Answer:
Specific heat at constant pressure is =  1.005 kJ/kg.K
Specific heat at constant volume is =  0.718 kJ/kg.K
Explanation:
given data 
temperature T1 =  50°C
temperature T2 = 80°C
solution
we know energy require to heat the air is express as
for constant pressure and volume 
Q  = m ×  c × ΔT     ........................1
here m is mass of the gas and c is specific heat of the gas and Δ
T is change in temperature of the gas 
here both Mass and temperature difference is equal and energy required is dependent on specific heat of air.
and here at constant pressure Specific heat  is greater than the specific heat at constant volume, 
so the amount of heat required to raise the temperature of one unit mass by one degree at constant pressure is
Specific heat at constant pressure is =  1.005 kJ/kg.K
and 
Specific heat at constant volume is =  0.718 kJ/kg.K
 
        
             
        
        
        
Answer:bienestar mental y social. La mayoría de los padres conocen los conceptos básicos para mantener
Explanation:bienestar mental y social. La mayoría de los padres conocen los conceptos básicos para mantener a los niños sanos, como alimentarlos con alimentos saludables, asegurarse de que duerman y hagan suficiente ejercicio, y garantizar su seguridad.
 
        
                    
             
        
        
        
Answer:
The crate's coefficient of kinetic friction on the floor is 0.23.
Explanation:
Given that,
Mass of the crate, m = 300 kg
One worker pushes forward on the crate with a force of 390 N while the other pulls in the same direction with a force of 320 N using a rope connected to the crate.
The crate slides with a constant speed. It means that the net force acting on it is 0. Net force acting on it is given by :

So, the crate's coefficient of kinetic friction on the floor is 0.23.
 
        
             
        
        
        
Answer:
<em>1108.464 N of force</em>
Explanation:
diameter of water hose = 70 cm = 0.7 m
radius = 0.7/2 = 0.35 m
volumetric flow rate Q = 420 L/min
1 L = 0.001 m^3
1 min = 60 s
therefore,
Q = 420 L/min = (420 x 0.001)/60 = 0.007 m^3/s
Area A of fire hose = π
 = 3.142 x 
 = 0.38 m^2
<em>From continuity equation, Q = AV</em>
where V1 is the velocity of the water through the pipe, and A1 is the area of the pipe.
Q = A1V1
0.007 = 0.38V1
V1 = 0.007/0.38 = 0.018 m/s.
Nozzle diameter = 0.75 cm = 0.0075 m
radius = 0.00375
Area = π
 = 3.142 x 
 = 4.42 x 
 m^2
velocity of water through the nozzle will be 
V2 = Q/A2 = 0.007 ÷ (4.42 x 
) = 158.37 m/s
From
<em>F = ρQ(v2 - v1)</em> 
Where,
F = force exerted
p = density of water = 1000 kg/m^3
F = 1000 x 0.007 x (158.37 - 0.018) = <em>1108.464 N of force</em>