Answer:
D. The primary source, because it was written by the researcher
Explanation:
One should learn to trust the primary source more because it is the real work of the researcher.
A primary work defines the work of research from his or her experimental findings.
- The primary source presents experimental data and other subordinates ones to reach a conclusion as seen from the view of the researcher.
- A secondary source implies someone else documenting their own opinion about the primary experimental set up.
- This can get dicey in the sense that results can be twisted to serve other ulterior motives.
Energy at top = U = mgh = 40 * 9.8 * 12 = 4704 J
Energy at bottom = 1/2 mv² = 1/2 * 40 * 10² = 4000 / 2 = 2000 J
Energy Lost = Final - Initial = 4704 - 2000 = 2704 J
In short, Your Answer would be 2704 Joules
Hope this helps!
Answer:
The work done by the frictional force is 600J.
Explanation:
The work
done by the frictional force is
.
Now,
and
; therefore,


Hence, the work done by friction is 660J.
Answer:
a) t = 3.35[s]; b) t = 1.386[s]
Explanation:
We can solve this problem by dividing it into two parts, for the first 55 [m] and then the second part with the remaining 55 [m].
We will take the initial velocity as zero, as the problem does not mention that the Rock was thrown at initial velocity.
And using kinematics equations:
![v_{f}^{2}= v_{o}^{2}+2*g*y\\where:\\v_{o}=0\\g=gravity = 9.81[m/s^2]\\y=55 [m]\\v_{f}^{2}=0+2*9.81*55\\v_{f}=\sqrt{2*9.81*55} \\v_{f}=32.85[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%5E%7B2%7D%3D%20v_%7Bo%7D%5E%7B2%7D%2B2%2Ag%2Ay%5C%5Cwhere%3A%5C%5Cv_%7Bo%7D%3D0%5C%5Cg%3Dgravity%20%3D%209.81%5Bm%2Fs%5E2%5D%5C%5Cy%3D55%20%5Bm%5D%5C%5Cv_%7Bf%7D%5E%7B2%7D%3D0%2B2%2A9.81%2A55%5C%5Cv_%7Bf%7D%3D%5Csqrt%7B2%2A9.81%2A55%7D%20%5C%5Cv_%7Bf%7D%3D32.85%5Bm%2Fs%5D)
Now we can calculate the time:
![v_{f}=v_{o}+g*t\\t=\frac{v_{f}-v_{o}}{g}\\ t=\frac{32.85-0}{9.81}\\ t=3.35[s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3Dv_%7Bo%7D%2Bg%2At%5C%5Ct%3D%5Cfrac%7Bv_%7Bf%7D-v_%7Bo%7D%7D%7Bg%7D%5C%5C%20t%3D%5Cfrac%7B32.85-0%7D%7B9.81%7D%5C%5C%20t%3D3.35%5Bs%5D)
Now we can calculate the second time, but using as a initial velocity 32.85[m/s].
The final velocity will be:
![v_{f}^{2}= v_{o}^{2}+2*g*y\\v_{f}=\sqrt{v_{o}^{2}+2*g*y} \\v_{f}=\sqrt{32.85^{2}+2*9.81*55 } \\v_{f}=46.45[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%5E%7B2%7D%3D%20v_%7Bo%7D%5E%7B2%7D%2B2%2Ag%2Ay%5C%5Cv_%7Bf%7D%3D%5Csqrt%7Bv_%7Bo%7D%5E%7B2%7D%2B2%2Ag%2Ay%7D%20%5C%5Cv_%7Bf%7D%3D%5Csqrt%7B32.85%5E%7B2%7D%2B2%2A9.81%2A55%20%7D%20%5C%5Cv_%7Bf%7D%3D46.45%5Bm%2Fs%5D)
Now we can calculate the second time:
![t=\frac{46.45-32.85}{9.81} \\t= 1.386[s]](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B46.45-32.85%7D%7B9.81%7D%20%5C%5Ct%3D%201.386%5Bs%5D)
Note: The reason the second time is shorter even though it is the same distance is that the acceleration of gravity increases the speed of the rock more and more as it falls.