A girl launches a toy rocket from the ground. The engine experiences an average thrust of 5.26 N. The mass of the engine plus fuel before liftoff is 25.0 g, which includes fuel mass of 12.3 g. The engine fires for a total of 1.90 s. A. What is the average exhaust speed of the engine?B. This engine is placed in a rocket body of mass 60.5 g. What is the magnitude of the final velocity of the rocket if it were to be fired from rest in outer space by an astronaut on a spacewalk?

The answer for A is - v = 786.93 m/s

The answer for B is - v = 122.40 m/s

Explanation:

a) To find the average exhaust speed (v) of the engine we can use the following equation:

F = vΔm

Where:

F: is the thrust by the engine = 5.26 N

Δm: is the mass of the fuel = 12.7 g

Δt: is the time of the burning of fuel = 1.90 s

v = F×ΔT/ΔT

b) To calculate the final velocity of the rocket we need to find the acceleration.

The acceleration (a) can be calculated as follows:

a = F/M

In the above equation, m is an average between the mass of the engine plus the rocket case mass and the mass of the engine plus the rocket case minus the fuel mass:

m = (m_{engine} + m_{rocket}) + (m_{engine} + m_{rocket} - m_{fuel})}{2} = {2*m_{engine} + 2*m_{rocket} - m_{fuel}}{2} = 2*25.0 g + 2*63.0 g - 12.7 g}{2} = 81.65 g

Now, the acceleration is:

a = 5.26 N/81.65-t 10^³kg} = 64.42 m*s^²

Finally, the final velocity of the rocket can be calculated using the following kinematic equation:

v= v_{0} + at = 0 + 64.42 m*s^{-2}*1.90 s = 122.40 m/s

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west

there are no contact forces mentioned on the list of choices that you've provided.