Answer:
The torque about the origin is 
Explanation:
Torque
is the cross product between force
and vector position
respect a fixed point (in our case the origin):

There are multiple ways to calculate a cross product but we're going to use most common method, finding the determinant of the matrix:
![\overrightarrow{r}\times\overrightarrow{F} =-\left[\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k}\\ F1_{x} & F1_{y} & F1_{z}\\ r_{x} & r_{y} & r_{z}\end{array}\right]](https://tex.z-dn.net/?f=%20%5Coverrightarrow%7Br%7D%5Ctimes%5Coverrightarrow%7BF%7D%20%3D-%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20%5Chat%7Bi%7D%20%26%20%5Chat%7Bj%7D%20%26%20%5Chat%7Bk%7D%5C%5C%20F1_%7Bx%7D%20%26%20F1_%7By%7D%20%26%20F1_%7Bz%7D%5C%5C%20r_%7Bx%7D%20%26%20r_%7By%7D%20%26%20r_%7Bz%7D%5Cend%7Barray%7D%5Cright%5D%20)



Answer:
The correct answer would be C, certain display rules are more appropriate than others.
Explanation:
I hope this helps you:)
Answer:
v = 3.08 m/s.
Explanation:
Given,
- Initial velocity of the block = u = 43 cm/s = 0.43 m/s
- mass of the block = m = 1.40 kg
- spring constant = k = 15.5 N/m.
- x = 0.650 m
At x = 0.650 m
Let 'a' be the acceleration of the block at x.
Total force due to the spring force on the blcok at x= kx

Hence the acceleration of the block at x is 7.19\ m/s^2.
Let 'v' be the velocity of the block at x.
From the kinematics,

Hence at the compression x = 0.650 m the velocity of the block is 3.08 m/s.