Here in crash test the two forces are acting on the dummy in two different directions
As we know that force is a vector quantity so we need to use vector addition laws in order to find the resultant force on it.
So here two forces are given in perpendicular direction with each other so as per vector addition law we need to use Pythagoras theorem to find the resultant of two vectors
so we can say
here given that
now we will plug in all data in the above equation
so it will have net force 4501.9 N which will be reported by sensor
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
Wound it be one that dissolves ?
Nothing is faster than light. Dark is the absence of light, or indeed anything at all. ... Darkness came before light, so light had to be created to get there. It's like turning on s flashlight in a dark room: the darkness is already there, and an outside source (you/the flashlight) needs to create light.
