Three wires are connected at a branch point. One wire carries a positive current of 18 A into the branch point, and a second wire carries a positive current of 7 A away from the branch point. Find the current carried by the third wire into the branch point.

Answer:

 law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time.

Answer:

The correct solution is "37.5 km".

Explanation:

Given:

Distance between the trains,  

d = 75 km

Speed of each train,

= 15 km/h

The relative speed will be:

= [tex]15 + (-15)[/tex]

= [tex]30 \ km/h[/tex]

The speed of the bird,

V = 15 km/h

Now,

The time taken to meet will be:

[tex]t=\frac{Distance}{Relative \ speed}[/tex]

  [tex]=\frac{75}{30}[/tex]

  [tex]=2.5 \ h[/tex]

hence,

The distance travelled by the bird in 2.5 h will be:

⇒ [tex]D = V t[/tex]

        [tex]=15\times 2.5[/tex]

        [tex]=37.5 \ km[/tex]

Answer:

About 37.70 radians.

Explanation:

1 revolution = 2[tex]\pi[/tex] radians

∴ 6 revolutions = (6)(2[tex]\pi[/tex] radians)

6 revolutions = 37.6991 or ≈ 37.70 radians

Answer:

If a neutral atom gains electrons, then it will become negatively charged. If a neutral atom loses electrons, then it become positively charged.

Answer:

[tex]v=(6ti+6k)\ m/s[/tex]

Explanation:

Given that,

The position of a particle is given by :

[tex]r(t) = (3.0 t^2 i + 5.0j+ 6.0 tk) m[/tex]

Let us assume we need to find its velocity.

We know that,

[tex]v=\dfrac{dr}{dt}\\\\=\dfrac{d}{dt}(3.0 t^2 i + 5.0j+ 6.0 tk) \\\\=(6ti+6k)\ m/s[/tex]

So, the velocity of the particle is [tex](6ti+6k)\ m/s[/tex].

Answer:

[tex]t_2=10[/tex]

Explanation:

From the question we are told that:

Velocity of both spaceships [tex]V=0.8c[/tex]

Time [tex]t_1=6[/tex]

Generally the equation for time of spaceship 2 through earth is mathematically given by

[tex]t_2=\frac{t_1}{\sqrt{1-v^2}}[/tex]

[tex]t_2=\frac{6}{\sqrt{1-0.8^2}}[/tex]

[tex]t_2=10[/tex]

Answer:

[tex]d=1.29*10^{-6}m[/tex]

Explanation:

From the question we are told that:

Distance of wall from CD [tex]D=1.4[/tex]

Second bright fringe [tex]y_2= 0.803 m[/tex]

Let

Strontium vapor laser has a wavelength \lambda= 431 nm=>431 *10^{-9}m

Generally the equation for Interference is mathematically given by

[tex]y=frac{n*\lambda*D}{d}[/tex]

Where

[tex]d=\frac{n*\lambda*D}{y}[/tex]

[tex]d=\frac{2*431 *10^{-9}m*1.4}{0.803}[/tex]

[tex]d=1.29*10^{-6}m[/tex]

Answer:

i think bro is becuse they want to get money also paying to this app  if i put paying right im latin but that is my opinion and also becuse they want we learn somethings

Explanation:

Answer:

the answer is push example: she thrust her hand into her pocket

Explanation:

some thing inside a resistor

Answer:

It will travel Vx * t = 30 m/s * 2 s = 60 m

Answer:

The mass of the second block=0.457 kg

Explanation:

We are given that

m1=1.5 kg

v1=1.3m/s

v2=4.3 m/s

V=2.0 m/s

We have to find the mass of the second block.

[tex]m_1v_1+m_2v_2=(m_1+m_2)V[/tex]

Let m2=m

Substitute the values

[tex]1.5(1.3)+m(4.3)=(1.5+m)(2)[/tex]

[tex]1.95+4.3m=3+2m[/tex]

[tex]4.3m-2m=3-1.95[/tex]

[tex]2.3m=1.05[/tex]

[tex]m=\frac{1.05}{2.3}[/tex]

[tex]m=0.457 kg[/tex]

Hence,  the mass of the second block=0.457 kg

Answer:

Magnitude = 140 x 10⁻¹² Cm

Direction = upwards

Explanation:

A pair of two equal and opposite point charges forms an electric dipole.

The magnitude of the moment of such dipole is the product of the magnitude of any of the charges (since the charges are the same in magnitude) and the distance of separation between them. i.e

p = q x d          ----------(i)

Where;

p = dipole moment

q = magnitude of any of the charges

d = distance between the charges.

The direction of the dipole moment is from the negative charge to the positive charge.

(a) From the question, the charges are +35 nC and -35 nC, and the distance between them is 4.00mm.

This implies that;

q = 35 nC = 35 x 10⁻⁹C

d = 4.00mm = 4.0 x 10⁻³ m

Substitute the values of q and d into equation (i) to give;

p = 35 x 10⁻⁹C x 4.00 x 10⁻³ m

p = (35 x 4.0) x (10⁻⁹ x 10⁻³) C m

p = 140 x 10⁻¹² Cm

The magnitude of the dipole moment is 140 x 10⁻¹² Cm

(b) From the question, the +35nC charge is above the -35nC charge on a vertical line as shown below;

                         o   +35nC

                          |

                          |

                          |

                          |

                          |

                          |

                         o    -35nC

Since the direction should point from the negative charge to the positive charge, this means that the direction of the dipole moment of the two charges is upwards (due North).

                         o   +35nC

                         ↑

                          |

                          |

                          |

                          |

                          |

                          |

                         o    -35nC

Answer:

It's electricity produced from hydropower. It's also a form of energy that controls the power of water motion.

Explanation:

One pro about hydroelectric power is that it's renewable energy. But one con about hydroelectric power is that it can impact the environment in a negative way.

Answer:

0.98 g/m

Explanation:

Note: Since Tension and frequency are constant,

Applying,

F₁²M₁ = F₂²M₂............... Equation 1

Where F₁ = Frequency of the G string, F₂ = Frequency of the A string, M₁ = mass density of the G string, M₂ = mass density of the A string.

make M₂ the subject of the equation

M₂ = F₁²M₁/F₂²............... Equation 2

From the question,

Given: F₁ = 196 Hz, M₁ = 0.31 g/m, F₂ = 110 Hz

Substitute these values into equation 2

M₂ = 196²(0.31)/110²

M₂ = 0.98 g/m

Answer:

[tex]\theta=30 \textdegree[/tex]

Explanation:

From the question we are told that:

Current [tex]I=2.0A[/tex]

Length [tex]L=0.60m[/tex]

Magnetic field [tex]B=0.30T[/tex]

Force [tex]F=0.18N[/tex]

Generally the equation for Force is mathematically given by

[tex]F = BIL sin\theta[/tex]

[tex]sin\theta=\frac{F}{BIL}[/tex]

[tex]\theta=sin^{-1}\frac{0.18}{0.3*2*0.6}[/tex]

[tex]\theta=30 \textdegree[/tex]

Answer:

opinion a

Explanation:

the si units of distance is metre (m)

Answer:

A

Explanation:

Answer:

  E = 1.25 MV / m

Explanation:

For this exercise let's use Newton's second law

          F = m a

where the force is electric

          F = q E

we substitute

          q E = m a

          E = m a / q

indicate there are 500,000 excess electrons

          q = 500000 e

          q = 500000 1.6 10⁻¹⁹

          q = 8 10⁻¹⁴ C

the mass is m = 10⁻¹¹ kg and the acceleration a = 10⁴ m / s²

let's calculate

          E = 10⁻¹¹ 10⁴ / 8 10⁻¹⁴

          E = 0.125 10⁷ V / m = 1.25 10⁶ V / m

          E = 1.25 MV / m

Answer:

35.047 m

Explanation:

The time it takes the lead ball to reach the surface of the water is

s = ut+gt²/2............. Equation 1

Where t = time it takes the lead ball to reach the surface of water, u = initial velocity of the lead ball, g = acceleration due to gravity, s = heigth.

From the question,

Given: s = 5.20 m, u = 0 m/s (dropped from a height)

Constant: g = 9.8 m/s²

5.2 = 0+9.8t²/2

t² = (5.2×2)/9.8

t² = 10.4/9.8

t² = 1.06

t = √(1.06)

t = 1.03 s

Hence, time taken for the lead ball to reach the bottom of the lake is

t' = 4.5-1.03

t' = 3.47 seconds

v² = u²+2gs............... Equation 2

Where v = final velocity of the lead ball

Substitute into equation 2

v² = 0+2(9.8)(5.2)

v² = 101.92

v = √(101.92)

v = 10.1 m/s

Therefore, depth of the lake is

D = vt'

D = 10.1(3.47)

D = 35.047 m

Answer:

A. Equals to that of the smaller sphere

B. 3 times less than that of the smaller sphere

Explanation:

(a) Equals to that of the smaller sphere

The potential of an isolated metal sphere, with charge Q and radius R, is kQ=R, so a sphere with charge 3Q and radius 3R has the same potential

b) 3 times less than that of the smaller sphere

However, the electric field at the surface of the smaller sphere is ?=? 0 = kQ=R2 , so tripling Q and R reduces the surface field by a factor of 1/3

Answer:

The answer is "Option b, c, and a".

Explanation:

Here that the earth pulls on the phone, as it will accelerate towards Earth when we drop it.

We now understand the effects of gravity:

[tex]F \propto M\\\\F\propto \frac{1}{r^2}\\\\or\\\\F \propto \frac{M}{r^2}\\\\Sun (\frac{M}{r^2}) = \frac{10^{28}}{(10^9)^2} = 10^{10}[/tex]

The force of the sun is, therefore, [tex]10^{10}[/tex] times greater and the proper sequence, therefore, option steps are:

b. Pull-on phone from earth

c. Pull-on phone from sun

a. Pull phone from you

Answer:

a)[tex]L=0.00142H[/tex]

b) [tex]C=2.65*10^{-12}[/tex]

Explanation:

From the question we are told that:

Frequency[tex]F=13kHz[/tex]

Current [tex]I=0.14A[/tex]

Capacitor[tex]C_e=1.4*10^{-5}J[/tex]

Generally the equation for Energy in the inductor is mathematically given by

Where L is now subject

[tex]L=\frac{2C_e}{I^2}[/tex]

[tex]L=\frac{2*1.4*10^{-5}}{(0.14)^2}[/tex]

[tex]L=0.00142H[/tex]

Generally the equation for Value of Capacitor is mathematically given by

[tex]C=\frac{1}{(2 \pi f)^2} L[/tex]

[tex]C=\frac{1}{(2 3.142 13*10^3Hz)^2} *0.00142[/tex]

[tex]C=2.65*10^{-12}[/tex]

Answer:

The answer would be 7.5 Hz.

Assuming ideal conditions, Boyle's law says that

P₁ V₁ = P₂ V₂

where P₁ and V₁ are the initial pressure and temperature, respectively, and P₂ and V₂ are the final pressure and temperature.

So you have

(455 mm Hg) (56.5 m³) = (632 mm Hg) V₂

==>   V₂ = (455 mm Hg) (56.5 m³) / (632 mm Hg) ≈ 40.7 m³

Answer:

The right answer is "0.273 m".

Explanation:

Given:

Power (P),

[tex]\frac{1}{f} = 2D[/tex]

Near point,

u = 0.6 m

As we know,

⇒ [tex]\frac{1}{v} -\frac{1}{u}=\frac{1}{f} = 2[/tex]

By substituting the values, we get

⇒ [tex]\frac{1}{v} -\frac{1}{0.6} =2[/tex]

            [tex]\frac{1}{v}=2+\frac{1}{0.6}[/tex]

            [tex]\frac{1}{v} =\frac{1.2+1}{0.6}[/tex]

            [tex]\frac{1}{v}=\frac{2.2}{0.6}[/tex]    

By applying cross-multiplication, we get

          [tex]0.6=2.2 \ v[/tex]

            [tex]v = \frac{0.6}{2.2}[/tex]

      [tex]S_{near} = 0.273 \ m[/tex]

The disadvantages of simple cell are: It is not rechargeable. The battery needs to be disposed of after all the power has been used up. It can't produce electricity anymore. That is why, why think before you use a simple cell.​

A battery designed to be used only once is called a simple cell.  Small gadgets used in the house are frequently powered by simple cells.

The benefits of a simple cell include:

Among the drawbacks of a simple cell are:

Learn more about cell here:

https://brainly.com/question/30046049

#SPJ2

a) The rotational kinetic energy of the airplane propeller is 1792152.287 joules.

b) The angular speed of the airplane propeller is approximately 3348.631 revolutions per minute.

Let consider the airplane propeller a rigid body, the rotational kinetic energy of the propeller (K), in joules, is described by the following formula:

K = 0.5 · I · ω²  (1)

Where:

In addition, the moment of inertia of a slender rod rotating around its center is:

I = 0.0833 · M · L²   (2)

Where:

a) If we know that M = 100 kg, L = 2.16 m and ω = 303.687 rad/s, then the rotational kinetic energy of the propeller is:

K = 0.5 · [0.0833 · (100 kg) · (2.16 m)²] · (303.687 rad/s)²

K = 1792152.287 J

The rotational kinetic energy of the airplane propeller is 1792152.287 joules. [tex]\blacksquare[/tex]

b) By (1) and (2) we know that the mass of the propeller is inversely proportional to the square of the angular speed. Therefore, we have the following relationship:

[tex]M_{o}\cdot \omega_{o}^{2} = M_{f}\cdot \omega_{f}^{2}[/tex]

[tex]\omega_{f} = \sqrt{\frac{M_{o}}{M_{f}} }\cdot \omega_{o}[/tex]   (3)

If we know that [tex]\omega_{o} = 2900\,\frac{rev}{min}[/tex], [tex]M_{o} = 100\,kg[/tex] and [tex]M_{f} = 75\,kg[/tex], then the angular speed of the airplane propeller is:

[tex]\omega_{f} = \left(2900\,\frac{rev}{min} \right)\cdot \sqrt{\frac{100\,kg}{75\,kg} }[/tex]

[tex]\omega_{f} \approx 3348.631\, \frac{rev}{min}[/tex]

The angular speed of the airplane propeller is approximately 3348.631 revolutions per minute. [tex]\blacksquare[/tex]

To learn more on rotational kinetic energy, we kindly invite to check this verified question: https://brainly.com/question/20261989

Answer:

K = 2 10⁻⁸ J

Explanation:

Let's solve this exercise in parts, we start by finding the charge on each plate of the capacitor

          C = Q / ΔV

           C = ε₀ A / d

          ε₀ A / d = Q / ΔV

          Q = ε₀ A ΔV / d        (1)

indicate the potential difference ΔV₁ = 12 V, the distance between the plates d₁ = 3 mm = 0.003 m,  

as the power supply is disconnected and the capacitor is ideal the charge remains constant

in the second part we separate the plates at d₂ = 5 mm = 0.005 m, using equation 1

          ΔV₂ = [tex]\frac{Q d_2}{ \epsilon_o A}[/tex]

we substitute the equation for Q

         ΔV₂ = [tex]\frac{d_2}{\epsilon_o A} \ \frac{\epsilon_o A \Delta V }{d_1}[/tex]

         ΔV₂ = [tex]\frac{d_2}{d_1} \ \Delta V_1[/tex]

in the third part we use the concepts of energy

starting point. Test charge near positive plate

          Em₀ = U = q ΔV₂

final point. Test charge near negative plate

          Em_f = K

energy is conserved

          Em₀ = Em_f

          q ΔV₂ = K

          K = q ΔV₁ [tex]\frac{d_2}{d_1}[/tex]

we calculate

          K = 1 10⁻⁹  12  0.005/0.003

          K = 2 10⁻⁸ J

The question is incomplete. The complete question is :

A wire 0.6 m long and with mass m = 11 g is positioned horizontally near the earth's surface and perpendicular to a horizontal magnetic field of magnitude B = 0.4 T. What current I must flow through the wire in order that the wire accelerate neither upwards nor downwards? The magnetic field is directed into the page.

Solution :

Given :

Length of the wire, L = 0.6 m

Mass of the wire length, m = 11 g

                                             = [tex]11 \times 10^{-3}[/tex] kg

Magnetic field , B = 0.4 T

Know we know that :

ILB = mg

or [tex]$I=\frac{mg}{BL}$[/tex]

 [tex]$I= \frac{(11 \times 10^{-3})(9.81)}{(0.4)(0.6)}$[/tex]

 [tex]I=0.44963\ A[/tex]

 [tex]I = 449.63 \ mA[/tex]