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
The relationship between inductance and frequency can be clearly described using the following equation of inductive reactance:
Xl = 2*pi*f*L ; simplifying:
L = Xl / 2*pi*f
Therefore, as what we saw, inductance and frequency are inversely proportional. To add up, when inductance increases the frequency would decrease.
3.11 is the answer I think
Answer:
if no mistaken then it is 300kN
Explanation:
Using SI units (the default if a system is not specified) 200T = 200000kg as the mass of the whale, it would weigh 1960kN in 1g Earth gravity.
On the Moon, the acceleration is 0.166g, to it would weigh about 330kN.
Answer:
The correct option is;
- 4x
Explanation:
From the inverse square law, as the distance from the source of a physical quantity increases, the intensity of the source is spread over an area proportional to the square of the distance of the object from the source
The inverse square law can be presented as follows;
As the distance, r, increases, the surface it covers also increases by the power of 2
Therefore, where the distance increases from r to 2·r, we have;
When, I, remain constant
The surface increases to 4·S by the inverse square law
Therefore, the correct option is 4 × x.
