A space vehicle is traveling at 3760 km/h relative to Earth when the exhausted rocket motor is disengaged and sent backward. The relative speed between the motor and the command module is then 90 km/h. The mass of the motor is four times the mass of the module. What is the speed of the command module relative to Earth just after the separation?

The command module is traveling at 3832 km/h after the separation.

Explanation:

Let's call the mass of the module M, and the mass of the motor is 4M. The total mass of the vehicle, then, is M + 4M = 5M.

Now, momentum = force x distance.

Thus, momentum of the vehicle before separation 3760 x 5M = 18800M kg-km/h.

Let's call the velocity of the motor v. So the velocity of the module must be v + 93.

Now, after the separation, the momentum of the motor will be 4Mv, and the momentum of the module is M(v + 90) = Mv + 90M. The total momentum is 4MV + Mv + 90M = 5Mv + 90M.

From conservation of momentum that says initial momentum equals final momentum, we have;

18800M = 5Mv + 90M

Divide both sides by M to get;

18800 = 5v + 90

18800 - 90 = 5v

18710 = 5v

v = 18710/5

v = 3742 km/h

But v is the velocity of the motor, so we have to add 90 to get the velocity of the module, thus, velocity of module = 3742 + 90 = 3832

So the command module is traveling at 3832 km/h after the separation.

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .