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STALIN [3.7K]
3 years ago
7

Examine the resistor network. The answers to each of the questions can be either "none" or the numbers of one or more resistors.

Which resistors are connected in parallel with resistor 2? Which resistors are connected in parallel with resistor 9? Which resistors are connected in parallel with resistor 11? Which resistors are not connected in parallel with any other resistors?

Physics
1 answer:
makkiz [27]3 years ago
6 0

Answer:

  • parallel with 2: 1
  • parallel with 9: 5, 6, 8
  • parallel with 11: 4, 7
  • not in a parallel branch: 3

Explanation:

The attachments show the labeling of the network nodes. We found it convenient to do this so we could identify the specific nodes any given branch was attached to. Then the connections for each resistor were identified, and the list sorted to make it easier to see parallel connections.

Resistors 1 and 2 are in parallel

Resistors 5, 6, 8, 9 are in parallel

Resistors 4, 7, 11 are in parallel

No resistor has a parallel connection to resistor 3.

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A block of ice with mass 5.50 kg is initially at rest on a frictionless, horizontal surface. A worker then applies a horizontal
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Answer:

A) 3.13 m/s

B) 5.34 N

C) W = 26.9 J

Explanation:

We are told that the position as a function of time is given by;

x(t) = αt² + βt³

Where;

α = 0.210 m/s² and β = 2.04×10^(−2) m/s³ = 0.0204 m/s³

Thus;

x(t) = 0.21t² + 0.0204t³

A) Velocity is gotten from the derivative of the displacement.

Thus;

v(t) = x'(t) = 2(0.21t) + 3(0.0204t²)

v(t) = 0.42t + 0.0612t²

v(4.5) = 0.42(4.5) + 0.0612(4.5)²

v(4.5) = 3.1293 m/s ≈ 3.13 m/s

B) acceleration is gotten from the derivative of the velocity

a(t) = v'(t) = 0.42 + 2(0.0612t)

a(4.5) = 0.42 + 2(0.0612 × 4.5)

a(4.5) = 0.9708 m/s²

Force = ma = 5.5 × 0.9708

F = 5.3394 N ≈ 5.34 N

C) Since no friction, work done is kinetic energy.

Thus;

W = ½mv²

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Answer and Explanation:

NOTE: Magnetism means the magnetic property of a material that causes it to create a magnetic field, hence getting it attracted to a magnet.

EXPERIMENTAL PROCEDURE

1. Use a tape to attach a permanent magnet to the end of a ruler so that the magnet is facing away from the ruler. Don't cover the magnetic surface with the tape. ( Leave the magnet in its decorative casing.)

2. Place your metal objects in a row, and make predictions of which one of them will be attracted to the magnet and which will not.

3. Hold the magnet over each metals, and record which metals are attracted to the magnet. Go back over the

objects that were not affected by the magnet at least one more time to be sure you didn't miss any.

In this experiment, the independent variable is the magnetism of the magnet used. This is the independent variable because it remained unchanged and unaffected by the metals' magnetic properties all through the experiment.

While the dependent variable is the magnetism of the metals used. This is so because the magnetism of these metals varied and also because it is what is been measured in the experiment. Some were attracted to the magnet from very close range while others were attracted even at some centimeters away from the magnet which indicates that those metals have strong metallic properties.

7 0
3 years ago
What is the final temperature in degrees centigrade of 100g of water at 30 c if it is mixed with 50 g of water at 0?
natka813 [3]

Answer:

<em> The final temperature = 293 K or 20 °C</em>

Explanation:

Heat gained by water at 0°C = heat lost by water at 30°C

c₁m₁(T₃ - T₁) = c₂m₂(T₂-T₃).......................... Equation 1

Making T₃ the subject of the equation,

T₃ = (c₂m₂T₂ + c₁m₁T₁)/(c₁m₁+c₂m₂)............. Equation 2

Where m₁ =mass of water at 0°C, c₁ = specific heat capacity of water at 0°C , c₂ = specific heat capacity of water at 30°C,  m₂ = mass of water at 30°C, T₁ = initial temperature of water at 0°C, T₂ = initial temperature of water at 30°C, T₃ = final temperature.

<em>Given: m₁ = 50 g = (50/1000) kg = 0.05 kg, m₂ = 100 g = (100/1000) kg = 0.1 kg., T₁ = 0°C = 273 K, T₂ = 30°C = (30+273 )= 203 K</em>

<em>Constants: c₁ = c₂ = 4200 J/kg.K</em>

<em>Substituting these values into equation 2,</em>

<em>T₃ = (4200×0.1×303 + 4200×0.05×273)/(4200×0.1 + 4200×0.05)</em>

T₃ = (127260 + 57330)/(420+210)

T₃ = 184590/630

T₃ = 293 K.

<em>Therefore the final temperature = 293 K or 20 °C</em>

3 0
3 years ago
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