Answer:
 W
 W
Explanation:
 = Temperature of the room = 22.0 °C = 22 + 273 = 295 K
 = Temperature of the room = 22.0 °C = 22 + 273 = 295 K 
 = Temperature of the skin = 33.0 °C = 33 + 273 = 306 K
 = Temperature of the skin = 33.0 °C = 33 + 273 = 306 K
 = Surface area = 1.50 m²
 = Surface area = 1.50 m²
  = emissivity = 0.97
 = emissivity = 0.97 
 = Stefan's constant = 5.67 x 10⁻⁸ Wm⁻² K⁻⁴
 = Stefan's constant = 5.67 x 10⁻⁸ Wm⁻² K⁻⁴
Rate of heat transfer is given as 


 W
 W
 
        
             
        
        
        
Responder:
Explicación:
Para que podamos calcular el tiempo que le tomará al esquiador permanecer en el aire, encontraremos el tiempo de vuelo como se muestra;
T = 2Usin theta / 2
theta = 90 grados
U = 25 m / s
T = 25sin90 / 2 (9,8)
T = 25 / 19,62
T = 1,27 segundos
Por lo tanto, los cielos usarán 1.27 segundos en el aire.
La distancia horizontal es el rango;
Rango R = U√2H / g
R = 25√2 (80) /9,8
R = 25√160 / 9,8
R = 25 * √16,326
R = 25 * 4.04
R = 101,02 m
Por tanto, la distancia horizontal recorrida por el esquiador es 101,02 m
 
        
             
        
        
        
Sodium (Na) is very reactive because it does not have a full valence shell.
        
             
        
        
        
Of the cliff?
Projectile motion
In the problem we are asked to find a height of certain cliff when a motorcycle stunt driver zoom out horizontally at the end the cliff at an initial velocity. So we will use one of the kinematics equation for projectile motion,
y
=
v
o
y
t
+
1
2
g
t
where 
v
o
y
 is just equal to zero since we can assume that the driver zooms out horizontally, 
g
=
9.8
m
/
s
2
 and 
t
 is time after
        
             
        
        
        
Answer:
55.96kJ
Explanation:
Energy = mass of diethyl ether × enthalpy of vaporization of diethyl ether
Volume (v) = 200mL, density (d) = 0.7138g/mL
Mass = d × v = 0.7138 × 200 = 142.76g
Enthalpy of vaporization of diethyl ether = 29kJ/mol
MW of diethyl ether (C2H5)2O = 74g/mol
Enthalpy in kJ/g = 29kJ/mol ÷ 74g/mol = 0.392kJ/g
Energy = 142.76g × 0.392kJ/g = 55.96kJ