Answer:
![a=4.19 m/s^(2)](https://img.qammunity.org/2022/formulas/physics/college/pdh096r0ajrsv4wmrkx7vjxe9fqjup6r16.png)
Step-by-step explanation:
In order to solve this problem, we must first draw a free body diagram of the situation. (See attached picture)
from this diagram, we can do the following sum of forces:
![\sum F=ma](https://img.qammunity.org/2022/formulas/physics/college/qdnjw0ns6vroki4mae92i4tnjk253z8c68.png)
where m is the mass of the rocket and a is its acceleration.
![F_(e)-mg=ma](https://img.qammunity.org/2022/formulas/physics/college/bbv3qz46tyu5jr5v63kznwooh6h0rg79vl.png)
where
is the force of exhaust. We can solve this for the acceleration, so we get:
![a=(F_(e)-mg)/(m)](https://img.qammunity.org/2022/formulas/physics/college/khvoztrn65n519rkwbaoj0sqpkrho4v79z.png)
we can find the force of exhaust by using the momentum formula:
Ft=mv
so we can solve this for the force:
![F_(e)=(mv)/(t)](https://img.qammunity.org/2022/formulas/physics/college/f8djk0fizm0utpcd1eax5ytxxjnpnpdtyl.png)
the problem already tells us what m/t is equal to, so we can directly substitute:
![F_(e)=(1.75x10^(4)kg/s)(2.40x10^3m/s)](https://img.qammunity.org/2022/formulas/physics/college/f3cwybxsrt0x50g2l9v7rxycykcxjjtnsd.png)
which yields:
![F_(e)=42x10^6 N](https://img.qammunity.org/2022/formulas/physics/college/3inwlzfk9sagcojd41rhx0p1ytxzfcgalt.png)
So we can now substitute:
![a=(42x10^6N-(3x10^6kg)(9.81m/s^(2)))/(3x10^6kg)](https://img.qammunity.org/2022/formulas/physics/college/i0k22u57mkcjkfrt5ah8vusnd69mtndcr8.png)
so:
![a=4.19 m/s^(2)](https://img.qammunity.org/2022/formulas/physics/college/pdh096r0ajrsv4wmrkx7vjxe9fqjup6r16.png)