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Reil [10]
1 year ago
6

A circuit with resistive elements of 220, 100, 57, and 43 produce what total resistance?

Physics
1 answer:
Korvikt [17]1 year ago
8 0

The total resistance is 420 ohm.

A circuit with resistive elements of 220, 100, 57, and 43 produce what total resistance  

 R= 220+ 100+ 57+ 43

    = 420 Ω

What is resistance and its types?

Resistance is a measure of the opposition to current flow in an electrical circuit also known as ohmic resistance or electrical resistance. Ohms are measured as resistance, symbolized by the Greek letter omega (Ω). The ratio of the applied voltage to the current through the material is then known as resistance.

What causes resistance?

An electric current flows when electrons move through a conductor, such as a metal wire. The moving electrons can collide with the ions in the metal. This makes it more difficult for the current to flow, and causes resistance.

Learn more about resistance:

brainly.com/question/17563681

#SPJ4

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Answer:

Explanation:

The work required to push will be equal to work done by friction . Let  d be the displacement required .

force of friction = mg x μ where m is mass of the suitcase , μ be the coefficient of friction

work done by force of friction

mg x μ x d   = 660

80 x 9.8 x .272 x d = 660

d = 3 .1 m .

8 0
3 years ago
The initial temperature of a bomb calorimeter is 28.50°
solniwko [45]
The answer is 5,170 J
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3 years ago
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B.A certain piece of copper wire is determined to have a mass of 2.00 g per meter.
masha68 [24]

Answer:

14cm

Explanation:

Mass per gram of the piece of wire;

         2g of the wire is found in 1m

 Since

       100cm  = 1m;

 

So;

       100cm of the wire contains 2g of the wire

To provide 0.28g

 Since;

        2g of wire is made up of 100cm

      0.28g of wire will be contained in \frac{100 x 0.28}{2}   = 14cm

14cm of the wire will contain 0.28g

8 0
3 years ago
When discussing magnetic fields we could say that
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Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
6 0
3 years ago
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A conducting spherical shell with inner radius a and outer radius b has a positive point charge +Q at the center in the empty re
Andreas93 [3]

Answer:

a) q_inner = -Q

, q_outer = -2Q

b)    E₁ = k Q / r²        r<a

       E₂ = 0               a<r<b

       E₃ = - k 2Q/r²     r>b

d)   the charge continues inside the spherical shell, the results do not change

Explanation:

a) The point load in the center induces a load on the inner surface of the shell with constant opposite sign  

q_inner = -Q  

the outer shell of the shell the load is  

q_outer = -3 Q + Q  

q_outer = -2Q

b) To find the electric field again, use Gauss's law,  

We define as a Gaussian surface a sphere  

Ф = E. dA = q_{int}/ε₀

in this case the electric field lines and the radii of the sphere are parallel, so the sclar product is reduced to the algegraic product  

E A = q_{int}/ε₀

the area of ​​a sphere is  

A = 4 π r²  

E = 1 / 4πε₀  Q/ r²  

k = 1 / 4πε₀  

let's apply this expression to the different radii

i) r <a  

in this case the load inside is the point load  

q_{int}= + Q  

E₁ = k Q / r²

ii) the field inside the shell  

a <r <b  

As the sphere is conductive, so that it is in electrostatic equilibrium, there can be no field within it.  

E₂ = 0

iii)  r>b

   q_{int}  = Q- 3Q = -2Q

    E₃ = k (-2Q/r²)

     E₃ = - k 2Q/r²

c) see  attached

d) as the charge continues inside the spherical shell, the results do not change, since the lcharge inside remains the same and it does not matter its precise location, but remains within the Gaussian surface

6 0
3 years ago
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