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saw5 [17]
3 years ago
10

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

y state.
a. current
b. voltage
c. resistance
Physics
1 answer:
Fed [463]3 years ago
3 0

Answer:

a) Current.

Explanation:

The "junction rule" (Kirchhoff's first law) says that the value of the electric current entering any point in the circuit must be the same as the one leaving it. This comes as a consequence of the conservation of charge, the electric charge that comes in must come out (when speaking about currents, per unit time).

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It have 5 valence electrons

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Two forces, F? 1 and F? 2, act at a point, as shown in the picture. (Figure 1) F? 1 has a magnitude of 9.20 N and is directed at
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Answer:

a. Fx = -8.089 N b. Fy = 3.525 N c. 8.824 N d. 336.45°

Explanation:

Since F₁ = 9.2 N and acts at 57° above the negative axis in the second quadrant, its x-component is -F₁cos57° and its y- component is F₁sin57°

Since F₁ = 5.2 N and acts at 53.7° below the negative axis in the third quadrant, its x-component is -F₂cos53.7° and its y- component is -F₂sin53.7°

Part A

What is the x component Fx of the resultant force?

The x component of the resultant force Fx = -F₁cos57° + -F₂cos53.7° = -9.2cos57° + (-5.2cos53.7°) = (-5.011 - 3.078) N = -8.089 N

Part B

What is the y component Fy of the resultant force?

The y component Fy of the resultant force = F₁sin57° + -(F₂sin53.7°) = 9.2sin57° - 5.2sin53.7° = (7.716 - 4.191) N = 3.525 N

Part C  

What is the magnitude F of the resultant force?

The magnitude F of the resultant force = √(Fx² + Fy²)

F = √(-8.089² N + 3.525² N) = √65.432 + 12.426 = √77.858 = 8.824 N

Part D

What is the angle ? that the resultant force forms with the negative x axis?

The angle the resultant force makes with the negative x axis is given by

θ = tan⁻¹(Fy/Fx) = tan⁻¹(3.525/-8.089) = tan⁻¹-0.4358 = -23.55°.

To measure it from the negative x axis, we add 360. So, our angle = 360 -23.55 = 336.45°

7 0
3 years ago
A flat loop of wire consisting of a single turn of cross-sectional area 8.00 cm2 is perpendicular to a magnetic field that incre
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Complete Question

A flat loop of wire consisting of a single turn of cross-sectional area 8.00 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 1.60 T in 0.99 s. What is the resulting induced current if the loop has a resistance of 1.20  \ \Omega

Answer:

The current is   I =  0.0007 41 \ A

Explanation:

From the question we are told that

   The  area is  A = 8.00 \ cm^2  = 8.0 *10^{-4} \  m^2

   The initial magnetic field at t_o = 0 \ seconds  is B_i = 0.500 \ T

   The magnetic field at t_1 = 0.99 \ seconds is  B_f  = 1.60 \ T

     The resistance is  R = 1.20  \ \Omega

Generally the induced emf is mathematically represented as

      \epsilon  =  A * \frac{B_f - B_i }{ t_f - t_o }

=>   \epsilon  =  8.0 *10^{-4} * \frac{1.60  - 0.500 }{ 0.99- 0  }

=>   \epsilon  = 0.000889 \ V

Generally the current induced is mathematically represented as

     I = \frac{\epsilon}{R }

=>  I = \frac{0.000889}{ 1.20 }  

=>  I =  0.0007 41 \ A  

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