A binomial nomenclature, more commonly referred to as a scientific name.
Answer:

Explanation:
The total force on the particle is given by

Then, by replacing we have:
![q\vec{v}\ X \vec{B}=q[7\hat{k}-9\hat{j}-\hat{k}]\\\\q\vec{E}=q[5\hat{i}-\hat{j}-2\hat{k}]\\\\\vec{F}=(9.61*10^{-19}C)[(7+9)\hat{i}+(-9-1)\hat{j}+(-1-2)\hat{k}]\\\\\vec{F}=(1.537*10^{-17}\hat{i}-9.61*10^{-19}\hat{j}-2.883*10^{-18}\hat{k})N](https://tex.z-dn.net/?f=q%5Cvec%7Bv%7D%5C%20X%20%5Cvec%7BB%7D%3Dq%5B7%5Chat%7Bk%7D-9%5Chat%7Bj%7D-%5Chat%7Bk%7D%5D%5C%5C%5C%5Cq%5Cvec%7BE%7D%3Dq%5B5%5Chat%7Bi%7D-%5Chat%7Bj%7D-2%5Chat%7Bk%7D%5D%5C%5C%5C%5C%5Cvec%7BF%7D%3D%289.61%2A10%5E%7B-19%7DC%29%5B%287%2B9%29%5Chat%7Bi%7D%2B%28-9-1%29%5Chat%7Bj%7D%2B%28-1-2%29%5Chat%7Bk%7D%5D%5C%5C%5C%5C%5Cvec%7BF%7D%3D%281.537%2A10%5E%7B-17%7D%5Chat%7Bi%7D-9.61%2A10%5E%7B-19%7D%5Chat%7Bj%7D-2.883%2A10%5E%7B-18%7D%5Chat%7Bk%7D%29N)
where the cross product can be made with the determinant method.
Hope this helps!!
Answer:
A) - 1.8 m/s
Explanation:
As we know that whole system is initially at rest and there is no external force on this system
So total momentum of the system must be conserved
so we will have

now plug in all data into above equation



so correct answer is
A) - 1.8 m/s
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