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:
9.21954 m/s
54 m/s²
Angle is zero
Explanation:
r = Radius of arm = 1.5 m
= Angular velocity = 6 rad/s
The horizontal component of speed is given by

The vertical component of speed is given by

The resultant of the two components will give us the velocity of hammer with respect to the ground

The velocity of hammer relative to the ground is 9.21954 m/s
Acceleration in the vertical component is zero
Net acceleration is given by

Net acceleration is 54 m/s²
As the acceleration is towards the center the angle is zero.
There is half the force that there was before it was split in half
If friction is acting along the plane upwards
then in this case we will have
For equilibrium of 100 kg box on inclined plane we have

also for other side of hanging mass we have

now we have




In other case we can assume that friction will act along the plane downwards
so now in that case we will have

also we have

now we have





<em>So the range of angle will be 23.45 degree to 37 degree</em>