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:
B
Explanation:
The impulse experienced by an object is the force•time.
Answer:
147.7 N
221.55 Nm
Explanation:
P = Pressure = 100000 Pa
= Mass-specific gas constant = 287.015 J/kg k
T = Temperature = 10+273 = 283 K
C = Drag coefficient = 1.1
A = Area
r = Radius = 0.2 m
v = Speed of wind = 
L = Length of pole
Density

Drag force

Force on the circular sign is 147.7 N

Bending moment at the bottom of the pole is 221.55 Nm
Answer:
Explanation:
= 4190 J/kg.K
= 910 J/Kg. K
= 1.50 kg
= 1.80 kg

ΔT +
ΔT
= (1.50)(910)(85.0-20)+(1.80)(4190)(85.0-20)
= 578,955 J
= 579 kJ
Answer:
31m/s
23.17°
Explanation:
Given the following :
From the diagram attached :
AB = 15m/s, BC = 18m/s, AC = a = resultant
Resolving Velocity into both vertical and horizontal component.
Kindly see attached picture for detailed explanation.