The question is asking to calculate the object's speed v1, v2, v3 at the certain time is the given of the problem, in my calculation, I would say that the speed would be 2m/s, 1.5m/s, 0.22m/s. I hope you are satisfied with my answer and feel free to ask for more if you have question and further clarification
Via the half-life equation:
![A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h}}](https://tex.z-dn.net/?f=A_%7Bfinal%7D%3DA_%7Binitial%7D%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7B%5Cfrac%7Bt%7D%7Bh%7D%7D)
Where the time elapse is 11,460 year and the half-life is 5,730 years.
![A_{final}=A_{initial}(\frac{1}{2})^\frac{11460}{5730} \\\\A_{final}=A_{initial}\frac{1}{4} \\\\A_{final}=\frac{1}{4}A_{initial}](https://tex.z-dn.net/?f=A_%7Bfinal%7D%3DA_%7Binitial%7D%28%5Cfrac%7B1%7D%7B2%7D%29%5E%5Cfrac%7B11460%7D%7B5730%7D%20%5C%5C%5C%5CA_%7Bfinal%7D%3DA_%7Binitial%7D%5Cfrac%7B1%7D%7B4%7D%20%5C%5C%5C%5CA_%7Bfinal%7D%3D%5Cfrac%7B1%7D%7B4%7DA_%7Binitial%7D)
Therefore after 11,460 years the amount of carbon-14 is one fourth (1/4) of the original amount.
1) Force = m*a = 1.00 g * (1kg / 1000 g) * 225 m/s^2 = 0.225 N
2) Charge
Force = K (charge)^2 /(distance)^2 => charge = √ [Force * distance^2 / k]
k = 9.00 * 10^9 N*m^2 / C^2
charge = √ [0.225 N * (0.02 m)^2 / 9.00* 10^9 N*m^2 / C^2 ]
charge = 0.0000001 C = 0.0001 mili C
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;
![L = 4.00m](https://tex.z-dn.net/?f=L%20%3D%204.00m)
- Distance of support from the left end;
![x = 3.00m](https://tex.z-dn.net/?f=x%20%3D%203.00m)
- First mass;
![m1 = 31.3 kg](https://tex.z-dn.net/?f=m1%20%3D%2031.3%20kg)
- Distance of beam from the left end( m₁ is attached to );
![x_1 = ?](https://tex.z-dn.net/?f=x_1%20%3D%20%3F)
- Second mass;
![m_2 = 61.7 kg](https://tex.z-dn.net/?f=m_2%20%3D%2061.7%20kg)
- Distance of beam from the right of the support( m₂ is attached to );
![x_1 = 0.273m](https://tex.z-dn.net/?f=x_1%20%3D%200.273m)
Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.
Hence, ![m_1g( x-x_1) = m_2gx_2](https://tex.z-dn.net/?f=m_1g%28%20x-x_1%29%20%3D%20m_2gx_2)
we divide both sides by ![g](https://tex.z-dn.net/?f=g)
![m_1( x-x_1) = m_2x_2](https://tex.z-dn.net/?f=m_1%28%20x-x_1%29%20%3D%20m_2x_2)
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)
![x_1 = 3.00m - 0.538m](https://tex.z-dn.net/?f=x_1%20%3D%203.00m%20-%200.538m)
![x_1 = 2.46m](https://tex.z-dn.net/?f=x_1%20%3D%202.46m)
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