True because the picture below proves this....
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The cart's acceleration to the right after the mass is released is determined as 7.54 m/s².
<h3>
Acceleration of the cart</h3>
The acceleration of the cart is determined from the net force acting on the mass-cart system.
Upward force = Downward force
ma = mg
13a = 10(9.8)
13a = 98
a = 98/13
a = 7.54 m/s²
Thus, the cart's acceleration to the right after the mass is released is determined as 7.54 m/s².
Learn more about acceleration here: brainly.com/question/14344386
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Answer:
1. 
2. 
Explanation:
1. According to Newton's law of motion, the puck motion is affected by the acceleration, which is generated by the push force F.
In Newton's 2nd law: F = ma
where m is the mass of the object and a is the resulted acceleration. So in the 2nd experiment, if we double the mass, a would be reduced by half.

Since the puck start from rest, in the 1st experiment, to achieve speed of v it would take t time

Now that acceleration is halved:


You would need to push for twice amount of time 
2. The distance traveled by the puck is as the following equation:

So if the acceleration is halved while maintaining the same d:

As
, then
. Also 



So t increased by 1.14
Mass is the physical quantity
Given parameters:
Displacement = 8km
Velocity = 3.8km/h
Unknown:
time = ?
Solution:
Velocity is displacement divided by time.
Velocity =
Displacement = velocity x time
Input the parameters:
8 = 3.8 x time
Time =
= 2.1s
The time taken is 2.1s