Respuesta :
calculate the volume, that is easy:
4 x 5 x 2.5 = 50m3
now, as you did not provided if you use nitrogen in
ntp (normal temperature and pressure - 20c, 1atm) weight is 1.165kg/m3
or:
stp (standard temperature and pressure - 0c, 1atm) weight is 1.2506
these are in case of pure nitrogen, as in presented case nitrogen is only 79% we need to change weights to represent that:
ntp - 1.165kg/m3 / 100% x 79% = 0.92035kg/m3
stp - 1.2506kg/m3 / 100% x 79% = 0.987974kg/m3
therefore in case of:
ntp, total weight of n2 will be 50m3 x 0.92035kg/m3 = 46.0175kg
stp, total weight of n2 will be 50m3 x 0.987974kg/m3 = 49.3987kg
4 x 5 x 2.5 = 50m3
now, as you did not provided if you use nitrogen in
ntp (normal temperature and pressure - 20c, 1atm) weight is 1.165kg/m3
or:
stp (standard temperature and pressure - 0c, 1atm) weight is 1.2506
these are in case of pure nitrogen, as in presented case nitrogen is only 79% we need to change weights to represent that:
ntp - 1.165kg/m3 / 100% x 79% = 0.92035kg/m3
stp - 1.2506kg/m3 / 100% x 79% = 0.987974kg/m3
therefore in case of:
ntp, total weight of n2 will be 50m3 x 0.92035kg/m3 = 46.0175kg
stp, total weight of n2 will be 50m3 x 0.987974kg/m3 = 49.3987kg
If we draw a sketch based on the problem, we would come up with two similar right triangles. We can then solve the problem by ratio and proportion. If we let x as the width of the illuminated part of the floor, then we have the equation:
0.5 / (0.5 + x) = x tan 83.5 / (4 + x tan 83.5)
Solving for x
x = 0.477 ft
Therefore, the area of the illuminated part of the floor is:
0.477 ft x 6.5 ft = 3.10 sqr. ft.
0.5 / (0.5 + x) = x tan 83.5 / (4 + x tan 83.5)
Solving for x
x = 0.477 ft
Therefore, the area of the illuminated part of the floor is:
0.477 ft x 6.5 ft = 3.10 sqr. ft.