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notka56 [123]
3 years ago
8

The junction rule describes the conservation of which quantity? Note that this rule applies only to circuits that are in a stead

y state.
Physics
2 answers:
denis-greek [22]3 years ago
5 0

Explanation:

The junction rule says that the sum of the currents going into a junction must equal the sum of the currents leaving a junction.  This describes the conservation of current.

Reptile [31]3 years ago
5 0

"Kirchhoff's Junction Rule" and "Kirchhoff's Current Law" are the same thing.

The rule is that the amount of current flowing IN to any point in a circuit is equal to the amount of current flowing OUT of the same point.  EVEN if the point is a complicated 'junction' with a thousand different paths in and out of it  !

This whole idea is pretty obvious if you understand the law of <em>Conservation of Electric Charge</em>.  Charge can't be created out of nothing, and it can't disappear or be destroyed.  So whenever you see charge appear, it had to come from somewhere; and whenever you see charge disappear, it had to go somewhere.

That's exactly what's happening at the 'junction' in an electric circuit.  Current into the junction brings charge into it.  The charge can't pile up there, and it can't just disappear, so it has to leave the junction in a current that's exactly the same size as the current that brought it there.

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Why is acceleration not constant near the speed of light
dmitriy555 [2]

Answer:

because when an object approaches the speed of light, it's mass starts to increase exponentially, and would be infinite at the speed of light. It would therefore require MORE than an infinite amount of energy to accelerate even a single electron to the speed of light

4 0
3 years ago
A 1000 kg elevator accelerates upward at 1.0 m/s2 for 10 m, starting from rest. a. How much work does gravity do on the elevator
cricket20 [7]

Answer:

a)= 98kJ

b)=108kJ

c) = 10kJ

Explanation:

a. The work that is done by gravity on the elevator is:

Work = force * distance  

= mass * gravity * distance

= 1000 * 9.81 * 10  

= 98,000 J

= 98kJ

b)The net force equation in the cable

T - mg = ma

T = m(g+a)

T = 1000(9.8 + 10)

T = 10800N

The work done by the cable is

W = T × d

= 10800N × 10

= 108000

=108kJ

c) PE at 10m = 1000 * 9.81 * 10 = 98,100 J  

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108,100 J = KE + 98,100 J  

KE = 10,000 J

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=

7 0
3 years ago
Read 2 more answers
Having aced your Physics 2111 class, you get a sweet summer-job working in the International Space Station. Your room-mate, Cosm
Sphinxa [80]

Answer:

a

The speed is   s =  5.857 m/s

b

The distance is  D = 22.4  \  m

Explanation:

From the question we are told that

     The speed of the banana is  v =  16 \ m/s

   The distance from my  location is  d =  8.2  \ m  

     The time taken is  t = 1.4 \ s

The speed of the ice cream is

          s =  \frac{d}{t}

substituting values

        s =  \frac{8.4}{1.4}

        s =  5.857 m/s

The distance of separation between i and Valdimir is the same as the distance covered by the banana

   So  

          D =  v * t

substituting values

        D = 16 *  1.4

        D = 22.4  \  m

     

3 0
3 years ago
Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist
Leno4ka [110]

Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

i(t)=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}}), where imax=\frac{E}{R}

Given that, at i(2)=\frac{imax}{2} =\frac{E}{2R}

⇒\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})

⇒\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}

⇒\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}

Applying logarithm on both sides,

⇒log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}

⇒log(2)=\frac{2}{\frac{L}{R}}

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Now substitute i(t)=\frac{3}{4}imax=\frac{3E}{4R}

⇒\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})

⇒\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}

⇒\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}

Applying logarithm on both sides,

⇒log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}

⇒log(4)=\frac{t}{\frac{L}{R}}

⇒t=log4\frac{L}{R}

now subs. \frac{L}{R}=\frac{2}{log2}

⇒t=log4\frac{2}{log2}

also log4=log2^{2}=2log2

⇒t=2log2\frac{2}{log2}

⇒t=4

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3 years ago
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