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
Hello.
The solor system is organized by the smallest star to the largest and by the 8 planets. it is also organized by how much gas a planet has. But it is also orgainzed becasue of gravity.
The different divisions are <span>planets, moons, asteroids, comets and meteoroids.
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Average speed is defined as the ratio of total distance covered in total given time

here we know that total distance that man moved is


so total distance is



now here total time of the motion is


total time will be given as


now by above formula


so his average speed is 30 km/h
This is a non testable question because it cannot be answered by doing an experiment. But it could be modified for example Dogs are more obedient then cats.
Answer:
The Internal energy of the gas did not change
Explanation:
In this situation the Internal energy of the gas did not change and this is because according the the first law of thermodynamics
Δ U = Q - W ------ ( 1 )
Δ U = change in internal energy
Q = heat added
W = work done
since Q = W. the value of ΔU will be = zero i.e. No change