Answer:


Explanation:
Given that.
Force acting on the particle, 
Position of the particle, 
To find,
(a) Torque on the particle about the origin.
(b) The angle between the directions of r and F
Solution,
(a) Torque acting on the particle is a scalar quantity. It is given by the cross product of force and position. It is given by :




So, the torque on the particle about the origin is (32 N-m).
(b) Magnitude of r, 
Magnitude of F, 
Using dot product formula,




Therefore, this is the required solution.
Answer:
Yes, you are pulling on Earth. Reasoning. Third Newton's law of motion, action and reaction law, sates that for every action force, there is an equal (in magnitude) and opposite reaction force.
Explanation:
goo gle.
Answer:
Option (A)
Explanation:
A flowchart can be described as a representation of a process or an algorithm in a sequential manner (stepwise) in the form of diagrams. These steps are made of symbols that are connected to one another with the help of arrows. These arrows are used to show the direction of the process.
There are mainly four symbols that are used to describe a flow chart, which includes-
- Oval or Pill Shape- This type of symbol is used to depict the start or the end of a process.
- Rectangle Shape- This type of symbols is used to describe the process
.
- Diamond Shape- This type of symbol is used to represent decision
s.
- Parallelogram- This type of symbol is used for the representation of an input or an output.
Thus, the correct answer is option (A).
The black circles are the solvent and the open circles are the solute