Answer:
4.8L ( i.e 4.8 x 10^-3 m3)
Explanation:
Step 1:
Data obtained from the question.
Initial volume (V1) = 4.2L
Initial temperature (T1) = 0°C
Final temperature (T2) = 37°C
Final volume (V2) =?
Step 2:
Conversion of celsius temperature to Kelvin temperature. This is illustrated below
K = °C + 273
T1 = 0°C = 0°C + 273 = 273K
T2 = 37°C = 37°C + 273 = 310K
Step 3:
Determination of the final volume.
Since the pressure is constant,
Charles' Law equation will be applied as shown below:
V1 /T1 = V2/T2
4.2/273 = V2 /310
Cross multiply to express in linear form
273 x V2 = 4.2 x 310
Divide both side by 273
V2 = (4.2 x 310)/273
V2 = 4.8L ( i.e 4.8 x 10^-3 m3)
Therefore, the volume of the air in the lungs at that point is 4.8L ( i.e 4.8 x 10^-3 m3)
Answer : The partial pressure of 
is, 67.009 atm
Solution : Given,
Partial pressure of 
at equilibrium = 30.6 atm
Partial pressure of 
at equilibrium = 13.9 atm
Equilibrium constant = 
The given balanced equilibrium reaction is,
The expression of 
will be,
Now put all the values of partial pressure, we get
Therefore, the partial pressure of is, 67.009 atm
Answer:
E means energy
M= Mass
C=speed of light squared (the exponent means squared)
Sounds like the shingle/ball is thrown from the roof horizontally, so that the distance it travels <em>x</em> after time <em>t</em> horizontally is
<em>x</em> = (7.2 m/s) <em>t</em>
The object's height <em>y</em> at time <em>t</em> is
<em>y</em> = 9.4 m - 1/2 <em>gt</em>²
where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity, and its vertical velocity is
<em>v</em> = -<em>gt</em>
(a) The object hits the ground when <em>y</em> = 0:
0 = 9.4 m - 1/2 <em>gt</em>²
<em>t</em>² = 2 * (9.4 m) / (9.80 m/s²)
<em>t</em> ≈ 1.92 s
at which time the object's vertical velocity is
<em>v</em> = -<em>g</em> (1.92 s) = -18.8 m/s ≈ -19 m/s
(b) See part (a); it takes the object about 1.9 s to reach the ground.
(c) The object travels a horizontal distance of
<em>x</em> = (7.2 m/s) * (1.92 s) ≈ 13.8 m ≈ 14 m
