Arocket ship flies past the earth at 91.0 % of the speed of light. inside, astronaut who is undergoing measured while he is lying down parallel to the direction in which the ship is moving. (a) if his height is measured to be 2.00 m by his doctor inside the ship, what height would a person watching this from the earth measure? an physical examination is having his height a (b) if the earth-based person had measured 2.00 m, what would the doctor in the spaceship have measured for the astronaut's height? is this a reasonable height? (c) suppose the astronaut in part (a) gets up after the examination and stands with his body perpendicular to the direction of motion. what would the doctor in with rest mass m= 1.67* \ reeds in opposite directi eston that also producn mass of the h0 is all at rest after th fractiqn

That would be the corect answer hope this

The answer is trial 3.

Answer: a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same sign as the  -4.0  µc charge.======electricstatic force between two point charges  is given by    where k is coulomb's constant,  8.99×10⁹  n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ =  6.0 µc, q₂ = -4.0  µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1  µc = 10⁻⁶ c)    the force from q₂    the force from q₃    the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃    isolating:     a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same as the  -4.0  µc charge