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
The ball will decelerate as it moves upwards.
The magnitude of the ball's acceleration is 0.3 m/s² and it directed backwards.
The given parameters;
- initial velocity of the ball, u = 1.25 m/s
- time of motion of the ball, t = 4.22 s
As the ball rolls up the inclined plane, the velocity decreases and eventually becomes zero when the ball reaches the highest point of the plane.
Thus, the ball decelerate as it moves upwards.
The acceleration of the ball is calculate as;

<em>at the highest point on the incline plane, the final velocity </em>
<em> is zero</em>

Thus, the magnitude of the ball's acceleration is 0.3 m/s² and it directed backwards.
Learn more here:brainly.com/question/23860763
Answer:
I think its C and D
Explanation:Force is like a energy that moves stuff. :)
Answer:
(a) 10 m/s
(b) 22.4 m/s
Explanation:
(a) Draw a free body diagram of the car when it is at the top of the loop. There are two forces: weight force mg pulling down, and normal force N pushing down.
Sum of forces in the centripetal direction (towards the center):
∑F = ma
mg + N = mv²/r
At minimum speed, the normal force is 0.
mg = mv²/r
g = v²/r
v = √(gr)
v = √(10 m/s² × 10.0 m)
v = 10 m/s
(b) Energy is conserved.
Initial kinetic energy + initial potential energy = final kinetic energy
½ mv₀² + mgh = ½ mv²
v₀² + 2gh = v²
(10 m/s)² + 2 (10 m/s²) (20.0 m) = v²
v = 22.4 m/s