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Mama L [17]
3 years ago
8

Two concentric current loops lie in the same plane. The smaller loop has a radius of 3.6 cm and a current of 12 A. The bigger lo

op has a current of 20 A . The magnetic field at the center of the loops is found to be zero. What is the radius of the bigger loop?
Physics
1 answer:
puteri [66]3 years ago
7 0

Answer:

the radius of bigger loop = 6 cm

Explanation:

given,

two concentric current loops

smaller loop radius = 3.6 cm

]current in smaller loop = 12 A

current in the bigger loop = 20 A

magnetic field at the center of loop = 0

Radius of the bigger loop = ?

B_t = B_1 + B_2

0 = \dfrac{\mu_0I_1}{2R_1} +\dfrac{\mu_0I_2}{2R_2}

now, on solving

\dfrac{I_1}{R_1} = \dfrac{I_2}{R_2}

R_2 = I_2\dfrac{R_1}{I_1}

       = 20\times \dfrac{3.6}{12}

       = 6 cm

hence, the radius of bigger loop = 6 cm

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s2008m [1.1K]

Answer:

A. Power generated by meteor = 892857.14 Watts

Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.

B. Workdone = 981000 J

Power required = 19620 Watts

Note: The question is incomplete. A similar complete question is given below:

A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's atmosphere. As the drag force slows the meteor down, its kinetic energy is converted to thermal energy, leaving a glowing trail across the sky. A typical meteor has a surprisingly small mass, but what it lacks in size it makes up for in speed. Assume that a meteor has a mass of 1.5 g and is moving at an impressive 50 km/s, both typical values. What power is generated if the meteor slows down over a typical 2.1 s? Can you see how this tiny object can make a glowing trail that can be seen hundreds of kilometers away? 61. a. How much work does an elevator motor do to lift a 1000 kg elevator a height of 100 m at a constant speed? b. How much power must the motor supply to do this in 50 s at constant speed?

Explanation:

A. Power = workdone / time taken

Workdone = Kinetic energy of the meteor

Kinetic energy = mass × velocity² / 2

Mass of meteor = 1.5 g = 0.0015 kg;

Velocity of meteor = 50 km/s = 50000 m/s

Kinetic energy = 0.0015 × (50000)² / 2 = 1875000 J

Power generated = 1875000/2.1 = 892857.14 Watts

Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.

B. Work done by elevator against gravity = mass × acceleration due to gravity × height

Work done = 1000 kg × 9.81 m/s² × 100 m

Workdone = 981000 J

Power required = workdone / time

Power = 981000 J / 50 s

Power required = 19620 Watts

Therefore, the motor must supply a power of 19620 Watts in order to lift a 1000 kg to a height of 100 m at a constant speed in 50 seconds.

6 0
3 years ago
A block with a mass of 33.0 kg is pushed with a horizontal force of 150 N. The block moves at a constant speed across a level, r
lord [1]

The work done is given by 742.5 J while the coefficient of kinetic friction between the block and the surface is 0.46.

<h3>What is the work done?</h3>

The work done is given by the use of the formula;

W = F * x

Where;

F = force applied

x = distance covered

W = 150 N *  4.95 m = 742.5 J

Now;

The coefficient of kinetic friction is given by;

μ = F/mg

μ = 150/ 33 * 9.8

μ = 0.46

Learn more about work done:brainly.com/question/13662169

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4 0
2 years ago
Jaiden is writing a report about the structure of the atom. In her report, she says that the atom has three main parts and two s
natulia [17]
No, because an atom consists of two main parts and three subatomic particles - protons, neutrons, electrons. Each one is smaller than an atom, therefore they are subatomic particles. An atom only requires protons and electrons to be an atom - Hydrogen has 1 proton and 1 electron. Neutrons DO NOT affect the overall charge of the atom, and only increase the atomic mass. 



4 0
3 years ago
Read 2 more answers
1 A football player runs directly down the field for 35 m before turning to the right at
ehidna [41]

Answer:

The magnitude of displacement is 56.54 m

The direction of the displacement is along the line joining the two vectors.

Explanation:

The resultant displacement is always the line joining the initial and final position of the vectors.

As in figure,

              the vector AB = 35 m

              the vector BC =  15 m

              the angle between AB and AC = 25'  (minutes)

              the resultant vector AC = ?

The resultant vector is given by the formula

             AC² = AB² + BC² + 2 AB BC Cos θ

Substituting the values in the equations,

             AC²  = 35² + 15² + 2 x 35 x 15 x Cos 25'

                      = 56.54

Therefore, the magnitude of displacement is 56.54 m

The direction of the displacement is along the line joining the two vectors.

7 0
3 years ago
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72) What is the freezing point (°C) of a solution prepared by dissolving 11.3 g of Ca(NO3)2 (formula weight = 164 g/mol) in 115
Rina8888 [55]

<u>Answer:</u> The freezing point of solution is -3.34°C

<u>Explanation:</u>

Vant hoff factor for ionic solute is the number of ions that are present in a solution. The equation for the ionization of calcium nitrate follows:

Ca(NO_3)_2(aq.)\rightarrow Ca^{2+}(aq.)+2NO_3^-(aq.)

The total number of ions present in the solution are 3.

  • To calculate the molality of solution, we use the equation:

Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

Where,

m_{solute} = Given mass of solute (Ca(NO_3)_2) = 11.3 g

M_{solute} = Molar mass of solute (Ca(NO_3)_2) = 164  g/mol

W_{solvent} = Mass of solvent (water) = 115 g

Putting values in above equation, we get:

\text{Molality of }Ca(NO_3)_2=\frac{11.3\times 1000}{164\times 115}\\\\\text{Molality of }Ca(NO_3)_2=0.599m

  • To calculate the depression in freezing point, we use the equation:

\Delta T=iK_fm

where,

i = Vant hoff factor = 3

K_f = molal freezing point depression constant = 1.86°C/m.g

m = molality of solution = 0.599 m

Putting values in above equation, we get:

\Delta T=3\times 1.86^oC/m.g\times 0.599m\\\\\Delta T=3.34^oC

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

\Delta T=\text{freezing point of water}-\text{freezing point of solution}

\Delta T = 3.34 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

3.34^oC=0^oC-\text{Freezing point of solution}\\\\\text{Freezing point of solution}=-3.34^oC

Hence, the freezing point of solution is -3.34°C

4 0
4 years ago
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