A133.7 ft3 volume of liquid hydrogen rocket fuel has a mass of 268 kg. calculate: the weight of the fuel at standard (earth) sea level conditions in n and ibr, the density of the fuel in kg/m3, and the specific volume in ft3/lbm (ibm=pound-mass, which is not the same as a a lbf or a slug)

given,

volume = 133.7 ft³

1 ft³ = 0.0283 m³

133.7 ft³ =  133.7 × 0.0283 = 3.784 m³

mass = 268 kg

1 kg = 2.204 lb

268 kg = 268 × 2.204 = 590.672 lb

weight of the fuel = 268 × 9.8 = 2626.4 N

1 N = 0.2248 lbf

2626.4 N =  2626.4 × 0.2248 = 590.414 lbf

ρ = 70.82 kg/m³

specific volume =

                           = 0.226 ft³/lbm

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

i think it's 35 meters