The value of the final speed depends on the mass of the ore.
Let's call m the mass of the ore. We can solve the exercise by requiring the conservation of momentum, which must be the same before and after the ore is loaded.
Initially, there is only the cart, so the momentum is

After the ore is loaded, the new mass will be (1200 kg+m), and the new speed is

. The momentum p is conserved, so it is still 12960 kg m/s. Therefore, we have

and so the final speed is
Answer:
Explanation:
Given
Launch angle =u
Initial Speed is 
Horizontal acceleration is 
At maximum height velocity is zero therefore



Total time of flight 
During this time horizontal range is


For maximum range 

![\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2}{g}\left [ \cos 2u-\frac{a}{g}\sin 2u\right ]=0](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmathrm%7Bd%7D%20R%7D%7B%5Cmathrm%7Bd%7D%20u%7D%3D%5Cfrac%7B2v_0%5E2%7D%7Bg%7D%5Cleft%20%5B%20%5Ccos%202u-%5Cfrac%7Ba%7D%7Bg%7D%5Csin%202u%5Cright%20%5D%3D0)


(b)If a =10% g

thus 

Answer:
3. Fructose
Explanation:
Fructose is a sugar found naturally in fruits, fruit juices, some vegetables and honey.
Answer:
option C
Explanation:
Final velocity of the object is 114 m / s. Hence, final velocity of the object is 114 m / s.
Failed experiments, uncontrolled variables, invalid data, and generalized human error