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ipn [44]
2 years ago
10

Explain why it is not advisable to wear Ornament like ring at work shop​

Engineering
1 answer:
disa [49]2 years ago
8 0

The reason that it is not advisable to wear Ornament like ring at work shop​ is that rings can be easily get caught and fingers or the hands of the person can be injured, cut, scared etc.

<h3>Why should you not wear a ring at work?</h3>

Jewelry is known to bring about a lot of safety hazards for people working around chemicals and others.

Note that the reason that it is not advisable to wear Ornament like ring at work shop​ is that rings can be easily get caught and fingers or the hands of the person can be injured, cut, scared etc.

Learn more about Ornament from

brainly.com/question/24286720

#SPJ1

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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
7.4 A pretimed four-timing-stage signal has critical lane group flow rates for the first three timing stages of 200, 187, and 21
Irina18 [472]

Answer:

16 seconds

Explanation:

Given:

C = 60

L = 4 seconds each = 4*4 =16

In this problem, the first 3 timing stages are given as:

200, 187, and 210 veh/h.

We are to find the estimated effective green time of the fourth timing stage. The formula for the estimated effective green time is:

g = (\frac{v}{s}) (\frac{C}{X})

Let's first find the fourth stage critical lane group ratio \frac{v}{s} , using the formula:

C = \frac{1.5L +5}{1 - ( \frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800}) + ( \frac{v}{s})}

60 = \frac{1.5*16 + 5}{1 - ( \frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800}) + ( \frac{v}{s})}

60 = \frac{24+5}{1 - (0.332 + ( \frac{v}{s}))}

Solving for (\frac{v}{s}), we have:

(\frac{v}{s}) = 0.185

Let's also calculate the volume capacity ratio X,

X = (\frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800} + 0.185)(\frac{60}{60-16}

X = 0.704

For the the estimated effective green time of the fourth timing stage, we have:

g_4 = (\frac{v}{s}) (\frac{C}{X})

Substituting figures in the equation, we now have:

g_4 = (0.185) (\frac{60}{0.704})

g_4 = 15.78 seconds

15.78 ≈ 16 seconds

The estimated effective green time of the fourth timing stage is 16 seconds

8 0
2 years ago
Select the correct statement(s) regarding frequency modulation (FM). a. baseband amplitude changes (information) are captured as
mario62 [17]
<h3>All of the above statement are correct statement.</h3>

<u>Explanation</u>

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scoundrel [369]

Answer:

An airfoil creates lift by exerting a downward force on the air as it flows past

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Free_Kalibri [48]

Answer:

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2) precipitation and sunshine

3) air and wind movement.

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