Since this is a distance/time graph, the speed at any time is the slope
of the part of the graph that's directly over that time on the x-axis.
At time t1 = 2.0 s
That's in the middle of the first segment of the graph,
that extends from zero to 3 seconds.
Its slope is 7/3 . v1 = 7/3 m/s .
At time t2 = 4.0 s
That's in the middle of the horizontal part of the graph
that runs from 3 to 6 seconds.
Its slope is zero.
v2 = zero .
At time t3 = 13 s.
That's in the middle of the part of the graph that's sloping down,
between 11 and 16 seconds.
Its slope is -3/5 . v3 = -0.6 m/s .
W=ΔKE , W=-5000j
KEinitial=(1/2)mv² , KEfinal=0j
ΔKE=-(1/2)mv²
-5000=-(1/2)(100kg)v²
v=10 m/s
Answer:
The answer is not correct.
Explanation:
Stu's answer is not correct, the equation to use is known as the law of ohm. In which the voltage is defined as the product of the current by the resistance, then we will see this equation.
![V = I*R\\where:\\I = current [amp]\\R = resistance [ohm]\\V = voltage [volts]\\](https://tex.z-dn.net/?f=V%20%3D%20I%2AR%5C%5Cwhere%3A%5C%5CI%20%3D%20current%20%5Bamp%5D%5C%5CR%20%3D%20resistance%20%5Bohm%5D%5C%5CV%20%3D%20voltage%20%5Bvolts%5D%5C%5C)
In order to find resistance, this term is found multiplying the current on the right side of the equation, therefore the current will be divided on the left side of the equation.
![R=\frac{V}{I} \\replacing:\\R=\frac{4}{0.5} \\R=8[ohms]](https://tex.z-dn.net/?f=R%3D%5Cfrac%7BV%7D%7BI%7D%20%5C%5Creplacing%3A%5C%5CR%3D%5Cfrac%7B4%7D%7B0.5%7D%20%5C%5CR%3D8%5Bohms%5D)
That is the reason that the result found by Stu is not correct.
10,000 units of momentum.
p=mv
20,000=m(2v)
10,000=mv