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Assoli18 [71]
3 years ago
13

Two different conducting objects, initially separated, are connected by a conducting wire. When the wire is removed, can you con

clude that the two objects: (a) carry the same net charge; (b) are at the same potential; (c) have the same charge and the same potential; or (d) have neither the same charge nor potential? Explain.
Physics
1 answer:
Alex Ar [27]3 years ago
8 0

The two objects after (b) are at the same potential

Explanation:

When the two conducting objects are connected together, charge will be transferred from one object ot another until the two objects reach the same potential.

In fact, if there is still some residual potential difference between the two objects, there would still be a net electric field pushing the charges from one object to the other. Therefore, equilibrium is reached only when the net electric field between the two objects becomes zero, and this occurs only when the potential difference is zero, so when the two objects are at the same potential.

At the equilibrium, however, the charge on the two objects can be different. In fact, the capacity of an object is given by

C=\frac{Q}{V}

where Q is the charge stored by an object and V is its potential. It can be re-arranged as

Q=CV

The two objects are at same potential, V, however their  capacity C can be different since the two objects are not identical: therefore, their stored charge can be different.

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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3 years ago
A mass-spring system has k = 56.8 N/m and m = 0.46 kg.
andriy [413]

Answer:

A. \omega=11.1121\ rad.s^{-1}

B. f=1.7685\ Hz

C. T=0.5654\ s

Explanation:

Given:

  • spring constant, k=56.8\ N.m^{-1}
  • mass attached, m=0.46\ kg

A)

for a spring-mass system the frequency is given as:

\omega=\sqrt{\frac{k}{m} }

\omega=\sqrt{\frac{56.8}{0.46}}

\omega=11.1121\ rad.s^{-1}

B)

frequency is given as:

f=\frac{\omega}{2\pi}

f=\frac{11.1121}{2\pi}

f=1.7685\ Hz

C)

Time period of a simple harmonic motion is given as:

T=\frac{1}{f}

T=0.5654\ s

7 0
3 years ago
In Milgram's experiment:
SpyIntel [72]

Answer:

B. The "Learner" was working with Milgram.

Explanation:

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3 0
2 years ago
A steel pot has a bottom with a cross sectional area of .1m2 and has a thickness of 1cm. The pot is filled with boiling water an
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Answer:

840000 J/min

Explanation:

Area = A = 0.1 m²

Bottom of pot temperature = 200 °C

Thermal conductivity = k = 14 J/sm°C

Thickness = L = 1 cm = 0.01 m

Temperature of boiling water = 100 °C

From the law of heat conduction

Q = kAΔT/L

⇒ Q = 14×0.1×(200-100)/0.01

⇒ Q = 14000 J/s

Converting to J/minute

Q = 14000×60 = 840000 J/min

∴ Heat being conducted through the pot is 840000 J/min

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