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Sergeu [11.5K]
3 years ago
8

A neutron has

Physics
2 answers:
zavuch27 [327]3 years ago
6 0

Answer:

D

Explanation:

a neutron does not have a positive nor negative charge it remains neutral

Bad White [126]3 years ago
6 0
A neutron had neither a positive or negative charge so the answer is D
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An inductor in an LC circuit has a maximum current of 2.4 A and a maximum energy of 56 mJ.
Harrizon [31]

Answer:

The energy stored in the capacitor, when the current in the inductor is 1.2 A, is 41.6 mJ.

Explanation:

In a LC oscillating circuit, the energy is stored in the electric field (between the plates of the capacitor) and in the magnetic field (surrounding the wires of the inductor).

At any time, the sum of both energies can be expressed as follows:

E = 1/2 Q² / C   +  1/2 L I²

In this type of circuit, energy oscillates, which means that it is exchanging between both fields all time.

When the capacitor is completely discharged, all the energy is stored in the magnetic field, and at that time, the current is maximum.

The total energy, when I is maximum, can be written as follows:

E = 1/2 L I² (1)

In our case, when I= 2.4A, E= 56 mJ.

So, we can find out the value of L, which will allow us to know the value of the magnetic energy at any time, having the value of the instantaneous current.

Solving for L in (1):

L = 2 *.56 mJ / (2.4)² A² = 20 mH

The next step is getting the value of the energy stored in the inductor, when I = 1.2 A, as follows:

Em = 1/2 *20 mH.* (1.2)² A² = 14.4 mJ

As the total energy must be always the same, i.e., 56 mJ, the energy stored in the capacitor, assuming no losses, must be the difference between the total energy and the one stored in the magnetic field:

Ec = 56 mJ - 14.4 mJ = 41.6 mJ

3 0
3 years ago
A kettle is rated at 1 kW, 220 V. Calculate the working resistance of the kettle.
Anna [14]

Explanation:

Power of electric kettle, P = 1 kW

Voltage, V = 220 V

(a) Electric power is given by the formula as follows :

P=\dfrac{V^2}{R}

R is resistance

R=\dfrac{V^2}{P}\\\\R=\dfrac{(220)^2}{10^3}\\\\R=48.4\ \Omega

(b) When connected to a 220 V supply, it takes 3 minutes for the water in the kettle to reach boiling point.

Energy supplied is given by :

E=P\times t

P is power, P=\dfrac{V^2}{R}

E=\dfrac{V^2}{R}t\\\\E=\dfrac{(220)^2}{48.4}\times 180\\\\E=180000\ J\\\\E=180\ kJ

5 0
3 years ago
Which of the following is true about a planet orbiting a star in uniform circular
Mnenie [13.5K]

The velocity vector of the planet points toward the center of the  circle is the following is true about a planet orbiting a star in uniform circular  motion.

A. The velocity vector of the planet points toward the center of the  circle.

<u>Explanation:</u>

Motion of the planet around the star is mentioned to be uniform and around a circular path. Objects in uniform circular motion motion has constant angular speed but the velocity of the object will not remain constant. Since the planet is in circular motion the direction of velocity vector at a particular point is tangential to the circular path at that particular point.

Thus at every point, the direction of velocity vector changes and this means the velocity is never constant. The objects in uniform circular motion has centripetal acceleration which means that velocity vector of the planet points toward the center of the  circle.

7 0
3 years ago
Select the correct answer.
Simora [160]

Answer:

A. 2.36 Newtons

Explanation:

F = GmM/d²

F = 6.673 x 10⁻¹¹(1)(5.98 x 10²⁴) / (1.3 x 10⁷)²

F = 2.36121...

Very poor question design.

  mass of box... 1 significant digit

        distance... 2 significant digits

mass of earth... 3 significant digits

     value of G... 4 significant digits

Answer precision to 3 significant digits is not justifiable

7 0
3 years ago
The ground state energy of an oscillating electron is 1.23 eV. How much energy must be added to the electron to move it to the t
Vikki [24]

Answer:

  • The energy that must be added to the electron to move it to the third excited state is  -1.153 eV
  • The energy that must be added to the electron to move it to the fourth excited state is  -1.181 eV

Explanation:

Given;

Energy of electron in ground state (n = 1 ) = 1.23 eV

E₁ = 1.23 eV

Eₙ = E₁ /n²

where;

E₁ is the energy of the electron in ground state

n is the energy level,

For third excited state, n = 4

E₄ = E₁ /4²

E₄ = (1.23 eV) / 16

E₄ = 0.077 eV

Change in energy level, = E₄ - E₁ = 0.077 eV - 1.23 eV = -1.153 eV

The energy that must be added to the electron to move it to the third excited state is  -1.153 eV

For fourth excited state, n = 5

E₅ = E₁ /5²

E₄ = (1.23 eV) / 25

E₄ = 0.049 eV

Change in energy level, = E₅ - E₁ = 0.049 eV - 1.23 eV = -1.181 eV

The energy that must be added to the electron to move it to the fourth excited state is  -1.181 eV

5 0
3 years ago
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