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Mnenie [13.5K]
3 years ago
7

Technician A says that the tinnerman nuts are used to hold the brake drum on and should be reinstalled when the drum is replaced

. Technician B says that a drum should be removed inside a sealed vacuum enclosure or washed with water or solvent to prevent possible asbestos dust from being released into the air. Which technician is correct?A. Technician A onlyB. Technician B onlyC. Both Technicians A and BD. Neither A nor B
Physics
1 answer:
Ahat [919]3 years ago
5 0

Answer:

B. Technician B only

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4 0
2 years ago
A parallel-plate capacitor has square plates that are 8.00cm on each side and 4.20mm apart. The space between the plates is comp
Sergeu [11.5K]

Answer:

U=1.29\times 10^{-7}\ J

Explanation:

Given that

a= 8 cm (square)

A= a ² = 64 cm²

d= 4.2 mm

d₁= 2.1 mm  ,K₁= 4.7

d₂=2.1 mm  , K₂=2.6

We know that capacitance given as

C_1=\dfrac{K_1\varepsilon _oA}{d_1}

C_1=\dfrac{4.7\times 8.85\times 10^{-12}\times 64\times 10^{-4}}{2.1\times 10^{-3}}

C_1=1.26\times 10^{-10}\ F

C_2=\dfrac{K_2\varepsilon _oA}{d_2}

C_2=\dfrac{2.6\times 8.85\times 10^{-12}\times 64\times 10^{-4}}{2.1\times 10^{-3}}

C_2=0.701\times 10^{-10}\ F

Net capacitance

C=\dfrac{C_1C_2}{C_1+C_2}

C=\dfrac{1.26\times 10^{-10}\times 0.701\times 10^{-10}}{1.26\times 10^{-10}+0.701\times 10^{-10}}\ F

C=4.5\times 10^{-11}\ F

We know that stored energy given as

U=\dfrac{CV^2}{2}

V= 76 V

U=\dfrac{4.5\times 10^{-11}\times 76^2}{2}\ J

U=1.29\times 10^{-7}\ J

3 0
2 years ago
A parallel-plate capacitor with circular plates of radius R is being charged by a battery, which provides a constant current. At
NikAS [45]

To solve this problem it is necessary to apply the concepts related to the magnetic field.

According to the information, the magnetic field INSIDE the plates is,

B=\frac{1}{2} \mu \epsilon_0 r

Where,

\mu =Permeability constant

\epsilon_0 =Electromotive force

r = Radius

From this deduction we can verify that the distance is proportional to the field

B \propto r

Then the distance relationship would be given by

\frac{r}{R} = \frac{B}{B_{max}}

r =\frac{B}{B_{max}} R

r = \frac{0.5B_{max}}{B_{max}}R

r = 0.5R

On the outside, however, it is defined by

B = \frac{\mu_0 i_d}{2\pi r}

Here the magnetic field is inversely proportional to the distance, that is

B \not\propto r

Then,

\frac{r}{R} = \frac{B_{max}{B}}

r = \frac{B_{max}{B}}R

r = \frac{B_{max}{0.5B_{max}}}R

r = 2R

7 0
2 years ago
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