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trapecia [35]
3 years ago
15

A toroidal solenoid has an inner radius of 12.0 cm and an outer radius of 15.0 cm . It carries a current of 1.50 A . Part A How

many equally spaced turns must it have so that it will produce a magnetic field of 3.75 mT at points within the coils 14.0 cm from its center? Enter your answer numerically.
Physics
2 answers:
julia-pushkina [17]3 years ago
4 0

Answer:

The number of turns are 1750.

Explanation:

Given that,

The inner radius of a toroid, r = 12 cm

Outer radius, r' = 15 cm

Magnetic field, B=3.75\ mT=3.75\times 10^{-3}\ T

To find,

It is required to find the number of turns that will produce the magnetic field of 3.75 mT at points within the coils 14.0 cm from its center.

The formula for the magnetic field at some distance from its center is given by :

B=\dfrac{\mu_o NI}{2\pi R}\\\\N=\dfrac{2\pi R B}{\mu_o I}\\\\N=\dfrac{2\pi \times 0.14\times 3.75\times 10^{-3}}{4\pi \times 10^{-7}\times 1.5}

N=1750

So, the number of turns are 1750.

AleksAgata [21]3 years ago
3 0

Answer:

The number of turns, N = 1750

Explanation:

It is given that,

The inner radius of a toroid, r = 12 cm

Outer radius, r' = 15 cm

The magnetic field at points within the coils 14 cm from its center is, B=3.75\ mT=3.75\times 10^{-3}\ T

R = 14 cm = 0.14 m

Current, I = 1.5 A

The formula for the magnetic field at some distance from its center is given by :

B=\dfrac{\mu_o NI}{2\pi R}

N=\dfrac{2\pi R B}{\mu_o I}

N=\dfrac{2\pi \times 0.14\times 3.75\times 10^{-3}}{4\pi \times 10^{-7}\times 1.5}

N = 1750

So, the number of turns must have in a toroidal solenoid is 1750. Hence, this is the required solution.

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Kobotan [32]

Answer:

Final velocity of white ball is 0m/s

Final velocity of black ball is 3.09m/s

Explanation:

An elastic collision is one that conserves internal kinetic energy

An internal kinetic energy is the sum of kinetic energies of objects in the system

Initial kinetic energy of white ball is Vi1 = 3.09m/s

Final kinetic energy of white ball is Vf1 = ?

Initial kinetic energy of black ball is Vi2 = 0m/s

Final kinetic energy of black ball is Vf2 = ?

m1 = 1.49kg mass of white ball

m2 = 1.49kg mass of black ball

The formula to calculate internal kinetic energy is

1/2m1Vf1^2 + 1/2m2Vf2^2 = 1/2m1Vi1^2

Solving the equation

1.Vf1 = (m1 - m2)Vi1/m1+m2

Vf1 = (1.49-1.49)*3.09/1.49+1/49

Vf1 = 0m/s

2. Vf2 = 2m1Vi1/m1+m2

Vf2 = 2*1.49*3.09/1.49+1.49

Vf2 = 3.09m/s

N:B following the general principle of collision when 2 bodies of same masses collide in elastic collision they exchange velocities.

7 0
4 years ago
Type the correct answer in the box. Express your answer to three significant figures.
Marrrta [24]

The mass of calcium phosphate the reaction can produce is <u>6.076</u> grams.

<h3>What is mass?</h3>

Mass is a numerical measure of inertia, which is a basic feature of all matter. It is, in effect, a body of matter's resistance to a change in speed or position caused by the application of a force.

In the International System of Units (SI), the kilogram is the unit of mass.

Given;

Mass of calcium nitrate (Ca(NO₃)₂) = 96.1 g

Mass of calcium phosphate=?

The chemical equation is found as;

3Ca(NO₃)₂ + 2Na₃PO₄ → 6NaNO₃ + Ca₃(PO4)₂

3 moles of Ca(NO₃)₂ produces 1 mole of Ca₃(PO4)molar mass of calcium phosphate is 164 g/mol

Moles are found as the ratio of the mass of the compound and its molar mass. Moles of Ca(NO₃)₂ is n.

n = 96.1/164 = 0.5859 moles

Moles of Ca3(PO4)2 produced ;

N=0.0589 ×(1/3) = 0.0196 moles

The molar mass of calcium phosphate is 310 g/mol and the mass of calcium phosphate produced will be;

M=0.0196×310

M = 6.076 g

Hence,the mass of calcium phosphate the reaction can produce is <u>6.076</u> grams.

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brainly.com/question/13073862

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5 0
2 years ago
39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of th
tatuchka [14]

<u>Answer:</u> The final temperature of the solution is 80.14^oC

<u>Explanation:</u>

The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, \text{heat}_{absorbed}=\text{heat}_{released}

To calculate the amount of heat released or absorbed, we use the equation:  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

Also,

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]    ..........(1)

where,

q = heat absorbed or released

m_1 = mass of aluminium = 39 g

m_2 = mass of coffee = 166 g

T_{final} = final temperature = ?

T_1 = temperature of aluminium = 24^oC

T_2 = temperature of coffee = 83^oC

c_1 = specific heat of aluminium = 0.904J/g^oC

c_2 = specific heat of coffee= 4.1801J/g^oC

Putting all the values in equation 1, we get:

39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]

T_{final}=80.14^oC

Hence, the final temperature of the solution is 80.14^oC

4 0
4 years ago
A thin, uniform rod is bent into a square of side length a. If the total mass is M, find the moment of inertia about an axis thr
Papessa [141]

Answer:

The  moment of inertia about an axis through the center and perpendicular to the plane of the square is

    I_s =  \frac{Ma^2}{3}

Explanation:

From the question we are told that

   The length of one side of the square is  a

   The total mass of the square is  M

Generally the mass of one size of the square is mathematically evaluated as

    m_1 = \frac{M}{4}

Generally the moment of inertia of one side of the square is mathematically represented as

        I_g =  \frac{1}{12}  *  m_1 * a^2

Generally given that m_1 = m_2 = m_3 = m_4 = m it means that this moment inertia evaluated above apply to every side of the square  

Now substituting for  m_1

  So

       I _g=  \frac{1}{12}  *  \frac{M}{4} * a^2

Now according to  parallel-axis theorem the moment of inertia of one side of the square about an axis through the center and perpendicular to the plane of the square is mathematically represented as

      I_a =  I_g + m [\frac{q}{2} ]^2

=>    I_a =  I_g + {\frac{M}{4} }* [\frac{q}{2} ]^2

substituting for I_g

=>    I_a =  \frac{1}{12}  *  \frac{M}{4} * a^2 + {\frac{M}{4} }* [\frac{q}{2} ]^2

=>    I_a = \frac{Ma^2}{48} + \frac{Ma^2}{16}

=>    I_a = \frac{Ma^2}{12}

Generally the moment of inertia of the square about an axis through the center and perpendicular to the plane of the square is mathematically represented as

      I_s = 4 * I_a

=>   I_s = 4 * \frac{Ma^2}{12}

=>   I_s =  \frac{Ma^2}{3}

8 0
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Naddik [55]

Answer:

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