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HACTEHA [7]
2 years ago
14

Have my equation written out but struggling to solve. Can someone help me solve!

Physics
1 answer:
Nina [5.8K]2 years ago
6 0

The solution of this system is (i₁, i₂, i₃) = (- 10.852 A, 8.479 A, - 2.374 A). The <em>negative</em> signs indicate that <em>real</em> direction of the current is opposite than supposed.

<h3>How to find the missing current in a circuit</h3>

In this question we must make use of Kirchhoff's laws to find the values of the <em>missing</em> currents in the circuit presented in the picture. There are two rules according to Kirchhoff's laws:

  1. The sum of currents found at nodes of circuits is always equal to zero.
  2. The net sum of voltages in a <em>closed</em> loop of a circuit is always equal to zero.

Based on the information given by the picture, we have the following system of <em>linear</em> equations that describes the <em>entire</em> circuit:

i₃ = i₁ + i₂    

- i₁ · R₁ + ε₁ - i₁ · r₁ - i₁ · R₅ + ε₂ - i₂ · (r₂ + R₂) = 0      

ε₂ - i₂ · (r₂ + R₂) - i₃ · r₄ - ε₄ - i₃ · r₃ + ε₃ - i₃ · R₃ = 0      

- i₁ - i₂ + i₃ = 0                                                      (1)

(R₁ + r₁ + R₅) · i₁ + (r₂ + R₂) · i₂ = ε₁ + ε₂                (2)

(r₂ + R₂) · i₂ + (r₄ + r₃ + R₃) · i₃ = ε₂ + ε₃ - ε₄        (3)

If we know that R₁ = 5 Ω, r₁ = 0.10 Ω, R₅ = 20 Ω, r₂ = 0.50 Ω, R₂ = 40 Ω, r₄ =  0.20 Ω, r₃ = 0.05 Ω, R₃ = 78 Ω, ε₁ = 22 V, ε₂ = 49 V, ε₃ = 10.5 V and ε₄ = 33 V, then the currents flowing in the circuit are:

- i₁ - i₂ + i₃ = 0

25.1 · i₁ + 40.5 · i₂ = 71

25.1 · i₂ + 78.25 · i₃ = 26.5

The solution of this system is (i₁, i₂, i₃) = (- 10.852 A, 8.479 A, - 2.374 A). The <em>negative</em> signs indicate that <em>real</em> direction of the current is opposite than supposed.

To learn more on Kirchhoff's laws: brainly.com/question/6417513

#SPJ1

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Answer:

 F = - k (x-xo) a graph of the weight or applied force against the elongation obtaining a line already proves Hooke's law.

Explanation:

The student wants to prove hooke's law which has the form

          F = - k (x-xo)

To do this we hang the spring in a vertical position and mark the equilibrium position on a tape measure, to simplify the calculations we can make this point zero by placing our reference system in this position.

Now for a series of known masses let's get them one by one and measure the spring elongation, building a table of weight vs elongation,

we must be careful when hanging the weights so as not to create oscillations in the spring

we look for the mass of each weight

         W = mg

          m = W / g

and we write them in a new column, we make a graph of the weight or applied force against the elongation and it should give a straight line; the slope of this line is sought, which is the spring constant.

The fact of obtaining a line already proves Hooke's law.

5 0
2 years ago
An object with mass 100 kg moved in outer space. When it was at location &lt;8, -30, -4&gt; its speed was 5.5 m/s. A single cons
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Answer:

v = ( 6.41 i^ + 8.43 j^ + 2.63 k^ ) m/s

Explanation:

We can solve this problem using the kinematic relations, we have a three-dimensional movement, but we can work as three one-dimensional movements where the only parameter in common is time (a scalar).

X axis.

They indicate the initial position x = 8 m, its initial velocity v₀ = 5.5 m / s, the force Fx₁ = 220 N x₁ = 14 m, now the force changes to Fx₂ = 100 N up to the point xf = 17 m. The final speed is wondered.

As this movement is in three dimensions we must find the projection of the initial velocity in each axis, for this we can use trigonometry

the angle fi is with respect to the in z and the angle theta with respect to the x axis.

               sin φ = z / r

                Cos φ = r_p / r

               z = r sin φ

               r_p = r cos φ

the modulus of the vector r can be found with the Pythagorean theorem

               r² = (x-x₀) ² + (y-y₀) ² + (z-z₀) ²

               r² = (14-8) 2 + (-21 + 30) 2+ (-7 +4) 2

               r = √126

               r = 11.23 m

Let's find the angle with respect to the z axis (φfi)

                φ = sin⁻¹ z / r

                φ = sin⁻¹ ( \frac{-7+4}{11.23} )

                φ = 15.5º

Let's find the projection of the position vector (r_p)

                r_p = r cos φ

                r_p = 11.23 cos 15.5

                r_p = 10.82 m

This vector is in the xy plane, so we can use trigonometry to find the angle with respect to the x axis.

                 cos θ = x / r_p

                 θ = cos⁻¹ x / r_p

                 θ = cos⁻¹ ( \frac{14-8}{10.82})  

                 θ = 56.3º

taking the angles we can decompose the initial velocity.

               sin φ = v_z / v₀

               cos φ = v_p / v₀

               v_z = v₀ sin φ

               v_z = 5.5 sin 15.5 = 1.47 m / z

               v_p = vo cos φ

               v_p = 5.5 cos 15.5 = 5.30 m / s

                 

               cos θ = vₓ / v_p

                sin θ = v_y / v_p

                vₓ = v_p cos θ

                v_y = v_p sin θ

                vₓ = 5.30 cos 56.3 = 2.94 m / s

                v_y = 5.30 sin 56.3 = 4.41 m / s

 

                 

we already have the components of the initial velocity

                v₀ = (2.94 i ^ + 4.41 j ^ + 1.47 k ^) m / s

let's find the acceleration on this axis (ax1) using Newton's second law

                Fₓx = m aₓ₁

                aₓ₁ = Fₓ / m

                aₓ₁ = 220/100

                aₓ₁ = 2.20 m / s²

Let's look for the velocity at the end of this interval (vx1)

Let's be careful if the initial velocity and they relate it has the same sense it must be added, but if the velocity and acceleration have the opposite direction it must be subtracted.

                 vₓ₁² = v₀ₓ² + 2 aₓ₁ (x₁-x₀)

                 

let's calculate

                 vₓ₁² = 2.94² + 2 2.20 (14-8)

                 vₓ₁ = √35.04

                 vₓ₁ = 5.92 m / s

to the second interval

they relate it to xf

                   aₓ₂ = Fₓ₂ / m

                   aₓ₂ = 100/100

                   aₓ₂ = 1 m / s²

final speed

                    v_{xf}²  = vₓ₁² + 2 aₓ₂ (x_f- x₁)

                    v_{xf}² = 5.92² + 2 1 (17-14)

                    v_{xf} =√41.05

                    v_{xf} = 6.41 m / s

We carry out the same calculation for each of the other axes.

Axis y

acceleration (a_{y1})

                      a_{y1} = F_y / m

                      a_{y1} = 460/100

                      a_[y1} = 4.60 m / s²

the velocity at the end of the interval (v_{y1})

                      v_{y1}² = v_{oy}² + 2 a_{y1{ (y₁ -y₀)

                      v_{y1}2 = 4.41² + 2 4.60 (-21 + 30)

                      v_{y1} = √102.25

                       v_{y1} = 10.11 m / s

second interval

acceleration (a_{y2})

                      a_{y2} = F_{y2} / m

                      a_{y2} = 260/100

                      a_{y2} = 2.60 m / s2

final speed

                     v_{yf}² = v_{y1}² + 2 a_{y2} (y₂ -y₁)

                     v_{yf}² = 10.11² + 2 2.60 (-27 + 21)

                      v_{yf} = √ 71.01

                      v_{yf} = 8.43 m / s

here there is an inconsistency in the problem if the body is at y₁ = -27m and passes the position y_f = -21 m with the relationship it must be contrary to the velocity

z axis

 

first interval, relate (a_{z1})

                      a_{z1} = F_{z1} / m

                      a_{z1} = -200/100

                      a_{z1} = -2 m / s

the negative sign indicates that the acceleration is the negative direction of the z axis

the speed at the end of the interval

                    v_{z1}² = v_{zo)² + 2 a_{z1} (z₁-z₀)

                    v_{z1}² = 1.47² + 2 (-2) (-7 + 4)

                    v_{z1} = √14.16

                    v_{z1} = -3.76 m / s

second interval, acceleration (a_{z2})

                    a_{z2} = F_{z2} / m

                    a_{z2} = 210/100

                    a_{z2} = 2.10 m / s2

final speed

                    v_{fz}² = v_{z1}² - 2 a_{z2} | z_f-z₁)

                    v_{fz}² = 3.14² - 2 2.10 (-3 + 7)

                     v_{fz} = √6.94

                     v_{fz} = 2.63 m / s

speed is     v = ( 6.41 i^ + 8.43 j^ + 2.63 k^ ) m/s

5 0
3 years ago
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