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sladkih [1.3K]
2 years ago
6

4. A lamp in a circuit has 4 Amps of current. If there is 32 12 of resistance

Physics
1 answer:
My name is Ann [436]2 years ago
8 0

Answer:

your mum because I don't know how to explain why she was doing it but I think I have to do it myself again but it was just the way

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What are conductors and insulators? Give at least five example of each ​
Leona [35]

Answer: conductors are substances that allow heat or electricity to pass through . It deals with only the flow of electrons.

Eg. water, copper wire, iron rod, some ceramic materials, metallic nail.

Insulators are materials that do not allow heat or electricity to pass through.

Eg. Book, plastic, rubber, glass, paper

Explanation:

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2 years ago
Measurements of two electric currents are shown in the chart. A 3-column table with 2 rows titled Electric Currents. The first c
Jobisdone [24]

The answer is B - Current Y has a greater potential difference, and the charges flow at a slower rate.

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3 years ago
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What is the main power and purpose of the legislative branch?<br> i always need help with physics
irina1246 [14]

Answer:

House and Senate.

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4 0
2 years ago
The parallel plates in a capacitor, with a plate area of 7.90 cm2 and an air-filled separation of 2.70 mm, are charged by a 7.90
s2008m [1.1K]

Answer:

A) 26V

Explanation:

(a) the potential difference between the plates

Initial capacitance can be calculated using below expresion

C1= A ε0/ d1

Where d1= distance between = 2.70 mm= 2.70× 10^-3 m

ε0= permittivity of space= 8.85× 10^-12 Fm^-1

A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2

If we substitute the values we

C1= A ε0/ d1

=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3

C1=2.589 ×10^-12 F= 2.59 pF

Initial charge can be determined using below expresion

q1= C1 × V1

V1=2.589 ×10^-12 F

V1= voltage=7.90 V

If we substitute we have

q1= 2.589 ×10^-12 × 7.90

q1= 20.45×10^-12C

20.45 pC

Final capacitance can be calculated as

C2= A ε0/ d2

d2=8.80 mm= /8.80× 10^-3

7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3

C1=0.794 ×10^-12 F= 0.794 pF

Final charge= initial charge

q2=q1 (since the battery is disconnected)

q2=q1= 20.45 pC

Final potential difference

V2= q/C2

= 20.45/0.794

= 26V

6 0
3 years ago
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