Answer:
The scattering angle is 0.80
Explanation:
As we know
Wave length = h/mc (1-cos α) --1
W = mc^2
m = E/c^2
Substituting this in equation 1 we get
Wavelength = hc/E (1-cos α)
Substituting the given values, we get -
0.95*10^-9 = [(6.626 *10^-34 * 2.998 *10^8)/5.111*1000*1.6*10^-19} (1-cos α)
(1-cos α) = 1.14 *10^-10
cos α = 1- 1.14 *10^-10
α = 0.80
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)
The answer is invention. Technology is vital to science for purposes
of dimension, data gathering, treatment of samples, calculation, transport to
research sites, defense from hazardous materials, and communication. So
technology is new tools and techniques that are being industrialized that make
it conceivable to progress various lines of scientific investigation.
Answer:
The tension in string P is 25 N, while that of Q is 85 N.
Explanation:
Considering the conditions for equilibrium,
i. Total upward force = Total downward force
+
= 110 N
ii. Taking moment about P,
clockwise moment = anticlockwise moment
110 × (2.5 - 0.8) =
× (3 - 0.8)
110 × 1.7 =
× 2.2
187 = 2.2![T_{Q}](https://tex.z-dn.net/?f=T_%7BQ%7D)
= ![\frac{187}{2.2}](https://tex.z-dn.net/?f=%5Cfrac%7B187%7D%7B2.2%7D)
= 85 N
From the first condition,
+
= 110 N
+ 85 N = 110 N
= 110 - 85
25 N
Therefore, the tension in string P is 25 N while that of Q is 85 N.