a jk flip flop has a condition of j=0, k=0, and both preset and clear are inactive. if a 100hz pulse is applied to the clock, the output of q will be
a.0
b.1
c.100 hz
d.50 hz
e.unpredictable

The reactions at the supports of the beam are [tex]R_1[/tex] = 1150 N and [tex]R_2[/tex] = 1150 N.

Let's denote the reactions at the supports as [tex]R_1[/tex]and [tex]R_2[/tex].

Since the beam is supported at both ends, we can assume that it is a simply supported beam.

The total downward load consists of the distributed load and the concentrated load.

The downward load due to the distributed load can be calculated by integrating the load intensity over the length:

Downward load due to distributed load = (500 N/m) * (3 m) = 1500 N

The total downward load due to the concentrated load is 800 N.

Now, let's consider the equilibrium equation:

[tex]R_1 + R_2[/tex] = 1500 N + 800 N

[tex]R_1 + R_2[/tex] = 2300 N

Since the beam is simply supported, we can assume that the reactions at the supports are equal. Therefore:

[tex]R_1 = R_2[/tex]= 2300 N / 2

[tex]R_1 = R_2[/tex]= 1150 N

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--The complete Question is, The internal loadings at a section of the beam are given as follows: a uniformly distributed load of 500 N/m over a length of 3 meters and a concentrated load of 800 N applied at the midpoint of the same section. Determine the reactions at the supports of the beam.--

The angle necessary to keep a 10 kg box motionless, given a coefficient of static friction of 0.55 between the box and the ramp, is 33.4°, which corresponds to Option A.

To determine the angle, we can use the relationship between the coefficient of static friction, the angle of the incline, and the gravitational force acting on the box. The maximum static friction force can be calculated using the formula:

Friction force = coefficient of static friction * Normal force

The Normal force can be found by decomposing the gravitational force acting on the box into components parallel and perpendicular to the incline. The perpendicular component (Normal force) is equal to the weight of the box (mass * gravitational acceleration).

Since the box is motionless, the friction force must be equal to the component of the gravitational force acting parallel to the incline:

Friction force = Component of weight parallel to incline

By substituting the given values and solving for the angle, we find:

coefficient of static friction = tan(angle)

angle = arctan(coefficient of static friction)

angle = arctan(0.55) ≈ 33.4°

Therefore, the correct answer is Option A, 33.4°.

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Answer:

Yes.

Explanation: the magnitude of the force is extremely small because the masses of the students are small relative to Earth's mass.

The gravitational force between two students sitting in classroom exists but its effect are not visible because the distance between these students is very very very small.

We have two students sitting in a classroom.

We have to investigate whether or not there is gravitational force between these two students.

Newton's law of universal gravitation states that every particle attracts every other particle in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. Mathematically -

[tex]$F=\frac{GMm}{r^{2} }[/tex]

According to the question, we have - Two students sitting in a classroom.

Take a close look and you will identify that the distance between any two students in classroom is very small. Gravitational force is also the weakest force since the value of Universal gravitation constant (G) = 6.674 x [tex]10^{-11}[/tex] [tex]m^{3}kg^{-1}s^{-2}[/tex] is very small. Moreover, the force of gravitation is inversely proportional to square of distance between two bodies -

[tex]$F \alpha \frac{1}{r^{2} }[/tex]

This will make the gravitational force even more weak. This is the main reason why you don't observe any effects of this force. The gravitational force exists between any two bodies that have mass and energy, but its magnitude might depend upon other factors to.

Hence, the gravitational force between two students sitting in classroom exists but its effect are not visible because the distance between these students is very very very small.

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The given statement is false, because doubling the number of units of a bottleneck resource does not necessarily double the process capacity.

The capacity of a process is determined by its bottleneck, which is the resource or step with the lowest capacity. Increasing the capacity of the bottleneck resource may improve the overall process capacity, but it depends on the specific circumstances and the nature of the process. Other factors such as dependencies, synchronization, and overall process design can also impact the process capacity. Therefore, simply doubling the units of a bottleneck resource does not guarantee a doubling of the process capacity.

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To estimate the temperature change from room temperature to water hot enough for a hot shower, we need more information such as the initial room temperature and the desired temperature of the hot water.

Assuming a typical room temperature of around 20°C and a desired hot water temperature for a shower of around 40-45°C, we can estimate the temperature change as follows: Temperature change = Desired hot water temperature - Initial room temperature. Let's assume the desired hot water temperature is 45°C: Temperature change = 45°C - 20°C = 25°C. Therefore, the estimated temperature change to go from room temperature to hot water for a shower would be approximately 25°C.

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The percent of acid dissociation of a certain acid in an aqueous solution was calculated to be 3.45 %.

When HCl molecules break down, they break down into H⁺ ions (HCl) and Cl⁻ ions (Cl-Cl). Because HCl is a nearly insoluble acid, it dissolves well in water. Acetic acid, on the other hand, dissolves poorly in water because many H⁺ons are trapped inside the molecule.

Acid Dissociation Constant (Ka) Acid Dissociation constant Ka is calculated as H⁺ = (A⁻) / (HA).

Given that

pKa = 4.60 = -logKa

Ka = 2.51 *10⁻⁵

pH = 3.16 = -log[H⁺]

[H+] = 6.92 *10⁻⁴

Let the Acid be HA with an initial concentration of x

                               [tex]HA \rightleftharpoons H^+ A^-[/tex]

Initial (M)                       x  0  0

Change (M)-6.92 *10⁻⁴    +6.92 *10⁻⁴ +6.92 *10⁻⁴

Equilibrium (M)x-6.92 *10⁻⁴  6.92 *10⁻⁴  6.92 *10⁻⁴

Ka = [H⁺][A⁻] / [HA]

2.51 *10⁻⁵ = (6.92 *10⁻⁴)2 / (x-6.92 *10⁻⁴)

(2.51 *10⁻⁵ )x - (1.74 *10⁻⁸) = 4.79 *10⁻⁷

(2.51 *10⁻⁵ )x = 4.96 *10⁻⁷

The initial concentration of acid (x) = 0.0198 M.

Percent of acid dissociated

=([H⁺] / [HA]initial)*100%  

= ((6.92 *10⁻⁴)/0.0198)*100% = 3.45 %

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False. A converging lens produces an enlarged real image when the object is placed beyond its focal point.

A converging lens is thicker at the center and thinner at the edges. When an object is placed beyond the focal point of a converging lens, a real and inverted image is formed on the opposite side of the lens.

This image is larger than the object, hence producing an enlarged real image. The position of the image depends on the object's distance from the lens and follows the rules of image formation by lenses.

On the other hand, a virtual image is formed when an object is placed between the lens and its focal point. The virtual image is upright and larger than the object. However, for a converging lens, this virtual image is not enlarged but rather diminished compared to the object.

Therefore, a converging lens does not produce an enlarged virtual image when the object is placed just beyond its focal point.

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Complete question here:

Choose true or false for each statement regarding a converging lens.

true false A converging lens produces an enlarged virtual image when the object is placed just beyond its focal point.

Approximately 7.5679 x 10^11 photons are emitted in this pulse of a ruby laser with a wavelength of 694.3 nm, an average power of 50 mW, and a duration of 4.6 ns.

To calculate the number of photons emitted in the pulse, we can use the formula:

Number of photons = (Average power of the pulse) / (Energy per photon)

First, let's calculate the energy per photon using the formula:

Energy per photon = Planck's constant (h) x Speed of light (c) / Wavelength

Given:

Wavelength (λ) = 694.3 nm = 694.3 x 10^-9 m

Average power = 50 mW = 50 x 10^-3 W

The Planck's constant (h) is approximately 6.626 x 10^-34 J·s, and the speed of light (c) is approximately 3 x 10^8 m/s.

Calculating the energy per photon:

Energy per photon = (6.626 x 10^-34 J·s) x (3 x 10^8 m/s) / (694.3 x 10^-9 m)

Now, we can calculate the number of photons using the formula mentioned earlier:

Number of photons = (Average power of the pulse) / (Energy per photon)

Substituting the values:

Number of photons = (50 x 10^-3 W) / [(6.626 x 10^-34 J·s) x (3 x 10^8 m/s) / (694.3 x 10^-9 m)]

Number of photons ≈ 7.5679 x 10^11 photons

Approximately 7.5679 x 10^11 photons are emitted in this pulse of a ruby laser with a wavelength of 694.3 nm, an average power of 50 mW, and a duration of 4.6 ns.

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The estimated temperature of the radiant wall heater when operating is approximately 257 Kelvin.

To estimate the temperature of the radiant wall heater, we can use the Stefan-Boltzmann law, which relates the power radiated by an object to its temperature.

The formula for power radiated is given by:

P = σ * A * T^4

where P is the power radiated, σ is the Stefan constant (6 x 10^-8 Wm^-2K^-4), A is the surface area of the heater, and T is the temperature in Kelvin.

First, we need to calculate the surface area of the silica cylinder. The formula for the surface area of a cylinder is:

A = 2πrh + 2πr^2

where r is the radius and h is the height (length) of the cylinder.

Given:

Radius, r = 6 mm = 6 x 10^-3 m

Length, h = 0.6 m

Plugging in these values, we can calculate the surface area:

A = 2π(6 x 10^-3 m)(0.6 m) + 2π(6 x 10^-3 m)^2

 = 0.072π m^2

Now, we can rearrange the Stefan-Boltzmann law to solve for temperature T:

T^4 = P / (σ * A)

Given:

Power, P = 1.5 kW = 1500 W

Stefan constant, σ = 6 x 10^-8 Wm^-2K^-4

Surface area, A = 0.072π m^2

Plugging in these values, we get:

T^4 = (1500 W) / (6 x 10^-8 Wm^-2K^-4 * 0.072π m^2)

T^4 ≈ 3.1 x 10^9 K^4

Taking the fourth root of both sides, we find:

T ≈ 257 K

Therefore, the estimated temperature of the radiant wall heater when operating is approximately 257 Kelvin.

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Answer:

Explanation:

it's D b/c only 25% of the energy is making light... the rest is probably heat.. :/

The coefficient of kinetic friction between block A and the tabletop is 0.336.

The weight of block A = 43.2 N

The weight of block B = 29.0 N

(a) The downward motion of block B is constant
(b) The acceleration of block B is  -0.00069 m/s²

(a)

The net force acting on the block B will be,

F_net = T - f_fric = m_b × a

Where

T is the tension in the string,

f_fric is the frictional force acting on the block A,

m_b is the mass of block B and

a is the acceleration of block B.

Also,

T = m_b × g = 29.0 N

where g is the acceleration due to gravity.

And as the block is moving with constant velocity, the acceleration of block B is zero.

So, F_net = 0

T - f_fric = 0

f_fric = T

The frictional force f_fric can be expressed as

f_fric = μ_k × N

where N is the normal force.

The normal force on block A is the weight of block A + the weight of the cat,

so,

N = m_Ag + m_catg

The mass of the cat is also 43.2 N.

Thus, N = 43.2 N + 43.2 N = 86.4 N

Therefore,

μ_k × N = T

μ_k = T/N

μ_k = 29.0/86.4

μ_k = 0.336

The coefficient of kinetic friction between block A and the tabletop is 0.336.

(b)

The net force acting on the block B is F_net = T - f_fric

F_net = m_b × a

Where T is the tension in the string,

f_fric is the frictional force acting on the block A,

m_b is the mass of block B and

a is the acceleration of block B.

T = 29.0 N

f_fric = μ_k × N

f_fric = 0.336 × 86.4

f_fric = 29.02 N

F_net = T - f_fric

F_net = 29.0 - 29.02

F_net = -0.02 N

Thus, F_net = m_b × a

-0.02 N = 29.0 N × a

a = -0.02/29.0

a = -0.00069 m/s²

The acceleration of block B is negative and it is slowing down.

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In the emission spectrum of the quantum system with allowed energies of 0.0 evev, 1.5 evev, 3.0 evev, and 6.0 evev, there are three different wavelengths that appear.

The allowed energies of the quantum system determine the possible transitions that can occur. A transition between two energy levels corresponds to the emission or absorption of a photon with a specific wavelength. In this case, we have four energy levels: 0.0 evev, 1.5 evev, 3.0 evev, and 6.0 evev.

To determine the wavelengths that appear in the emission spectrum, we need to consider the possible transitions between these energy levels. The differences between the energy levels correspond to the energies of the emitted photons. We have three energy differences: 1.5 evev (from 0.0 evev to 1.5 evev), 1.5 evev (from 1.5 evev to 3.0 evev), and 3.0 evev (from 3.0 evev to 6.0 evev).

Using the relationship between energy and wavelength (E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength), we can calculate the corresponding wavelengths. By substituting the energy differences into the equation, we find the wavelengths to be approximately 827 nm, 554 nm, and 277 nm.

Therefore, in the emission spectrum of this quantum system, three different wavelengths appear: approximately 827 nm, 554 nm, and 277 nm. Each of these wavelengths corresponds to a specific energy transition between the allowed energy levels of the system.

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Benzene is a cyclic compound with six carbon atoms and six hydrogen atoms with alternating double bonds. Aldehyde functional group has a -CHO group, where C is the carbonyl carbon, and it is attached to one hydrogen atom and one R group. The structural formula for benzene is C6H6.

To draw the molecule produced after replacing one hydrogen atom of benzene with an aldehyde functional group, we first need to remove that hydrogen atom. The aldehyde functional group (-CHO) replaces the hydrogen atom.

This replaces the valency of carbon and makes it the centre of the functional group. The carbon atom in the aldehyde functional group is attached to two other groups - a hydrogen atom (H) and a carbon atom (C).

The carbon atom of the functional group is attached to the carbon atom of the benzene ring, which is then connected to two other carbon atoms with alternating double bonds and single bonds.

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Answer: its flowing, reaction

Explanation: this is because currents in a device have a flowing object inside

Answer:

60-34+56×22

Explanation:

that's cuz imma teacher THOY!!!!!!

Answer:

60-34+56×22

Explanation:

The magnetic field inside the solenoid is approximately 0.051 T or 51 mT , magnetic flux through each turn of the solenoid is approximately 4.421 × 10^(-6) T·m^2 and inductance of the solenoid is approximately 1.573 mH or 1573 μH.

The magnetic field inside a solenoid can be calculated using the formula:

B = μ₀ * (N * I) / L

where B is the magnetic field, μ₀ is the permeability of free space (4π × 10^(-7) T·m/A), N is the number of turns, I is the current, and L is the length of the solenoid.

Plugging in the values:

N = 475 turns

I = 42.0 mA = 42.0 × 10^(-3) A

L = 13.0 cm = 13.0 × 10^(-2) m

B = (4π × 10^(-7) T·m/A) * (475 * 42.0 × 10^(-3) A) / (13.0 × 10^(-2) m)

B ≈ 0.051 T (or 51 mT)

Therefore, the magnetic field inside the solenoid is approximately 0.051 T or 51 mT.

The magnetic flux through each turn of the solenoid can be calculated using the formula:

Φ = B * A

where Φ is the magnetic flux, B is the magnetic field, and A is the cross-sectional area of the solenoid.

The cross-sectional area of a solenoid can be approximated as:

A = π * (d/2)^2

where d is the diameter of the solenoid.

Plugging in the values:

d = 10.5 mm = 10.5 × 10^(-3) m

A = π * (10.5 × 10^(-3)/2)^2

A ≈ 8.660 × 10^(-5) m^2

Φ = (0.051 T) * (8.660 × 10^(-5) m^2)

Φ ≈ 4.421 × 10^(-6) T·m^2

Therefore, the magnetic flux through each turn of the solenoid is approximately 4.421 × 10^(-6) T·m^2.

The inductance of a solenoid can be calculated using the formula:

L = μ₀ * (N^2 * A) / L

where L is the inductance, μ₀ is the permeability of free space, N is the number of turns, A is the cross-sectional area, and L is the length of the solenoid.

Plugging in the values:

N = 475 turns

A = 8.660 × 10^(-5) m^2

L = 13.0 cm = 13.0 × 10^(-2) m

L = (4π × 10^(-7) T·m/A) * (475^2 * 8.660 × 10^(-5) m^2) / (13.0 × 10^(-2) m)

L ≈ 1.573 mH (or 1573 μH)

Therefore, the inductance of the solenoid is approximately 1.573 mH or 1573 μ

The quantities that depend on the current are:

Magnetic field inside the solenoid: The magnetic field is directly proportional to the current (B ∝ I).

Magnetic flux through each turn: The magnetic flux is directly proportional to the current (Φ ∝ I).

Inductance of the sol

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Answer:

1476 J

Explanation:

From the question,

Net Work done = Net force× distance moved by net force.

W' = (F-F')×d................... Equation 1

Where W' = Net work done, F = force applied, F' = Frictional force, d = distance moved.

Given: F = 150 N, F' = 37 N, d = 12 m

Substitute these values into equation 1

W' = (150-37)×12

W' = 123×12

W' = 1476 J.

hence the Net Work done by the object is 1476 J

Answer:

Earth: [tex]t = 498.667\,s[/tex], Mercury: [tex]t = 193\,s[/tex], Pluto: [tex]t = 19700\,s[/tex]

Explanation:

The light travels at a constant speed of approximately [tex]3\times 10^{8}[/tex] meters per second. The time ([tex]t[/tex]), in seconds, required for light to travel a given distance is:

[tex]t = \frac{x}{v_{l}}[/tex] (1)

Where:

[tex]x[/tex] - Travelled distance, in meters.

[tex]v_{l}[/tex] - Speed of light, in meters per second.

Now, we calculate the time for light to travel to each planet:

Earth ([tex]v_{l} = 3\times 10^{8}\,\frac{m}{s}[/tex], [tex]x = 1.496\times 10^{11}\,m[/tex])

[tex]t = \frac{x}{v_{l}}[/tex]

[tex]t = 498.667\,s[/tex]

Mercury ([tex]v_{l} = 3\times 10^{8}\,\frac{m}{s}[/tex], [tex]x = 5.79\times 10^{10}\,m[/tex])

[tex]t = \frac{x}{v_{l}}[/tex]

[tex]t = 193\,s[/tex]

Pluto ([tex]v_{l} = 3\times 10^{8}\,\frac{m}{s}[/tex], [tex]x = 5.91\times 10^{12}\,m[/tex])

[tex]t = \frac{x}{v_{l}}[/tex]

[tex]t = 19700\,s[/tex]

For the series of RLC circuits has components,

a. The resonant frequency is 179.1 kHz.

b. The current amplitude at resonance is 6.67 A.

c. The Q of the circuit is 1.35.

d. The voltage amplitude across the inductor is 135 V (or 0.135 kV).

(a) To find the resonant frequency of the circuit, we can use the formula:

ω = 1 / √(LC)

Given:

L = 18.0 mH = 18.0 × [tex]10^{(-3)[/tex] H

C = 80.0 nF = 80.0 × [tex]10^{(-9)[/tex] F

Substituting the values into the formula:

ω = 1 / √((18.0 × [tex]10^{(-3)[/tex]) × (80.0 × [tex]10^{(-9)[/tex]))

Calculating the value of ω:

ω ≈ 1123.6 rad/s

To convert the angular frequency to frequency in kHz, we divide ω by 2π:

f = ω / (2π)

Substituting the value of ω:

f ≈ 1123.6 / (2π) ≈ 179.1 kHz

Therefore, the resonant frequency of the circuit is approximately 179.1 kHz.

(b) At the resonant frequency, the impedance of the circuit is at a minimum, and the current amplitude is at maximum. The current amplitude can be calculated using the formula:

I = ΔVmax / R

Given:

ΔVmax = 100 V

R = 15.0 Ω

Substituting the values into the formula:

I = 100 / 15.0 ≈ 6.67 A

Therefore, the amplitude of the current at the resonant frequency is approximately 6.67 A.

(c) The Q of the circuit can be calculated using the formula:

Q = ωL / R

Given:

L = 18.0 mH = 18.0 × [tex]10^{(-3)[/tex] H

R = 15.0 Ω

ω = 1123.6 rad/s

Substituting the values into the formula:

Q = (1123.6 × (18.0 × [tex]10^{(-3)[/tex])) / 15.0 ≈ 1.35

Therefore, the Q of the circuit is approximately 1.35.

(d) The amplitude of the voltage across the inductor at resonance can be calculated using the formula:

VL = Q × VR

Given:

Q = 1.35

VR = ΔVmax = 100 V

Substituting the values into the formula:

VL = 1.35 × 100 ≈ 135 V

Therefore, the amplitude of the voltage across the inductor at resonance is approximately 135 V (or 0.135 kV).

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Answer:

The kinetic energy of a moving object is directly proportional to its mass and directly proportional to the square of its velocity. This means that an object with twice the mass and equal speed will have twice the kinetic energy while an object with equal mass and twice the speed will have quadruple the kinetic energy.

Answer:

It would be B.

Explanation:

The correct order of electromagnetic radiation from lowest to highest energy per photon is- Radio waves < Microwaves < Infrared radiation > Visible light < Ultraviolet radiation < and x-rays. So the order is 1,2,4,3.

Radio waves contain low-energy photons; microwave photons have slightly higher energy than radio waves; infrared photons have more energy than visible, ultraviolet, and x-rays.

Gamma irradiation is very penetrating, and it interacts with matter by ionization in three ways; photoelectric effects, Compton scattering, or pair generation. These radiations are referred to as non-ionizing radiations as they can ionize the molecules due to high penetration power.

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Answer:

3

Explanation:

mass number minus the atomic number

35-32

3

Answer: the answer is d

Explanation:

I got it right

On a turntable, two pennies spin, (a) The angular velocity of coin A is (d) twice that of coin B. (b) The centripetal acceleration of coin A is (a) one-fourth ([tex]\frac{1}{4}[/tex]) that of coin B.

Let's denote the angular velocity of coin A as ω[tex]_A[/tex] and the angular velocity of coin B as ω[tex]_B[/tex].

The relationship between the angular velocities of two objects rotating about the same axis is determined by the ratio of their radial distances from the axis.

In this case, it is given that coin B is twice as far from the axis as coin A. Let's denote the radial distance of coin A as [tex]r_A[/tex] and the radial distance of coin B as [tex]r_B[/tex]. Therefore, we have:

[tex]r_B[/tex] = 2[tex]r_A[/tex]

Now, the relationship between the angular velocities is:

[tex]\begin{equation}\frac{\omega_A}{\omega_B} = \frac{r_B}{r_A}[/tex]

Substituting the given values, we have:

[tex]\begin{equation}\frac{\omega_A}{\omega_B} = \frac{2r_A}{r_A}[/tex]

Simplifying the expression, we find:

[tex]\[\frac{\omega_A}{\omega_B} = 2\][/tex]

Therefore, the angular velocity of coin A is twice that of coin B.

Now let's consider the centripetal accelerations of coin A and coin B.

The centripetal acceleration is given by the formula:

a = ω²r

where a is the centripetal acceleration, ω is the angular velocity, and r is the radial distance from the axis.

For coin A, the centripetal acceleration is:

[tex]a_A[/tex] = ω[tex]_A[/tex]² * [tex]r_A[/tex]

For coin B, the centripetal acceleration is:

[tex]a_B[/tex] = ω[tex]_B[/tex]² * [tex]r_B[/tex]

We know that [tex]\begin{equation}\frac{\omega_A}{\omega_B} = 2[/tex], and [tex]r_B[/tex] = 2[tex]r_A[/tex]. Substituting these values into the equation for [tex]a_B[/tex], we have:

[tex]a_B[/tex] = ω[tex]_B[/tex]² * (2[tex]r_A[/tex])

Simplifying the expression, we find:

[tex]a_B[/tex] = 4ω[tex]_B[/tex]² * [tex]r_A[/tex]

Now let's compare the centripetal accelerations of coin A and coin B:

[tex]\begin{equation}\frac{a_A}{a_B} = \frac{\omega_A^2 r_A}{4\omega_B^2 r_A}[/tex]

Simplifying the expression, we find:

[tex]\begin{equation}\frac{a_A}{a_B} = \frac{1}{4}[/tex]

Therefore, the centripetal acceleration of coin A is one-fourth ([tex]\frac{1}{4}[/tex]) that of coin B.

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Answer:

 q1 = q₂=  -3

therefore each sphere has the same charge of -3 untis

Explanation:

The metallic spheres have mobile charge, so when the two spheres come into contact the total charge

           Q_total = q₁ + q₂

           Q_total = -2 -4

          Q_total = -6 units

it is distributed in between the two spheres evenly since the charges of the same sign repel each other.

When the spheres separate each one has

            q₁ = -6/2

            q1 = q₂=  -3

therefore each sphere has the same charge of -3 untis

The formula relating the difference in decibel levels (b1 and b2) of a sound source to the ratio of its distances (r1 and r2) from the receivers is

b1 - b2 = 20 * log10(r2 / r1)

The difference in decibel levels (b1 and b2) of a sound source can be related to the ratio of its distances (r1 and r2) from the receivers using the inverse square law. The inverse square law states that the intensity of sound decreases proportionally to the square of the distance from the source.

The formula for the difference in decibel levels can be expressed as

b1 - b2 = 10 * log10(I1 / I2)

Where:

b1 and b2 are the decibel levels at distances r1 and r2 respectively.

I1 and I2 are the intensities of sound at distances r1 and r2 respectively.

According to the inverse square law, the relationship between the intensities and distances is:

I1 / I2 = [tex][(r2 / r1)^2][/tex]

Substituting this into the formula for the difference in decibel levels:

b1 - b2 = 10 * log10[tex][(r2 / r1)^2][/tex]

Using the logarithmic property log(a^b) = b * log(a), we can simplify further:

b1 - b2 = 20 * log10(r2 / r1)

Therefore, the formula relating the difference in decibel levels (b1 and b2) of a sound source to the ratio of its distances (r1 and r2) from the receivers is:

b1 - b2 = 20 * log10(r2 / r1)

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