The slope of a speed-time graph is the acceleration represented by the graph.
All other parts of this question refer to a lab experiment or exercise
where I was not present, but Zeesam16 was. Therefore I have no data
with which to answer the rest of the question, and hope that Zeesam can
handle it.
Answer:
1 angstrom = 0.1nm
5000 angstrom = 5000/1 × 0.1nm
<h3>= 500nm</h3>

5000 angstrom = 5000 × 1 × 10^-10
<h3>= 5 × 10^-7 m</h3>
Hope this helps you
The resulting change in momentum of the system will be +18.6 Ns. The momentum is conserved.
<h3>What is the law of conservation of momentum?</h3>
According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.
The given data in the problem is;
m is the mass =6.0 kg
t is the time interval=2 second
From Newton's second law;

From the graph;

The change in the momentum is;

Hence, the resulting change in momentum of the system will be +18.6 Ns.
To learn more about the law of conservation of momentum, refer;
brainly.com/question/1113396
#SPJ1
Answer:
noble gases are basically a group of gases that are similar in their chemical compounds, theres six of them : helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and the radioactive radon (Rn).
~batmans wife dun dun dun.....
In the frame of reference of anybody in the car.