Answer:
274N 0.41
Explanation:
As he is sliding down in a constant speed then the force that accelerates him (weight) and the force that slows his down (friction) are equal.
then
<em>friction=mass x gravity x sin(21)</em>
Fr=78kg x 9.8m/s2 x sin(21)=274N
<em>friction= coefficient of kinetic friction x normal force of from the slope</em>
Fr= u x 78kg x 9.8m/s2 x cos(21)=274N
Fr= u x 78kg x 9.8m/s2 x cos(21)=274Nu=274/677=0.41
Answer:
0.306mm
Explanation:
The radius of the conductor is 3mm, or 0.003m
The area of the conductor is:
A = π*r^2 = π*(.003)^2 = 2.8*10^-5 m^2
The current density is:
J = 130/2.8*10^-5 = 4.64*10^6 A/m
According to the listed reference:
Vd = J/(n*e) = 4.64*10^6 / ( 8.46*10^28 * 1.6*10^-19 ) = 0.34*10^-6 m/s = 0.34mm/s
The distance traveled is:
x = v*t = 0.34 * .90 = 0.306 mm
Answer:
Follows are the solution to this question:
Explanation:
Please find the correct question in the attachment file.
Let:
Calculating the value of 
![\to \left | \begin{array}{ccc}\hat{i}&\hat{j}&\hat{K}\\R_i&R_j&R_k\\S_i&S_j&S_k\end{array}\right | = \hat{i}[R_j S_k-S_jR_k]-\hat{j}[R_i S_k-S_iR_k]+\hat{k}[R_i S_j-S_iR_j]](https://tex.z-dn.net/?f=%5Cto%20%5Cleft%20%7C%20%5Cbegin%7Barray%7D%7Bccc%7D%5Chat%7Bi%7D%26%5Chat%7Bj%7D%26%5Chat%7BK%7D%5C%5CR_i%26R_j%26R_k%5C%5CS_i%26S_j%26S_k%5Cend%7Barray%7D%5Cright%20%7C%20%3D%20%5Chat%7Bi%7D%5BR_j%20S_k-S_jR_k%5D-%5Chat%7Bj%7D%5BR_i%20S_k-S_iR_k%5D%2B%5Chat%7Bk%7D%5BR_i%20S_j-S_iR_j%5D)
Calculating the value of 
![\to (R_i\hat{i}+R_j\hat{j}+R_k\hat{k}) \cdot ( \hat{i}[R_j S_k-S_jR_k]-\hat{j}[R_i S_k-S_iR_k]+\hat{k}[R_i S_j-S_iR_j])](https://tex.z-dn.net/?f=%5Cto%20%28R_i%5Chat%7Bi%7D%2BR_j%5Chat%7Bj%7D%2BR_k%5Chat%7Bk%7D%29%20%5Ccdot%20%28%20%5Chat%7Bi%7D%5BR_j%20S_k-S_jR_k%5D-%5Chat%7Bj%7D%5BR_i%20S_k-S_iR_k%5D%2B%5Chat%7Bk%7D%5BR_i%20S_j-S_iR_j%5D%29)
by solving this value it is equal to 0.
