Answer:
induced current
Explanation:
intentionally manipulated.
D. March because it is just below the 1 million marker on the graph and it is the only one that low.
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 / Explanation:
The result of the Training and Exercise Planning Workshop (TEPW) is to set the foundation for the strategy and pattern for a proposed exercise program. The TEPW purpose is to engage elected and selected officials in identifying exercise program priorities and planning a schedule of training and exercise events to meet those priorities.
An essential factor for the exercise management process is to create a collaborative environment where a whole community stakeholders can engage in a forum to discuss and coordinate training and exercise activities across local organizations to maximize the use of available resources and prevent duplication of effort.
