Answer:
1) Final Temperature of the gas in A will be GREATER than the temperature in B
2) Diagram of both processes on a single PV has been uploaded below
3) The Initial pressures in containers A and B is 3039.87 J/liters
4) the final volume of container B is 923.36 cm³
Explanation:
Given that;
Temperature = 20°C = 293 K
mass of piston = 10 kg
Area = 100cm³
Volume V = 800 cm³ = 0.8 L
ideal gas constant R = 8.3 J/K·mol
1)
Final Temperature of the gas in A will b GREATER than the temperature in B
2)
Diagram of both processes on a single PV has been uploaded below,
3)
Initial pressures in containers A and B
PV = nRT
P = RT/V
we substitute
P = (8.3 × 293) / 0.8
P = 2431.9 / 0.8
P = 3039.87 J/liters
Therefore, The Initial pressures in containers A and B is 3039.87 J/liters
4)
Given that;
power = 25 W
time t = 15s
the final volume of container B = ?
we know that;
work done = power × time
work done = 25 × 15 = 375
Also work done = P( V₂ - V₁ )
so we substitute
375 = 3039.87 ( V₂ - 0.8 )
( V₂ - 0.8 ) = 375 / 3039.87
V₂ - 0.8 = 0.12336
V₂ = 0.12336 + 0.8
V₂ = 0.92336 Litres
V₂ = 923.36 cm³
Therefore, the final volume of container B is 923.36 cm³
(a) 0.249 (24.9 %)
The maximum efficiency of a heat engine is given by
where
Tc is the low-temperature reservoir
Th is the high-temperature reservoir
For the engine in this problem,
Therefore the maximum efficiency is
(b-c) 0.221 (22.1 %)
The second steam engine operates using the exhaust of the first. So we have:
is the high-temperature reservoir
is the low-temperature reservoir
If we apply again the formula of the efficiency
The maximum efficiency of the second engine is
Answer:
The ball is in the air for 3.5 seconds
The initial horizontal component of velocity is 28.6 m/s
The vertical component of the final velocity is 34.3 m/s downward
The final velocity is 44.7 m/s in the direction 50.2° below the horizontal
Explanation:
A ball is thrown horizontally
That means the vertical component of the initial velocity
The initial velocity is the horizontal component
The ball is thrown from the top of a 60 m
That means the vertical displacement component y = 60 m
→ y = t +
gt²
where g is the acceleration of gravity and t is the time
y = -60 m , g = -9.8 m/s² ,
Substitute these values in the rule
→ -60 = 0 + (-9.8)t²
→ -60 = -4.9t²
Divide both sides by -4.9
→ 12.2449 = t²
Take √ for both sides
∴ t = 3.5 seconds
* <em>The ball is in the air for 3.5 seconds </em>
The initial velocity is the horizontal component
The ball lands 100 meter from the base of the building
That means the horizontal displacement x = 100 m
→ x = t
→ t = 3.5 s , x = 100 m
Substitute these values in the rule
→ 100 = (3.5)
Divide both sides by 3.5
→ = 28.57 m/s
<em>The initial horizontal component of velocity is 28.6 m/s</em>
The vertical component of the final velocity is
→ =
+ gt
→ = 0 , g = -9.8 m/s² , t = 3.5 s
Substitute these values in the rule
→ = 0 + (-9.8)(3.5)
→ = -34.3 m/s
<em>The vertical component of the final velocity is 34.3 m/s downward</em>
The final velocity v is the resultant vector of and
→ Its magnetude is
→ Its direction
→ = 28.6 ,
= -34.3
Substitute this values in the rules above
→
→ Its direction
The negative sign means the direction is below the horizontal
<em>The final velocity is 44.7 m/s in the direction 50.2° below the horizontal</em>
Explanation:
It is given that,
Focal length of the concave mirror, f = -13.5 cm
Image distance, v = -37.5 cm (in front of mirror)
Let u is the object distance. It can be calculated using the mirror's formula as :
u = -21.09 cm
The magnification of the mirror is given by :
m = -1.77
So, the magnification produced by the mirror is (-1.77). Hence, this is the required solution.