Answer:
please do well to ask questions in English. This will help people provide you answers ASAP. Thank you
Well,
When an object's velocity changes, we call it acceleration.
Acceleration: The time rate of change in an object's velocity
Answer:
71 rpm
Explanation:
Given that:
Angular momentum (L) = 0.26
Diameter = 25cm = 0.25 cm
Radius, r = (d/2) = 0.125m
Mass = 5.6 kg
Moment of inertia (I) = 2mr² / 5
I = (2 * 5.6 * 0.125^2) / 5
= 0.175
= 0.175 / 5
= 0.035 kgm²
Angular speed (w) ;
w = L / I
w = 0.26 / 0.035
= 7.4285714
= 7.429 rad/s
w = (7.429 * 60/2π)
w = 445.74 / 2π rpm
w = 70.941724
Angular speed = 70.94 rpm
= 71 rpm
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
