Answer:
the object has least potential energy at mean position of the SHM
Explanation:
If a block is connected with a spring and there is no resistive force on the system
In this case the total energy of the system is always conserved and it will change from one form to another form
So here we will say that
Kinetic energy + Potential energy = Total Mechanical energy
As we can say that total energy is conserved so here we have least potential energy when the system has maximum kinetic energy
So here we also know that at mean position of the SHM the system has maximum speed and hence maximum kinetic energy.
So the object has least potential energy at mean position of the SHM
The tension in the first and second rope are; 147 Newton and 98 Newton respectively.
Given the data in the question
- Mass of first block;

- Mass of second block,

- Tension on first rope;

- Tension on second rope;
To find the Tension in each of the ropes, we make use of the equation from Newton's Second Laws of Motion:

Where F is the force, m is the mass of the object and a is the acceleration ( In this case the block is under gravity. Hence ''a" becomes acceleration due to gravity
)
For the First Rope
Total mass hanging on it; 
So Tension of the rope;

Therefore, the tension in the first rope is 147 Newton
For the Second Rope
Since only the block of mass 10kg is hang from the second, the tension in the second rope will be;

Therefore, the tension in the second rope is 98 Newton
Learn More, brainly.com/question/18288215
Answer:
0.00091
Explanation:
(9x10^9) (2.6x10^-6) (1.4x10^-6) / 36
(9,000,000,000) (0.0000026) (0.0000014) /36
|
23,400(0.0000014) /36
|
0.03276 /36
|
0.00091
Biology, Chemistry and Physics
If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m
Given the data in the question;
- Length of the massless beam;

- Distance of support from the left end;

- First mass;

- Distance of beam from the left end( m₁ is attached to );

- Second mass;

- Distance of beam from the right of the support( m₂ is attached to );

Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.
Hence, 
we divide both sides by 

Next, we make
, the subject of the formula
![x_1 = x - [ \frac{m_2x_2}{m_1} ]](https://tex.z-dn.net/?f=x_1%20%3D%20x%20-%20%5B%20%5Cfrac%7Bm_2x_2%7D%7Bm_1%7D%20%5D)
We substitute in our given values
![x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]](https://tex.z-dn.net/?f=x_1%20%3D%203.00m%20-%20%5B%20%5Cfrac%7B61.7kg%5C%20%2A%20%5C%200.273m%7D%7B31.3kg%7D%20%5D)


Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m
Learn more; brainly.com/question/3882839