Clutches And Torque Converters both provide all of these EXCEPT:

A hollow steel composite door with an 18-guage metal facing, hung on butt hinges with nonremovable pins, often with ventilation

Nadya [2.5K]

Answer:

Average industrial personnel door

Explanation:

Going by the description in which the door has ventilation louvers or glass panel, this shows that it can never be any type of security door. This is because with louvers or glass panel, it can be easier for any intruder to look or break into the house or space that is intended to be protected fully.

Hence, an hollow steel composite door with an 18-guage metal facing, hung on butt hinges with nonremovable pins, often with ventilation louvers or glass panels is likely an Average industrial personnel door

5 0

3 years ago

Two 2.30 cm × 2.30 cm plates that form a parallel-plate capacitor are charged to ± 0.708 nC . Part A What is the electric field

Alchen [17]

Answer:

A. $E = 1.512\times 10^{5}\ V$

B. 151.2 V

C. $E = 1.512\times 10^{5}\ V$

D. V = 302.4 V

Solution:

As per the question:

Area of the plates of the parallel plate capacitors, A = $2.30\times 2.30 = 5.29\ cm^{2} = 5.29\times 10^{- 4}\ m^{2}$

Charge on the plates of the capacitor, $Q_{c} = \pm 0.708\ nC = \pm 0.708\times 10^{- 9} \C$

Now,

(A) To calculate the electric field strength, E when the separation distance, d = 1.00 mm = $10^{- 3}\ m$ :

$E = \frac{Q}{\epsilon_{o}A}$

$E = \frac{0.708\times 10^{- 9}}{8.85\times 10^{- 12}\times 5.29\times 10^{- 4}} = 1.512\times 10^{5}\ N/C$

(B) To calculate potential difference between the plates:

$V = Ed = 1.512\times 10^{5}\times 10^{- 3} = 151.2V$

(C) Electric field strength when spacing is 2 mm, i.e., $2\times 10^{- 3}\ m$ :

$E = \frac{Q}{\epsilon_{o}A}$

Since, the above expression of the electric field shows that it does not depend on the separation distance between the plates thus it will remain same, i.e., $1.512\times 10^{5}\ V$