calculated as:
where
n is mols
R is constant of gas
T is temperature in Kelvin
As you can see, kinetic energy and temperature are directly proportional: when tem perature increases, so does energy.
So, as temperature of an ideal gas increased 3 times, kinetic energy will increase 3 times.
For temperature and energy, the factor of change is 3.
(ii) Rms is root mean square velocity and is defined as
Calculating velocity for each temperature:
For 300K:
For 900K:
Comparing both veolcities:
For rms, factor of change is
(iii) Average momentum change of molecule depends upon velocity:
q = m.v
Since velocity has a factor of and velocity and momentum are proportional, average momentum change increase by a factor of
(iv) Collisions increase with increase in velocity, which increases with increase of temperature. So, rate of collisions also increase by a factor of .
(v) According to the Pressure-Temperature Law, also known as Gay-Lussac's Law, when the volume of an ideal gas is kept constant, pressure and temperature are directly proportional. So, when temperature increases by a factor of 3, Pressure also increases by a factor of 3.
Answer:
The lighter frog goes higher than the heavier frog.
The lighter frog is moving faster than the heavier frog
Explanation:
If both frogs have the same kinetic energy when they leave the ground, the following equality applies:
Now, if the only force acting on the frogs is gravity, when they reach to the maximum height, we can apply the following kinematic equation:
When h= hmax, the object comes momentarily to an stop, so vf =0
Solving for hmax:
As the lighter frog, in order to have the same kinetic energy than the heavier one, has a greater initial velocity, it will go higher than the other.
As a consequence of both having the same kinetic energy, the lighter frog will be moving faster than the heavier frog.