Answer:=14,160 kJ
Explanation: Let m1 and m2 be the initial and final amounts of mass within the tank, respectively. The steam properties are listed in the table below
Specific Internal SpecificTemp Pressure Volume Energy Enthalpy Quality Phase
C MPa m^3/kg kJ/kg kJ/kg
1 260 4.689 0.02993 2158 2298 0.7 Liquid Vapor Mixture
2 260 4.689 0.0422 2599 2797 1 Saturated Vapor
The mass initially contained in the tank is m1 = V/v1
m1 =0.85 m^3 /0.02993 m^3 /kg
= 28.4 kg
The mass finally contained in the tank is
m2 =V2/v
= 0.85 m^3 /0.0422 m^3 /kg
= 20.14 kg
The heat transfer is then
Qcv = m2u2 − m1u1 − he(m2 − m1)
Qcv = (20.14)(2599) − (28.4)(2158) − (2797)(20.14 − 28.4) = 14,160 kJ
Answer:
54mph
Explanation:
Speed is defined as the change in distance of a body with respect to time.
Let x be the speed of Eric on his way to the football game, if he used 11hours to the football game, his distance will be the product of his speed and time taken i.e
Distance = speed × time
Distance= x × 11
Distance = 11x... (1)
During his return journey, his speed increases by 12mph with a return drive in only 9hours i.e speed = x+12 and time is 9hours
Since distance = speed × time (return Journey)
Distance = (x+12)×9
Distance = 9x+84...(2)
Note that the distance to and fro the journey is the same as such we will equate equations 1 and 2 to get his initial speed "x" to have;
11x = 9x+84
11x-9x = 84
2x = 84
x = 42mph
This means that his initial speed on his way to the football game is 42mph. Since his speed increases by 12mph during his return drive, his return speed will be 42mph + 12mph which gives 54mph.
Let <em>v</em> denote the initial speed of the ball. The ball's position at time <em>t</em> is given by the vector
where <em>g</em> is the acceleration due to gravity with magnitude 9.80 m/s^2.
The ball reaches the goal 13 m away at time <em>t</em> such that
at which point it attains a height of 2.44 m, so that
<span>Answer:
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