52m/s is the answer because before it was gonna accelerate it was 52 m/s
The net force on the barge is 8000 N
Explanation:
In order to find the net force on the badge, we have to use the rules of vector addition, since force is a vector quantity.
In this problem, we have two forces:
- The force of tugboat A,
, acting in a certain direction - The force of tugboat B,
, also acting in the same direction
Since the two forces act in the same direction, this means that we can simply add their magnitudes to find the net combined force on the barge. Therefore, we get

and the direction is the same as the direction of the two forces.
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Answer:
D. 2.8 × 10⁹ N
Explanation:
The force between two charges is directly proportional to the amount of charges at the two points and inversely proportional to the square of distance between the two points.
Fe= k Q₁Q₂/r²
Q₁= -0.0045 C
Q₂= -0.0025 C
r= 0.0060 m
k= 9.00 × 10 ⁹ Nm²/C²
Fe= (9.00 × 10 ⁹ Nm²/C²×-0.0045 C×-0.0025 C)/0.0060²
=2.8 × 10⁹ N
Momentum would be the same before and after the collision
Before the collision:
Momentum of the single cart: 1 * 0.50 = 0.50
After the collision
velocity = 0.25m / s
1 * 0.25 + 1 * 0.25 =
0.25 * (1 + 1) =
0.25 * 2 =
0.50
Now new momentum will be 0.5
answer
the same before and after the collision