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Triss [41]
3 years ago
14

What type of diagrams are used to present the conditions that are available to control outputs based on inputs?

Physics
2 answers:
NISA [10]3 years ago
8 0

Answer:

a

Explanation:

I did this. already. good luck!

creativ13 [48]3 years ago
4 0

Answer:

a

Explanation:

because it shows how this gets in

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A parallel-plate capacitor in air has circular plates of radius 2.8 cm separated by 1.1 mm. Charge is flowing onto the upper pla
Hatshy [7]

Answer:

The time rate of change of the electric field between the plates is  \frac{E }{t} =  2.29 *10^{14} \   N \cdot C  \cdot  s^{-1}  

Explanation:

From the question we are told that

    The  radius is  r =  2.8 \ cm  =  0.028 \ m

     The distance of separation is  d =  1.1  \ mm  =  0.0011 \ m

      The  current is  I  =  5 \ A

Generally the electric field generated is mathematically represented as

         E = \frac{q }{ \pi  *  r^2  \epsilon_o  }

Where \epsilon_o is the permitivity of free space with a value

          \epsilon_o  =  8.85*10^{-12 }\   m^{-3} \cdot kg^{-1}\cdot  s^4 \cdot A^2

So the time rate of change of the electric field between the plates is mathematically represented as

        \frac{E }{t} =   \frac{q}{t} *   \frac{1 }{ \pi  *  r^2  \epsilon_o  }

But \frac{q}{t }  =  I

So  

       \frac{E }{t} =   *   \frac{I }{ \pi  *  r^2  \epsilon_o  }

substituting values  

        \frac{E }{t} =   *   \frac{5 }{3.142  *  (0.028)^2 *   8.85 *10^{-12}  }

        \frac{E }{t} =  2.29 *10^{14} \   N \cdot C  \cdot  s^{-1}

5 0
3 years ago
The Earth moves in predicable patterns. One of these patterns causes the seasons as
Rina8888 [55]

Answer:

A A A A A A A A

8 0
3 years ago
Read 2 more answers
A mixture can be classified as a solution, suspension, or coiled based on the: A. Number or particles it contains. B. Size of it
gladu [14]
According to chapter 1 in the mixture of particles Article a mixture has to contain a serten size to mix with another particle so i would go with B size of the largest particle
5 0
3 years ago
A long wire carries a current density proportional to the distance from its center, J=(Jo/ro)•r, where Jo and ro are constants a
IgorC [24]

Answer:

B = \mu_0(\frac{1}{3} \frac{J_0}{r_0} r^2)

Explanation:

As the current density is given as

J = \frac{J_0}{r_0}r

now we have current inside wire given as

i = \int J(2\pi r)dr

i = \int \frac{J_0}{r_0} r(2\pi r)dr

i = 2\pi \frac{J_0}{r_0} \int r^2 dr

i = \frac{2}{3} \pi \frac{J_0}{r_0} r^3

Now by Ampere's law we will have

\int B. dl = \mu_0 i

B. (2\pi r) = \mu_0(\frac{2}{3} \pi \frac{J_0}{r_0} r^3)

B = \mu_0(\frac{1}{3} \frac{J_0}{r_0} r^2)

4 0
3 years ago
A piece of aluminum has a volume of 4.89 x 10-3 m3. The coefficient of volume expansion for aluminum is = 69 x 10-6(C°)-1. The t
Leno4ka [110]

Answer:

11.515 Joule

Explanation:

Volume of aluminium = V = 4.89×10⁻³ m³

Coefficient of volume expansion for aluminum = α = 69×10⁻⁶ /°C

Initial temperature = 19.1°C

Final temperature = 357°C

Pressure of air = 1.01×10⁵ Pa

Change in temperature = ΔT= 357-19.1 = 337.9 °C

Change in volume

ΔV = αVΔT

⇒ΔV = 69×10⁻⁶×4.89×10⁻³×337.9

⇒ΔV = 114010.839×10⁻⁹ m³

Work done

W = PΔV

⇒W = 1.01×10⁵×114010.839×10⁻⁹

⇒W = 11.515 J

∴ Work is done by the expanding aluminum is 11.515 Joule

7 0
3 years ago
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