What would be the correct value of m in the interference equation if the growth process were started again and the diamond layer were grown to three times the thickness t

Answers

Answer 1

Answer:

To calculate the correct value of m if the thickness t is grown 3 times again we can deduce that:

2*(3u)*t = m* lambda

Making m the subject of the formula will give the formula:

m= 6*u*t/ Lambda

Given: Lambda= 633*10^9 while u and t are unknown

Therefore the value of m can be calculated given the formula below:

m= 6*u*t/ 633*10^9

Explanation:

To calculate the correct value of m if the thickness t is grown 3 times again we can deduce that:

2*(3u)*t = m* lambda

Making m the subject of the formula will give the formula:

m= 6*u*t/ Lambda

Given: Lambda= 633*10^9 while u and t are unknown

Therefore the value of m can be calculated given the formula below:

m= 6*u*t/ 633*10^9


Related Questions

The AC voltage source supplies an rms voltage of 146 V at frequency f. The circuit has R = 110 Ω, XL = 210 Ω, and XC = 110 Ω. At the instant the voltage across the generator is at its maximum value, what is the magnitude of the current in the circuit?

Answers

Answer:

1.03A

Explanation:

For computing the magnitude of the current in the circuit we need to do the following calculations

LCR circuit impedance

[tex]Z = \sqrt{R^2 + (X_L - X_c)^2} \\\\ = \sqrt{110^2 + (210 - 110)^2}[/tex]

= 148.7Ω

Now the phase angle is

[tex]\phi = tan^{-1} (\frac{X_L - X_C}{R}) \\\\ = tan^{-1} (\frac{210 - 110}{110})\\\\ = 42.3^{\circ}[/tex]

Now the rms current flowing in the circuit is

[tex]I_{rms} = \frac{V_{rms}}{Z} \\\\ = \frac{146}{148.7}[/tex]

= 0.98 A

The current flowing in the circuit is

[tex]I = I_{rms}\sqrt{2} \\\\ = (0.98) (1.414)[/tex]

= 1.39 A

And, finally, the current across the generator is

[tex]I'= I cos \phi[/tex]

[tex]= (1.39) cos 42.3^{\circ}[/tex]

= 1.03A

Hence, the magnitude of the circuit current is 1.03A

Force and distance are used to calculate work. Work is measured in which unit? joules watts newtons meters

Answers

Answer:

The unit of work is joules

Force and displacement are used to calculate the work done by an object. This work is measured in the units of Joules. Thus, the correct option is A.

What is Work?

Work can be defined as the force that is applied on an object which shows some displacement. Examples of work done include lifting an object against the Earth's gravitational force, and driving a car up on a hill. Work is a form of energy. It is a vector quantity as it has both the direction as well as the magnitude. The standard unit of work done is the joule (J). This unit is equivalent to a newton-meter (N·m).

The nature of work done by an object can be categorized into three different classes. These classes are positive work, negative work and zero work. The nature of work done depends on the angle between the force and displacement of the object. Positive work is done if the applied force displaces the object in its direction, then the work done is known as positive work. Negative work is opposite of positive work as in this work, the applied force and displacement of the object are in opposite directions to each other and zero work is done when there is no displacement.

Therefore, the correct option is A.

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water and air are both fluids. why is it easier to lift a rock in water rather thatn lifting a rock in air? a the force of gravity. b the bouyant force is greater on the rock in water. c the bouyant force is greater on the rock in air. d the force of gravity on the rock is less in water.

Answers

Answer:

The answer is option b.the buoyant force is greater on the rock in water.

The rotor of a gas turbine is rotating at a speed of 7000 rpm when the turbine is shut down. It is observed that 3.5 minutes is required for the rotor to coast to rest. Assuming uniformly accelerated motion, determine the number of revolutions that the rotor executes before coming to rest. Hint: there will be a large number of rotations.

Answers

Answer:

The rotor of the gas turbine rotates 12250 revolutions before coming to rest.

Explanation:

Given that rotor of gas turbine is decelerating at constant rate, it is required to obtained the value of angular acceleration as a function of time, as well as initial and final angular speeds. That is:

[tex]\dot n = \dot n_{o} + \ddot n \cdot t[/tex]

Where:

[tex]\dot n_{o}[/tex] - Initial angular speed, measured in revolutions per minute.

[tex]\dot n[/tex] - Final angular speed, measured in revolutions per minute.

[tex]t[/tex] - Time, measured in minutes.

[tex]\ddot n[/tex] - Angular acceleration, measured in revoiutions per square minute.

The angular acceleration is now cleared:

[tex]\ddot n = \frac{\dot n - \dot n_{o}}{t}[/tex]

If [tex]\dot n_{o} = 7000\,\frac{rev}{min}[/tex], [tex]\dot n = 0\,\frac{rev}{min}[/tex] and [tex]t = 3.5\,min[/tex], the angular acceleration is:

[tex]\ddot n = \frac{0\,\frac{rev}{min}-7000\,\frac{rev}{min} }{3.5\,min}[/tex]

[tex]\ddot n = -2000\,\frac{rev}{min^{2}}[/tex]

Now, the final angular speed as a function of initial angular speed, angular acceleration and the change in angular position is represented by this kinematic equation:

[tex]\dot n^{2} = \dot n_{o}^{2} + 2\cdot \ddot n \cdot (n-n_{o})[/tex]

Where [tex]n[/tex] and [tex]n_{o}[/tex] are the initial and final angular position, respectively.

The change in angular position is cleared herein:

[tex]n-n_{o} = \frac{\dot n^{2}-\dot n_{o}^{2}}{2\cdot \ddot n}[/tex]

If [tex]\dot n_{o} = 7000\,\frac{rev}{min}[/tex], [tex]\dot n = 0\,\frac{rev}{min}[/tex] and [tex]\ddot n = -2000\,\frac{rev}{min^{2}}[/tex], the change in angular position is:

[tex]n-n_{o} = \frac{\left(0\,\frac{rev}{min} \right)^{2}-\left(7000\,\frac{rev}{min} \right)^{2}}{2\cdot \left(-2000\,\frac{rev}{min^{2}} \right)}[/tex]

[tex]n-n_{o} = 12250\,rev[/tex]

The rotor of the gas turbine rotates 12250 revolutions before coming to rest.

A thin film with an index of refraction of 1.60 is placed in one of the beams of a Michelson interferometer. If this causes a shift of 8 bright fringes in the pattern produced by light of wavelength 580 nm, what is the thickness of the film

Answers

Answer:

3.867 μm

Explanation:

The index of refraction, μ = 1.6

Wavelength of the light, λ = 580 nm

N2 - N1 = (2L / λ) (n2 - n1), Making L subject of formula, we have

(N2 - N1) λ = 2L (n2 - n1)

L = [(N2 - N1) * λ] / 2(n2 - n1)

L = (8 * 580) / 2(1.6 - 1.0)

L = 4640 nm / 1.2

L = 3867 nm or 3.867 μm

Therefore we can come to the conclusion that the thickness of the film is 3.867 nm

Find the rms (a) electric and (b)magnetic fields at a point 2.00 m from a lightbulb that radiates 90.0 W of light uniformly in all directions.

Answers

Answer:

a) rms of electric field =

[tex]E_{rms}[/tex]= 25.97 V/m

b) rms of magnetic field

[tex]B_{rms}[/tex] = 8.655 × 10⁻⁸

[tex]B_{rms}[/tex] = 86.55nT

Explanation:

given

power p = 90.0W

distance d = 2.00m

Intensity = [tex]\frac{power}{area}[/tex]

I = [tex]\frac{p}{A}[/tex]

A = [tex]4\pi d^{2}[/tex]

I = [tex]\frac{p}{4\pi d^{2} }[/tex]

I = [tex]\frac{90}{4\pi(2^{2}) }[/tex]

I = 1.79 W/m²

a) [tex]I_{ave}[/tex] = ε₀ × [tex]E^{2} _{rms}[/tex] × c

where ε₀ is permittivity of free space = 8.85×10⁻¹²,  [tex]E^{2} _{rms}[/tex] is the root mean value and c is speed of light = 3×10⁸m/s

1.79 = 8.85×10⁻¹² × [tex]E^{2} _{rms}[/tex] × 3×10⁸

[tex]E^{2} _{rms}[/tex] = [tex]\frac{1.79}{8.85x10^{-12} x 3x10^{8} }[/tex]

[tex]E^{2} _{rms}[/tex]= 674.1996

[tex]E_{rms}[/tex]= 25.97 V/m

b)for rems magnetic field

[tex]E_{rms}[/tex]= c [tex]B_{rms}[/tex]

[tex]B_{rms}[/tex] = [tex]\frac{E_{rms} }{c}[/tex]

[tex]B_{rms}[/tex] = [tex]\frac{25.97 V/m}{3x10^{8} }[/tex]

[tex]B_{rms}[/tex] = 8.655 × 10⁻⁸

[tex]B_{rms}[/tex] = 86.55nT

Two uncharged metal spheres, #1 and #2, are mounted on insulating support rods. A third metal sphere, carrying a positive charge, is then placed near #2. Now a copper wire is momentarily connected between #1 and #2 and then removed. Finally, sphere #3 is removed.
In this final state
a) spheres #1 and #2 are still uncharged.
b) sphere #1 carries negative charge and #2 carries positive charge.
c) spheres #1 and #2 both carry positive charge.
d) spheres #1 and #2 both carry negative charge.
e) sphere #1 carries positive charge and #2 carries negative charge

Answers

Answer:

sphere #1 carries positive charge and #2 carries negative charge

This is because from the laws of static electricity, disconnecting the copper wire makes #1 to be positively charged and #2 to be negatively charged

An insulating hollow sphere has inner radius a and outer radius b. Within the insulating material the volume charge density is given by rho(r)=αr,where α is a positive constant.
A). What is the magnitude of the electric field at a distance r from the center of the shell, where a Express your answer in terms of the variables α, a, r, and electric constant ϵ0.
B) .A point charge
q is placed at the center of the hollow space, at r=0. What value must q have (sign and magnitude) in order for the electric field to be constant in the region a Express your answer in terms of the variables α, a, and appropriate constants.
C). What then is the value of the constant field in this region?
Express your answer in terms of the variable αand electric constant ϵ0.

Answers

Answer:

E   = α/2∈₀ [ 1 - a²/r² ]

Ф = α/2∈₀

Explanation:

Using Gauss Law:

    ρ(r) = a/r, dA

          = 4 π r²d r

    Ф = [tex]\int\limits^r_a[/tex] ρ(r')dA

    Ф[tex]_{encl}[/tex] = [tex]\int\limits^r_a[/tex] ρ(r')dA

             = 4πα [tex]\int\limits^r_a[/tex] r'dr'

Ф[tex]_{encl}[/tex]     = 4 π α 1/2(r²-a²)

E(4πr²) = [tex]2\pi\alpha (r^{2}-a^{2} )/[/tex]∈₀

           = [tex]2\pi\alpha (r^{2}-a^{2} )/[/tex]∈₀(4πr²)

           = α (r² - a²) / 2 ∈₀ (r²)

           = α/2∈₀ [ r²/r² - a²/r² ]

      E   = α/2∈₀ [ 1 - a²/r² ]

Electric field of the point charge:

E[tex]_{q}[/tex] = q / 4π∈₀r²

[tex]E_{total}[/tex] = α / 2 ∈₀ - (α / 2 ∈₀ )(a² / r²) + q / 4 π ∈₀ r²

For [tex]E_{total}[/tex]  to be constant:

- (αa²/ 2 ∈₀ ) + q / 4 π ∈₀ = 0 and q = 2παa²

-> α / 2 ∈₀ - αa²/ 2 ∈₀ + 2παa² / 4 π ∈₀

= α - αa² + αa² / 2 ∈₀

= α /2 ∈₀

Hence:

Ф = α/2∈₀

The 2-Mg truck is traveling at 15 m/s when the brakes on all its wheels are applied, causing it to skid for 10 m before coming to rest. The total mass of the boat and trailer is 1 Mg. Determine the constant horizontal force developed in the coupling C, and the friction force developed between the tires of the truck and the road during this time.

Answers

Answer:

constant horizontal force developed in the coupling C = 11.25KN

the friction force developed between the tires of the truck and the road during this time is 33.75KN

Explanation:

See attached file

The friction force between the tires of the truck and the road is 22500 N.

Calculating the friction force:

It is given that a 2 Mg truck ( m = 2000 Kg) is initially moving with a speed of u = 15 m/s.

Distance traveled before coming to rest, s = 10m

The final velocity of the truck will be zero, v = 0

When the breaks are applied, only the frictional force is acting on the truck and it is opposite to the motion of the truck.

The frictional force is given by:

f = -ma

the acceleration of the truck = -a

The negative sign indicates that the acceleration is opposite to the motion.

Applying the third equation of motion we get:

v² = u² -2as

0 = 15² - 2×a×10

225 = 20a

a = 11.25 m/s²

So the magnitude of frictional force is:

f = ma = 2000 × 11.25 N

f = 22500 N

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Suppose you have a lens system that is to be used primarily for 775-nm light. What is the second thinnest coating of fluorite (calcium fluoride) that would be non-reflective for this wavelength?

Answers

Answer:

406 nm

Explanation:

We are given;

Wavelength; λ = 775 nm

Refractive index of Calcium fluoride with wavelength of 775 nm as seen in the graph attached is approximately 1.4308.

n = 1.4308

Formula for the thickness of the film that would destruct the light is;

t = (m + 0.5)(λ/2n)

Where m is the order of the thickness.

The first smallest thickness is at m = 0 while the second smallest thickness is at m = 1.

Thus;

t = (1 + 0.5)(775/(2 × 1.4308))

t ≈ 406 nm

have an electrical charge of +1, while
have an electrical charge of -1.
A. Neutrons, electrons
B. Protons, electrons
C. Electrons, neutrons
D. Electrons, protons

Answers

Answer:

B

Explanation:

Protons have a positive electrical charge of +1,

Electrons have a negative charge of -1,

Neutrons have a neutral charge of about 0.

An optical fiber uses one glass clad with another glass. What is the critical angle? (Assume the glass in the fiber has an index of refraction of 1.69, and the glass in the cladding has an index of refraction of 1.50.)

Answers

Answer:

The critical angle is [tex]\theta _c = 62.57^o[/tex]

Explanation:

From the question we are told that

    The  index of refraction of the glass in the fiber is [tex]n_f = 1.69[/tex]

     The index of refraction of the glass in the cladding is [tex]n_c = 1.50[/tex]

The critical angle is mathematically evaluated as

       [tex]\theta_c = sin^{-1}[\frac{n_c }{n_f } ][/tex]

substituting values

       [tex]\theta_c = sin^{-1}[\frac{1.50 }{1.69 } ][/tex]

       [tex]\theta _c = 62.57^o[/tex]

A department store expects to have 225 customers and 20 employees at peak times in summer. Determine the contribution of people to the total cooling load of the store. The average rate of heat generation from people doing light work is 115 W, and 70% of it is in sensible form.

Answers

Answer:

The  contribution of people to the cooling load of the store is 19722.5 W

Explanation:

Total amount of customers = 225

Total amount of employees = 20

Total amount of people in the store at that instant n = 245 people

Average rate of heat generation Q = 115 W

percentage of these heat generated that is sensible heat = 70%

Sensible heat raises the surrounding temperature. Latent heat only causes a change of state.

The total heat generated by all the people in the store = n x Q

==> 245 x 115 = 28175 W

but only 70% of this heat is sensible heat that raises the temperature of the store, therefore, the contribution of people to the cooling load of the store = 70% of 28175 W

==> 0.7 x 28175 = 19722.5 W

If the current flowing through a circuit of constant resistance is doubled, the power dissipated by that circuit will

Answers

Assuming that the voltage is constant, and the resistance was doubled, since I=V/R, the current through the circuit will be halved. As P=IV, with the same voltage and halved current, the power dissipated by the circuit will be halved

A bowling ball of mass 5 kg rolls off the edge of a building 20 meters tall. What is the work done by gravity during the fall, in Joules

Answers

Answer:

1000j

Explanation:

work done = force x distance

w = 5 x 10 x 20 = 1000joules

Based on the graph below, what prediction can we make about the acceleration when the force is 0 newtons? A. It will be 0 meters per second per second. B. It will be 5 meters per second per second. C. It will be 10 meters per second per second. D. It will be 15 meters per second per second.

Answers

Answer:

Option A

Explanation:

From the graph, we came to know that Force and acceleration are in direct relationship.

Also,

Force = 0 when Acceleration = 0

Because Both are 0 at the origin.

Answer:

A. It will be 0 meters per second per second.

Explanation:

The force and acceleration is in a proportional relationship, that means the line goes through the origin.

On the graph, when the force is at 0, the acceleration is 0. The line passes through the origin.

The phenomenon of magnetism is best understood in terms ofA) the existence of magnetic poles.B)the magnetic fields associated with the movement of charged particles.C)gravitational forces between nuclei and orbital electrons.D) electrical fluid

Answers

Answer:

A) the existence of magnetic poles.

Explanation:

Magnetism is defined as the ability of a magnet to attract magnetic substance to itself. Such magnet has the ability of being magnetized. A magnet is known to possess poles which are the north poles and south poles. The presence of this poles is what makes them possess the properties of a magnet. An ordinary steel bar doesn't have the properties of a magnet unless it is magnetized and when you are trying to magnetize a steel bar, you are invariably introducing the magnetic poles.

According to the law of magnetism, like poles repel but unlike poles attract. From the above explanation, it can be concluded that the phenomenon of magnetism is best understood interns of existence of magnetic poles. This poles are called the north and the south poles.

Red light is bent the least of all colors as it passes through a prism. What does this tell you about red light? It has a short wavelength. It has a long wavelength. It has a high intensity. It has a low intensity.

Answers

Answer:

Longest wavelength, lowest intensity

Explanation:

Answer:

It has a long wavelength

Explanation:

GRADPOINT

Two parallel wires run in a north-south direction. The eastern wire carries 15.0 A northward while the western wire carries 6.0 A northward. If the wires are separated by 30 cm, what is the magnetic field magnitude and direction at a point between the wires at a distance of 10 cm from the western wire?

Answers

Answer:

The magnitude  and direction of the  magnetic field is 2.7 x 10⁻⁵ T upwards

Explanation:

Given;

current in the eastern wire, [tex]I_e[/tex] = 15 A

current in the western wire, [tex]I_w[/tex] = 6 A

distance between the wires, d = 30 cm = 0.3 m

The magnetic field at a distance R from a line current I, is given as;

[tex]B = \frac{\mu_o I }{2 \pi R}[/tex]

The magnetic field between the wires, are in opposite directions, and since the currents are also in opposite directions, the magnetic fields of the wires will be added.

The total field = magnetic field (east) + magnetic field (west);

[tex]B = \frac{\mu_o I_e}{2 \pi R_e} + \frac{\mu_0 I_w}{2 \pi R_w} \\\\B = \frac{\mu_o}{2\pi} (\frac{I_e}{R_e} + \frac{I_w}{R_w})[/tex]

where;

[tex]R_w[/tex] is the distance of the field from west = 10cm = 0.1 m

[tex]R_e[/tex] is the distance of the field on east from west = d - 10cm = 30cm - 10cm = 20cm = 0.2 m

The total magnetic field is;

[tex]B = \frac{\mu_o}{2\pi} (\frac{I_e}{R_e} + \frac{I_w}{R_w})\\\\B = \frac{4\pi *10^{-7}}{2\pi} (\frac{15}{0.2} + \frac{6}{0.1})\\\\B = 2*10^{-7}(75 + 60)\\\\B = 2*10^{-7}(135)\\\\B = 2.7*10^{-5} \ T[/tex]

Since total magnetic field is positive, the direction of the field is upwards (positive y direction)

Therefore, the magnitude  and direction of the  magnetic field is 2.7 x 10⁻⁵ T upwards

Two people, one of mass 85 kg and the other of mass 50 kg, sit in a rowboat of mass 90 kg. With the boat initially at rest, the two people, who have been sitting at opposite ends of the boat, 3.5 m apart from each other, now exchange seats. How far does the boat move?

Answers

Answer:

0.11m

Explanation:

let's assume the boat is of uniform construction

Ignoring friction losses

Also assume the origin is at the end of the boat originally with the heavier person

the center of mass of the whole system will not change relative to the water when the two swap ends

Originally, the center of mass is

85[0] + 90[3.5/2] + 50[3.5] / (85 + 90 + 50) = 1.14m from the origin

after the swap, the center of mass is

50[0] + 90[3.5/2] + 85[3.5] / (85 + 90+ 50) = 1.030m from the origin

The center of mass has shifted

1.14-1.030 = 0.11m

as no external force acted on the system, the center of mass relative to the water will not change. The boat will therefore shift towards the end where the heavier person originally sat

Sergio has made the hypothesis that "the more time that passes, the farther away a person riding a bike will be." Do the data in the table below support his hypothesis? A. Yes, the data support the hypothesis. B. No, the data support the opposite of the hypothesis. C. The data show no relationship between the time passed and the distance.

Answers

Answer:

Option A

Explanation:

Given that

Distance = Speed / Time

So, they are in inverse relation.

Such that when the time passes, the distance from the reacing point will become less and vice versa.

So, Yes! The more time that passes, the farther away a person riding a bike will be.

If radio waves were used to communicate with an alien spaceship approaching Earth at 10% of the speed of light c, Earth would receive their signals at a speed of

Answers

Answer:

Explanation:

speed of alien spaceship = .1 c

We shall apply formula of relativistic mechanics to solve the problem

relative velocity =

[tex]\frac{v+v_1}{1 -\frac{v\times v }{c^2} }[/tex]

Here v = v₁ = .1 c

relative velocity  = .1c + .1 c / 1 - .1²

= .2 c / .99

= .202 c

The earth would receive the signal at the speed of .202 c .

. A 24-V battery is attached to a 3.0-mF capacitor and a 100-ohm resistor. If the capacitor is initially uncharged, what is the voltage across the capacitor 0.16 seconds after the circuit is connected to the battery

Answers

Answer:

The  voltage is [tex]V_c = 9.92 \ V[/tex]

Explanation:

From the question we are told that

     The voltage of the battery is  [tex]V_b = 24 \ V[/tex]

     The capacitance of the capacitor is  [tex]C = 3.0 mF = 3.0 *10^{-3} \ F[/tex]

     The  resistance of the resistor is [tex]R = 100\ \Omega[/tex]

     The time taken is  [tex]t = 0.16 \ s[/tex]  

Generally the voltage of a charging charging capacitor after time t is mathematically represented as

       [tex]V_c = V_o (1 - e^{- \frac{t}{RC} })[/tex]

Here [tex]V_o[/tex] is the voltage of the capacitor when it is fully charged which in the case of this question is equivalent to the voltage of the battery so  

      [tex]V_c = 24 (1 - e^{- \frac{0.16}{100 * 3.0 *10^{-1}} })[/tex]

      [tex]V_c = 9.92 \ V[/tex]

g A particle (charge = +40 mC) is located on the x axis at the point x = -20 cm, and a second particle (charge = -50 mC) is placed on the x axis at x = +30 cm. What is the magnitude of the total electrostatic force on a third particle (charge = -4.0 mC) placed at the origin (x = 0)? Group of answer choices

Answers

Answer:

Explanation:

We shall find electric field at origin due to two given charges sitting   on the either side of origin .

Total field will add up due to their same direction .

Field due to a charge Q

= 9 x 10⁹ x Q / R²  ;  R is distance of point , Q is charge

Field due to first charge

= 9 x 10⁹ x 40 x 10⁻³ / 2² x 10⁻⁴

= 90 x 10¹⁰ N/C

Field due to second  charge

= 9 x 10⁹ x 50 x 10⁻³ / 2² x 10⁻⁴

= 112.5 x 10¹⁰ N/C

Total field

= 202.5 x 10¹⁰ N/C

Force on given charge at origin

= charge x field

= 4 x 10⁻³ x 202.5 x 10¹⁰

= 810 x 10⁷ N .

A conventional current of 3 A runs clockwise in a circular loop of wire in the plane, with center at the origin and with radius 0.093 m. Another circular loop of wire lies in the same plane, with its center at the origin and with radius 0.03 m. How much conventional current must run counterclockwise in this smaller loop in order for the magnetic field at the origin to be zero

Answers

Answer:

The current in the small radius loop must be 0.9677 A

Explanation:

Recall that the formula for the magnetic field at the center of a loop of radius R which runs a current I, is given by:

[tex]B=\mu_0\,\frac{I}{2\,R}[/tex]

therefore for the first loop in the problem, that magnetic field strength is:

[tex]B=\mu_0\,\frac{I}{2\,R} =\mu_0\,\frac{3}{2\,(0.093)} =16.129\,\mu_{0}\,[/tex]

with the direction of the magnetic field towards the plane.

For the second smaller loop of wire, since the current goes counterclockwise, the magnetic field will be pointing coming out of the plane, and will subtract from the othe field. In order to the addition of these two magnetic fields to be zero, the magnitudes of them have to be equal, that is:

[tex]16.129\,\,\mu_{0}=\mu_0\,\frac{I'}{2\,R'} =\mu_{0}\,\frac{I'}{2\,(0.03)} \\I'=16.129\,(2)\,(0.03)=0.9677\,\,Amps[/tex]

Equal charges, one at rest, the other having a velocity of 104 m/s, are released in a uniform magnetic field. Which charge has the largest force exerted on it by the magnetic field

Answers

Answer:

case 1 of physics is the answer

A 1.8 kg microphone is connected to a spring and is oscillating in simple harmonic motion up and down with a period of 3s. Below the microphone is 1.8 hz, calculate the spring constant

Answers

Answer:

230N/m

Explanation:

Pls see attached file

a figures skater rotating at 5 rads with arms extended has a moment of inertia of 2.25 kg. if the arms are pulled in so the moment of inertia decrease to 1.8 what is the final angular speed

Answers

Answer:

The final angular speed is 6.25 rad/s

Explanation:

Given;

initial angular speed, ω₁ = 5 rad/s

initial moment of inertia, I₁ = 2.25 kg.m²

Final moment of inertia, I₂ = 1.8 kg.m²

final angular speed, ω₂ = ?

Based on conservation of angular momentum, we will have the following expression;

ω₁I₁ = ω₂I₂

ω₂ = (ω₁I₁ ) / I₂

ω₂ = (5 x 2.25) / 1.8

ω₂ = 6.25 rad/s

Therefore, the final angular speed is 6.25 rad/s

A flat loop of wire consisting of a single turn of cross-sectional area 8.20 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 2.60 T in 1.02 s. What is the resulting induced current if the loop has a resistance of 2.70

Answers

Answer:

The  induced current is [tex]I = 6.25*10^{-4} \ A[/tex]

Explanation:

From the question we are told that  

    The number of turns is  [tex]N = 1[/tex]

     The  cross-sectional area is  [tex]A = 8.20 cm^2 = 8.20 * 10^{-4} \ m^2[/tex]

    The  initial magnetic field is  [tex]B_i = 0.500 \ T[/tex]

     The  magnetic field at time =  1.02 s  is  [tex]B_t = 2.60 \ T[/tex]

     The  resistance is  [tex]R = 2.70\ \Omega[/tex]

The  induced emf is mathematically represented as

       [tex]\epsilon = - N * \frac{ d\phi }{dt}[/tex]

The  negative sign tells us that the induced emf is moving opposite to the change in magnetic flux

      Here  [tex]d\phi[/tex] is the change in magnetic flux which is mathematically represented as

        [tex]d \phi = dB * A[/tex]

Where  dB  is the change in magnetic field which is mathematically represented as

        [tex]dB = B_t - B_i[/tex]

substituting values

        [tex]dB = 2.60 - 0.500[/tex]

        [tex]dB = 2.1 \ T[/tex]

Thus  

      [tex]d \phi = 2.1 * 8.20 *10^{-4}[/tex]

     [tex]d \phi = 1.722*10^{-3} \ weber[/tex]

So  

     [tex]|\epsilon| = 1 * \frac{ 1.722*10^{-3}}{1.02}[/tex]

     [tex]|\epsilon| = 1.69 *10^{-3} \ V[/tex]

The  induced current i mathematically represented as

      [tex]I = \frac{\epsilon}{ R }[/tex]

  substituting values

       [tex]I = \frac{1.69*10^{-3}}{ 2.70 }[/tex]

       [tex]I = 6.25*10^{-4} \ A[/tex]

The number of neutrons in the nucleus of zinc 65 Zn 30 is:
35
Need more data to answer
65
30

Answers

Explanation:

proton number + neutron number = atomic mass

30 + 35 = 65

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