The momentum of an object is given by the product between its mass and its velocity:
![p=mv](https://tex.z-dn.net/?f=p%3Dmv)
where m is the mass and v the velocity.
For the object in our problem, m=10 kg and v=10 m/s, therefore its momentum is
![p=mv=(10 kg)(10 m/s)=100 kg m/s](https://tex.z-dn.net/?f=p%3Dmv%3D%2810%20kg%29%2810%20m%2Fs%29%3D100%20kg%20m%2Fs)
So, the correct answer is B).
Answer:
a luminous ball of plasma
Answer:
hi there!
the correct answer to this question is: 6.67 mph
Explanation:
you convert minutes to hours
10 miles * 60 mins / 90 mins
Answer:
The torque about the origin is ![2.0Nm\hat{i}-8.0Nm\hat{j}-12.0Nm\hat{k}](https://tex.z-dn.net/?f=2.0Nm%5Chat%7Bi%7D-8.0Nm%5Chat%7Bj%7D-12.0Nm%5Chat%7Bk%7D%20)
Explanation:
Torque
is the cross product between force
and vector position
respect a fixed point (in our case the origin):
![\overrightarrow{\tau}=\overrightarrow{r}\times\overrightarrow{F}](https://tex.z-dn.net/?f=%5Coverrightarrow%7B%5Ctau%7D%3D%5Coverrightarrow%7Br%7D%5Ctimes%5Coverrightarrow%7BF%7D%20)
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)
![\overrightarrow{r}\times\overrightarrow{F} =-((F1_{y}r_{z}-F1_{z}r_{y})\hat{i}-(F1_{x}r_{z}-F1_{z}r_{x})\hat{j}+(F1_{x}r_{y}-F1_{y}r_{x})\hat{k})](https://tex.z-dn.net/?f=%5Coverrightarrow%7Br%7D%5Ctimes%5Coverrightarrow%7BF%7D%20%3D-%28%28F1_%7By%7Dr_%7Bz%7D-F1_%7Bz%7Dr_%7By%7D%29%5Chat%7Bi%7D-%28F1_%7Bx%7Dr_%7Bz%7D-F1_%7Bz%7Dr_%7Bx%7D%29%5Chat%7Bj%7D%2B%28F1_%7Bx%7Dr_%7By%7D-F1_%7By%7Dr_%7Bx%7D%29%5Chat%7Bk%7D%29%20)
![\overrightarrow{r}\times\overrightarrow{F} =-((0(2.0m)-0(-3.0m))\hat{i}-((4.0N)(2.0m)-(0)(0))\hat{j}+((4.0N)(-3.0m)-0(0))\hat{k})](https://tex.z-dn.net/?f=%5Coverrightarrow%7Br%7D%5Ctimes%5Coverrightarrow%7BF%7D%20%3D-%28%280%282.0m%29-0%28-3.0m%29%29%5Chat%7Bi%7D-%28%284.0N%29%282.0m%29-%280%29%280%29%29%5Chat%7Bj%7D%2B%28%284.0N%29%28-3.0m%29-0%280%29%29%5Chat%7Bk%7D%29%20)
![\overrightarrow{r}\times\overrightarrow{F}=-2.0Nm\hat{i}+8.0Nm\hat{j}+12.0Nm\hat{k}=\overrightarrow{\tau}](https://tex.z-dn.net/?f=%20%5Coverrightarrow%7Br%7D%5Ctimes%5Coverrightarrow%7BF%7D%3D-2.0Nm%5Chat%7Bi%7D%2B8.0Nm%5Chat%7Bj%7D%2B12.0Nm%5Chat%7Bk%7D%3D%5Coverrightarrow%7B%5Ctau%7D%20)
It really depends on the angle where you look at it from and what type of glass/shape they are in. Mine always appeared pretty close even when it wasn't.
Source: Had 5 fish of my own.
Have a lovely day! ~Pooch ♥