Answer:
The question is incomplete, below is the complete question "A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.00 N)k and the corresponding torque about the origin is vector tau = (4 N · m)i hat + (10 N · m)j + (11N · m)k.
Determine Fx."

Explanation:
We asked to determine the "x" component of the applied force. To do this, we need to write out the expression for the torque in the in vector representation.
torque=cross product of force and position . mathematically this can be express as

Where
and the position vector

using the determinant method to expand the cross product in order to determine the torque we have
![\left[\begin{array}{ccc}i&j&k\\2&-3&2\\ F_{x} &7&-5\end{array}\right]\\\\](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C2%26-3%262%5C%5C%20F_%7Bx%7D%20%267%26-5%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C)
by expanding we arrive at

since we have determine the vector value of the toque, we now compare with the torque value given in the question

if we directly compare the j coordinate we have

Answer:
Change in electric potential energy ∆E = 365.72 kJ
Explanation:
Electric potential energy can be defined mathematically as:
E = kq1q2/r ....1
k = coulomb's constant = 9.0×10^9 N m^2/C^2
q1 = charge 1 = -2.1C
q2 = charge 2 = -5.0C
∆r = change in distance between the charges
r1 = 420km = 420000m
r2 = 160km = 160000m
From equation 1
∆E = kq1q2 (1/r2 -1/r1) ......2
Substituting the given values
∆E = 9.0×10^9 × -2.1 ×-5.0(1/160000 - 1/420000)
∆E = 94.5 × 10^9 (3.87 × 10^-6) J
∆E = 365.72 × 10^3 J
∆E = 365.72 kJ
Answer:
The friction coefficient's minimum value will be "0.173".
Explanation:
The given query seems to be incomplete. Below is the attached file of the complete question.
According to the question,
(a)
The net friction force's magnitude will be:
⇒ 


(b)
For m₃,
⇒ 
Or,
⇒ 


When paper burns, some of the chemical compounds in the
paper combine with oxygen in the air and become different
chemical compounds. That's a chemical change.
When iron rusts, or copper or silver turns green, that's the result
of the metal at the surface combining with the oxygen in the air and
forming a new chemical compound. Those are chemical changes.
When water evaporates, H₂O in the liquid phase gains thermal
energy and changes to H₂O in the gaseous phase. No chemical
compounds are lost, gained, or changed to other compounds.
It's just a physical change.