He would have potential, he hasn’t gone down the mountain yet so he remains potential
High task repitition, forceful exertions, repetitive or sustained awkward posture
Answer:
4s.
Explanation:
Hello,
In this case, considering that the instantaneous acceleration is computed via the change in the velocity divided by the change in the time, at 4 s we can evidence that the instantaneous acceleration is -2m/s² since the initial velocity at t=0s is 8m/s and at t=4s the velocity is 0 m/s, meaning that such decrease accounts for a deceleration process which is represented by a negative acceleration, and the value is:
![a=\frac{0m/s-8m/s}{4s-0s}\\ \\a=-2m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B0m%2Fs-8m%2Fs%7D%7B4s-0s%7D%5C%5C%20%5C%5Ca%3D-2m%2Fs%5E2)
Best regards.
It should be 0.25kg because you converter from g to kg and since 1g<1kg so you move the decimal to the left
Answer:
![F_{net} = 6.879\times 10^{- 7}\ N](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%206.879%5Ctimes%2010%5E%7B-%207%7D%5C%20N)
Solution:
As per the question:
Mass of first object, m = 120 kg
Mass of second object, m' = 420 kg
Mass of the third object, M = 69.0 kg
Distance between the m and m', d = 0.380 m
Now,
To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m:
To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m':
To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m and m':
![F_{net} = F + F'](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20F%20%2B%20F%27)
![F_{net} = 1.529\times 10^{- 7} + 5.35\times 10^{- 7} = 6.879\times 10^{- 7}\ N](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%201.529%5Ctimes%2010%5E%7B-%207%7D%20%2B%205.35%5Ctimes%2010%5E%7B-%207%7D%20%3D%206.879%5Ctimes%2010%5E%7B-%207%7D%5C%20N)