Answer:
Yes.
Explanation:
A negative power would just represent a loss of power. So in your case it lost -1252.16 W
The weight of the meterstick is:

and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance

from the pivot.
The torque generated by the weight of the meterstick around the pivot is:

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:

from which we find the value of d2:

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
Answer:
final displacement lf = 0.39 m
Explanation:
from change in momentum equation:
![\delta p = m \sqrt(2g * y/x)* [\sqrt li + \sqrt lf]](https://tex.z-dn.net/?f=%5Cdelta%20p%20%3D%20m%20%5Csqrt%282g%20%2A%20y%2Fx%29%2A%20%5B%5Csqrt%20li%20%2B%20%5Csqrt%20lf%5D)
given: m = 0.4kg, y/x = 19/85, li = 1.9 m,
\delta p = 1.27 kg*m/s.
putting all value to get the final displacement value
![1.27 = 0.4\sqrt(2*9.81 *(19/85))* [\sqrt 1.9 + \sqrt lf]](https://tex.z-dn.net/?f=1.27%20%3D%200.4%5Csqrt%282%2A9.81%20%2A%2819%2F85%29%29%2A%20%5B%5Csqrt%201.9%20%2B%20%5Csqrt%20lf%5D)
final displacement lf = 0.39 m
Answer:
4.7 s
Explanation:
The complete question is presented in the attached image to this solution.
v(t) = 61 - 61e⁻⁰•²⁶ᵗ
At what time will v(t) = 43 m/s?
We just substitute 43 m/s into the equation for the velocity of the diver and solve for t.
43 = 61 - 61e⁻⁰•²⁶ᵗ
- 61e⁻⁰•²⁶ᵗ = 43 - 61 = -18
e⁻⁰•²⁶ᵗ = (18/61) = 0.2951
In e⁻⁰•²⁶ᵗ = In 0.2951 = -1.2205
-0.26t = -1.2205
t = (1.2205/0.26) = 4.694 s = 4.7 s to the nearest tenth.
Hope this Helps!!!