Draw perpendicular lines based on the info in the image.

When measuring a Brake Drum, the Brake Micrometer is set to a Base Drum Diameter of 10 Inches plus four notches, and the dial re

kozerog [31]

Answer:

10.5

Explanation:

7 0

2 years ago

What does Clay say will happen if the system is rejected?

juin [17]

Answer:

the nation will suffer terrible consequences

Explanation:

I did that and got it right

6 0

2 years ago

Read 2 more answers

For every 3 packages that Marcella wraps she uses 1/2 of a roll of wrapping paper if she uses 2 1/2 rolls of wrapping paper how

hram777 [196]

She wrapped 5 packages

6 0

2 years ago

Read 2 more answers

can you translate this me gusta el queso :sorry i would have put 300 points buut i used them all for my last question

Irina-Kira [14]

Translate in Spanish: lo siento, hubiera puesto 300 puntos pero los usé todos para mi última pregunta

If you meant a different language lmk

4 0

2 years ago

Given the vector current density J = 10rho2zarho − 4rho cos2 φ aφ mA/m2:

Xelga [282]

Answer:

(a) Current density at P is $J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\$ .

(b) Total current I is 3.257 A

Explanation:

Because question includes symbols and formulas it can be misunderstood. In the question current density is given as below;

$J=10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}\\$

where $\textbf{a}_{\rho}$ and $\textbf{a}_{\phi}$ unit vectors.

(a) In order to find the current density at a specific point <em>(P)</em>, we can simply replace the coordinates in the current density equation. Therefore

$J(P(\rho=3, \phi=30^o,z=2))=10.3^2.2.\textbf{a}_{\rho}-4.3.(\cos(30^o)^2).\textbf{a}_{\phi}\\\\J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\$

(b) Total current flowing outward can be calculated by using the relation,

$I=\int {\textbf{J} \, \textbf{ds}$

where integral is calculated through the circular band given in the question. We can write the integral as below,

$I=\int\{(10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}).(\rho.d\phi.dz.\textbf{a}_{\rho}})\}\\\\I=\int\{(10\rho^2z).(\rho.d\phi.dz)\}\\\\\\$

due to unit vector multiplication. Then,

$I=10\int\(\rho^3z.dz.d\phi$

where $\rho=3,\ 0$ . Therefore

$I=10.3^3\int_2^{2.8}\(zdz.\int_0^{2\pi}d\phi\\I=270(\frac{2.8^2}{2}-\frac{2^2}{2} )(2\pi-0)=3257.2\ mA\\I=3.257\ A$

4 0

3 years ago