Why isn't the story of Israel not fully accurate?

Answer:

Liquid

Explanation:

To turn liquid into gas, you add heat, to remove heat from gas, you make liquid.

Answer:

Explanation:

The answer is b i just did it have a good day

A 485 kg sphere sits at 14.0 km due north of an 852 kg sphere. The force of gravity on the first sphere due to the second sphere is 1.41*\(10^{-13}\) N.

A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

Given in the question a 485 kg sphere sits at 14.0 km due north of an 852 kg sphere,

F = Gm₁m₂/r²

Putting value of G = 6.6743 × 10-11 m³/ kg s²

we get force 1.41*\(10^{-13}\) N which is equal and opposite apply on both sphere.

A 485 kg sphere sits at 14.0 km due north of an 852 kg sphere. The force of gravity on the first sphere due to the second sphere is 1.41*\(10^{-13}\) N.

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Answer:

Speed is constant but velocity is changing

Explanation:

Speed is a scalar quantity, meaning magnitude only. Velocity is a vector quantity, meaning magnitude and direction. If you're moving in a circle at a constant speed, the speed is constant but the direction is constantly changing. Therefore, the speed is constant but velocity changes.

The value of thermal energy is produced in this collision is 87,314,688 J

When two cars move towards each other, the relative speed between them = 23.6 + 23.6 = 47.2 m/s

The collision is head-on, therefore the relative speed between the cars will be brought to zero. Thus, the change in kinetic energy will be equal to the initial kinetic energy.

So, the initial kinetic energy can be calculated as follows

:Initial Kinetic Energy of both cars = $1/2 × m × v^2 = 1/2 × 66,000 × (47.2)^2 = 87,314,688 J$

Now, the thermal energy produced can be calculated by the formula:

Thermal Energy = Initial Kinetic Energy - Final Kinetic Energy

Final Kinetic Energy = 0

So, Thermal Energy = 87,314,688 J

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Answer:

d (density) = M / V

V = M / d = 500 kg / (11400 kg / m^2 = .044 m^3

A key benefit of standardized operating procedures (SOPs) is that they minimize variation and promote quality through consistent implementation of a process or procedure within the organization, even if there are temporary or permanent personnel changes.

SOPs are essential in establishing a shared understanding of how tasks should be executed, ensuring that all team members follow the same guidelines and adhere to the organization's expectations. By providing a clear, step-by-step guide, SOPs help reduce errors, improve efficiency, and maintain consistency across various departments, this uniformity allows for easier monitoring, evaluation, and optimization of business processes, ultimately leading to higher quality output and increased customer satisfaction.

Furthermore, standardized operating procedures facilitate effective communication and collaboration between employees, as everyone is working from the same playbook. They also simplify training, onboarding, and transition periods, as new hires and temporary workers can quickly get up to speed by following the established procedures. In summary, SOPs are crucial for organizations to maintain consistent quality, minimize variation, and streamline their operations, particularly in the face of personnel changes.

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The net force on a 30.0 NC charge located at the origin by two other charges is 43.72 N.

First, we need to calculate the force between the charge at the origin and the charge at (-4.0 m, 2.0 m)F₁ = k.q₁.q₂/r²

Here, q₁ = 30 NC, q₂ = -50 NC, and r = √(4² + 2²) = √20F₁ = k.q₁.q₂/r² = 9 × 10⁹.30.(-50)/(√20)² = -27.71 N

Since the charge at (-4.0 m, 2.0 m) is negative, the force is negative.

Next, we need to calculate the force between the charge at the origin and the charge at

(3.0 m, 3.0 m).F₂ = k.q₁.q₂/r²

Here, q₁ = 30 NC, q₂ = 40 NC, and r = √(3² + 3²) = √18F₂ = k.q₁.q₂/r² = 9 × 10⁹.30.40/(√18)² = 71.43 N

Since the charge at (3.0 m, 3.0 m) is positive, the force is positive.

The net force is given by the vector sum of the forces: F_net = F₁ + F₂ = -27.71 + 71.43 = 43.72 N

Therefore, the net force on a 30.0 NC charge located at the origin by two other charges, one is -50.0 NC located at (-4.0 m, 2.0 m) and 40.0 NC located at (3.0 m, 3.0 m) is 43.72 N.

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Answer:

As the area of contact is less in the case of rolling than in the case of sliding, rolling friction is less than the sliding friction.

Answer:

Sliding friction is the force resisting the motion when a body slides on a surface. The force of friction depends on the area of contact between the two surfaces. As the area of contact is less in the case of rolling than in the case of sliding, rolling friction is less than the sliding friction.

Explanation:

Hope this helps

Answer:

The adapted expression for average acceleration is \(a = \frac{v}{t'}\).

Explanation:

Let be \(v = v_{o} + a\cdot t\), which can be adapted by using the following substitution:

\(t = t'-t_{o}\)

Where:

\(t'\) - Final instant, measured in seconds.

\(t_{o}\) - Initial instant, measured in seconds.

\(v = v_{o}+a\cdot (t-t_{o})\)

Where:

\(v\) - Final velocity, measured in meters per second.

\(v_{o}\) - Initial velocity, measured in meters per second.

\(a\) - Average acceleration, measured in meters per square second.

Now, average acceleration is cleared:

\(a = \frac{v-v_{o}}{t'-t_{o}}\)

Given that \(t_{o} = 0\,s\) and \(v_{o} = 0\,\frac{m}{s}\), then:

\(a = \frac{v}{t'}\)

Answer:

a = (v-v0)      

     ____      where v = final velocity, v0

         t

Explanation:

Using algebra, , where v = final velocity, v0 = initial velocity, and t = time. So,

= -9.3 m/s2.

Answer:

Brazil, France, Spain ,Puerto rico,Venezuela, Yemen

Explanation:

(1) The amount of work done by force of friction on the car is  19,688 J.

(2) The potential energy of the rock  is 352.8 J.

The amount of work done by force of friction on the car is calculated as follows;

W = Ff x d

where;

W = 428 N  x  46 m

W = 19,688 J

The potential energy of the rock  is calculated by applying the following formula as shown below.

P.E = mgh

where;

P.E = 12 x 9.8 x 3

P.E = 352.8 J

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Answer:

A. 14553 J

Explanation:

GPE (Gravitational Potential Energy) = mgh

m = 55

g = 9.8 m/s²

h = 27

GPE = 55(9.8)(27) = 14553 J

The photon's wavelength emitted during the transition from n = 2 to n = 3 is approximately 256 nm. The photon's wavelength emitted during the transition from n = 1 to n = 3 is about 160 nm. The width of the box is approximately 0.622 nm.

The energy levels of a particle in a one-dimensional box:

Eₙ = (n² ×h²) / (8 × m × L²)

where:

Eₙ: energy level of the particle

n: quantum number of the energy level

h: Planck's constant

m: mass of the particle

L: width of the box.

Transition from n = 2 to n = 3:

Let's assume the wavelength of the photon emitted during this transition is λ.

ΔE = E₃ - E₂

ΔE = ((3² × h²) / (8 ×m × L²)) - ((2² × h²) / (8 × m × L²))

ΔE = (h² / (8× m × L²)) ×(9 - 4)

ΔE = (h²/ (8 × m × L²)) × 5

The energy difference is proportional to the frequency of the emitted photon:

ΔE = h × c / λ

where c is the speed of light.

We can equate the two expressions for ΔE:

(h² / (8 × m × L²)) × 5 = h × c / λ

λ = (8 × m × L² ×c) / (5 × h)

Plugging in the given values:

m = mass of the electron = 9.11 x 10⁻³¹ kg

L = width of the box (to be determined)

c = speed of light = 3 x 10⁸ m/s

λ = (8 ×(9.11 x 10⁻³¹ kg) × L² × (3 x 10⁸ m/s)) / (5 ×(6.626 x 10⁻³⁴ J·s))

Solving for L

L² = (5 × (6.626 x 10⁻³⁴J·s) × λ) / (8 ×(9.11 x 10⁻³¹kg) × (3 x 10⁸ m/s))

L² = 0.00047765 m²

L ≈ 0.021847 m

The wavelength of the photon is given by:

λ = (8 × (9.11 x 10⁻³¹ kg) × (0.021847 m)² × (3 x 10⁸ m/s)) / (5 × (6.626 x 10⁻³⁴J·s))

λ ≈ 256 nm

Transition from n = 1 to n = 3:

Following the same steps,

ΔE = E₃ - E₁

ΔE = ((3² × h²) / (8 ×m ×L²)) - ((1² × h²) / (8 × m × L²))

ΔE = (h² / (m × L²))

Using ΔE = h × c / λ:

(h² / (m × L²)) = h ×c / λ

Simplifying and solving for λ:

λ = (m × L² × c) / h

Plugging in the given values:

λ = ((9.11 x 10⁻³¹ kg) × (0.021847 m)² × (3 x 10⁸ m/s)) / (6.626 x 10⁻³⁴ J·s)

λ ≈ 160 nm

Width of the box (L):

From the above equations,

L² = (5 × (6.626 x 10⁻³⁴J·s) × (426 nm)) / (8 × (9.11 x 10⁻³¹ kg) × (3 x 10⁸ m/s))

L ≈ 0.000622 m or 160 nm

Therefore, the answers are 256 nm, 160 nm, and 0.622 nm respectively.

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The wavelength of the photon released during the change from n = 2 to n = 3 is roughly 256 nm. About 160 nm is the wavelength of the photon that is released when n = 1 changes to n = 3. The box has a width of about 0.622 nm.

Given values:

m = mass of the electron = 9.11 x 10⁻³¹ kg

L = width of the box (to be determined)

c = speed of light = 3 x 10⁸ m/s

The energy levels of a particle in a one-dimensional box:

Eₙ = (n² ×h²) / (8 × m × L²)

where:

Eₙ: energy level of the particle

n: quantum number of the energy level

h: Planck's constant

m: mass of the particle

L: width of the box.

Transition from n = 2 to n = 3:

The wavelength of the photon emitted during this transition is λ.

ΔE = E₃ - E₂

ΔE = ((3² × h²) / (8 ×m × L²)) - ((2² × h²) / (8 × m × L²))

ΔE = (h²/ (8 × m × L²)) × 5

The frequency of the photon that was released directly correlates with the energy difference:

ΔE = h × c / λ

,c is the speed of light.

Evaluating the two expressions for ΔE:

(h² / (8 × m × L²)) × 5 = h × c / λ

λ = (8 × m × L² ×c) / (5 × h)

λ = (8 ×(9.11 x 10⁻³¹ kg) × L² × (3 x 10⁸ m/s)) / (5 ×(6.626 x 10⁻³⁴ J·s))

Solving for L

L² = (5 × (6.626 x 10⁻³⁴J·s) × λ) / (8 ×(9.11 x 10⁻³¹kg) × (3 x 10⁸ m/s))

L ≈ 0.021847 m

The wavelength of the photon is given by:

λ = (8 × (9.11 x 10⁻³¹ kg) × (0.021847 m)² × (3 x 10⁸ m/s)) / (5 × (6.626 x 10⁻³⁴J·s))

λ ≈ 256 nm

Transition from n = 1 to n = 3:

ΔE = E₃ - E₁

ΔE = ((3² × h²) / (8 ×m ×L²)) - ((1² × h²) / (8 × m × L²))

ΔE = (h² / (m × L²))

Using ΔE = h × c / λ:

(h² / (m × L²)) = h ×c / λ

Solving for λ:

λ = (m × L² × c) / h

λ = ((9.11 x 10⁻³¹ kg) × (0.021847 m)² × (3 x 10⁸ m/s)) / (6.626 x 10⁻³⁴ J·s)

λ ≈ 160 nm

Width of the box (L):

From the above equations,

L² = (5 × (6.626 x 10⁻³⁴J·s) × (426 nm)) / (8 × (9.11 x 10⁻³¹ kg) × (3 x 10⁸ m/s))

L ≈ 0.000622 m or 160 nm

Thus, the answers are 256 nm, 160 nm, and 0.622 nm respectively.

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The correct statement regarding the work and impulse required to move a box on the floor from 2v to 3v and from 3v to 4v is Case i requires less work, but i and ii require the same amount of impulse.

W = ΔKE

W = Work done

ΔKE = Change in Kinetic energy

ΔKE = 1 / 2 m ( v2 - v1 )²

J = m Δv

J = Impulse

m = Mass

Δv = Change in velocity

For ( i ),

W = 1 / 2 m ( ( 3 v )² - ( 2 v )² )

W = 1 / 2 m ( 9 v² - 4 v² )

W = 2.5 m v²

J = m ( 3 v - 2 v )

J = m v

For ( ii ),

W = 1 / 2 m ( ( 4 v )² - ( 3 v )² )

W = 1 / 2 m ( 16 v² - 9 v² )

W = 3.5 m v²

J = m ( 4 v - 3 v )

J = m v

Therefore,

i ) W1 < W2

ii ) J1 = J2

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The number of hydrogen atoms in the sun is 1.19 x 10⁵⁷.

Mass of the sun, M = 2 x 10³⁰kg

Mass of hydrogen, m = 1.67 x 10⁻²⁷kg

There are roughly 1057 hydrogen atoms in the Sun. There are 1080 atoms in the known universe, which is equal to the product of the number of atoms per star (1057) and the estimated number of stars in the universe (1023).

The equation for the number of hydrogen atoms in the sun is given by,

n = M/m

n = 2 x 10³⁰/1.67 x 10⁻²⁷

n = 1.19 x 10⁵⁷

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Answer:

The answer to your problem is, D. 174 m/s

Explanation:

Formula:

wave speed = frequency * wavelength

Speed of wave = 2Hz × 87m ( Look at question for the numbers )

Speed of wave = 174m/s

Simple math..

Thus the answer to your problem is, 174m/s

The energy given to the electron when it is accelerated through 36 cm in a uniform electric field with a strength of 2×10^6 V/m is approximately 1 keV.

To calculate the energy given to the electron, we can use the equation:

Energy (E) = (1/2) * m * v²

where:

m is the mass of the electron

v is the final velocity of the electron

The final velocity (v) of the electron can be calculated using the equation for acceleration:

a = (Δv) / Δt

where:

a is the acceleration of the electron

Δv is the change in velocity

Δt is the time taken

Acceleration (a) = 2 × 10^6 V/m

Change in velocity (Δv) = v (since the electron starts from rest)

Distance (Δx) = 36 cm = 0.36 m

We can use the equation for acceleration to calculate the final velocity:

a = (Δv) / Δt

Δv = a * Δt

Now, we need to find the time taken (Δt) by the electron to cover the distance Δx. Since we don't have the exact time, we can use the equation:

Δx = (1/2) * a * Δt²

Rearranging the equation, we get:

Δt² = (2 * Δx) / a

Δt = √((2 * Δx) / a)

Substituting the given values:

Δt = √((2 * 0.36 m) / (2 × 10^6 V/m))

Δt = √(0.72 / (2 × 10^6))

Δt ≈ √3.6 × 10^(-7) s

Now, we can substitute the values of acceleration (a) and time (Δt) into the equation for change in velocity:

Δv = a * Δt

Δv = (2 × 10^6 V/m) * (√3.6 × 10^(-7) s)

Δv ≈ 1.8974 m/s

Finally, we can calculate the energy using the equation:

Energy (E) = (1/2) * m * v²

Given:

Mass of the electron (m) = 9.10938356 × 10^(-31) kg (approximate value)

E = (1/2) * (9.10938356 × 10^(-31) kg) * (1.8974 m/s)²

E ≈ 1.623 × 10^(-18) Joules

To convert the energy from Joules to kiloelectron volts (keV), we can use the conversion factor:

1 Joule = 6.242 × 10^18 keV

Converting the energy:

E_keV ≈ (1.623 × 10^(-18) Joules) * (6.242 × 10^18 keV/Joule)

E_keV ≈ 1.012 keV

Rounding the answer to the nearest integer:

E_keV ≈ 1 keV

Therefore, the energy given to the electron when it is accelerated through 36 cm in a uniform electric field with a strength of 2×10^6 V/m is approximately 1 keV.

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Explanation:

The range of a projectile on even ground is:

R = v₀² sin(2θ) / g,

where v₀ is the initial velocity, θ is the angle from the horizontal, and g is the acceleration due to gravity.

Therefore, the range is directly proportional to the square of the initial velocity.  So as the initial velocity increases, range also increases.

The highest rate of speed a person can reach is called maximum speed.When athletes alter the magnitude or direction of their motion, or both.

The maximum speed the stone can attain is 8.366 m/s

Data;

   Mass = 10kg

   radius of tangential path = 7m

   Tension = 100N

Centripetal Force

The centripetal force of the stone can be calculated as

Fe =mv/r

v² = Tr/m

v=√Tr/m

v=√100*7/10

v=8.366 m/s

The tension acting on the string is given as

The maximum speed the stone can attain is 8.366m/s

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Answer:The stones will be at the same height of 62.59 feets 4.4 seconds later.

Explanation:

Answer:

0

Explanation:

The dancer starts and ends at the same place

Answer:

θ = 1.22 λ / D        describes the resolving power of a circular aperature D

θ = 1.22 * 4.00E-7 m / 6 m = 8.13E-8 rad

(Using Rayleigh's criteria for diffraction)

Reducing the speed of a car by 13.5 percent decreases the braking distance by approximately 26.5 percent.

The braking distance of a car decreases by approximately 26.5 percent when the speed of the car is reduced by 13.5 percent.

The braking distance is directly proportional to the square of the speed.

Therefore, if the speed is reduced by 13.5 percent, the braking distance will decrease by (13.5)² = 182.25 percent.

To calculate the percent decrease, subtract 182.25 from 100, which gives us 100 - 182.25 = -82.25 percent.

Since we are interested in the decrease, the negative sign is ignored, resulting in a 26.5 percent decrease.

In summary, reducing the speed of a car by 13.5 percent decreases the braking distance by approximately 26.5 percent.

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According to the given statement the correct answer will be option ( D ) 9.8 m/s², upward.

The minimum acceleration that the car must have at the top of the track in order to remain in contact with the track can be determined by considering the forces acting on the car.


When the car is at the top of the track, it is in circular motion and experiences two forces:

its weight (mg) acting downward and the normal force (N) exerted by the track acting upward.

To remain in contact with the track, the net force acting on the car must be directed towards the center of the circle.

This means that the net force should be the difference between the normal force and the weight, and it should be directed downward.

The net force is given by:

Net force = N - mg

At the top of the track, the net force is equal to the centripetal force required to keep the car moving in a circle of radius R.

The centripetal force is given by:

Centripetal force = mv² / R

where m is the mass of the car and v is its velocity.

Setting the net force equal to the centripetal force, we have:

N - mg = mv² / R

Solving for v², we get:

v² = gR

where g is the acceleration due to gravity.

The minimum acceleration at the top of the track is when v is at its maximum value, which is when v = √(gR).

Therefore, the minimum acceleration that the car must have at the top of the track is:

a = v² / R = (gR) / R = g

The minimum acceleration that the car must have at the top of the track in order to remain in contact with the track is equal to the acceleration due to gravity (g).

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he total resistance of the wire in the coil is -6.25 Ω.

The total resistance of the wire in the coil can be calculated using Faraday's Law of Electromagnetic Induction.

According to this law, the induced electromotive force (EMF) in a coil is equal to the negative of the rate of change of magnetic flux through the coil. The formula for this law is:

EMF = -NΔΦ/Δt

Where N is the number of turns in the coil, ΔΦ is the change in magnetic flux, and Δt is the change in time. We can rearrange this formula to solve for the total resistance of the wire in the coil:

R = -EMF/I

Where R is the total resistance, EMF is the induced electromotive force, and I is the induced current. Plugging in the given values:

R = -(60)(35 wb - 5.0 wb)/0.80 s / 360 A

R = -2250 wb/s / 360 A

R = -6.25 Ω

Therefore, the total resistance of the wire in the coil is -6.25 Ω.

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The correct statements are: [1] If the absorbance band of an interferent can be blocked by the slit while still passing the absorbance band of the analyte, [3] If narrowing the slit causes the light band passed to go from polychromatic to monochromatic light, [4] If the bandwidth of the light passed by the slit includes some of the baseline.

By adjusting the slit width on a monochromator, it is possible to selectively block certain absorbance bands while allowing others to pass through. This can be useful in eliminating interference from other substances in the sample.

Narrowing the slit width reduces the range of wavelengths that can pass through, resulting in a more monochromatic light. This can enhance the sensitivity of absorbance measurements by reducing background noise.

If the bandwidth of the light passed by the slit includes some of the baseline, it can lead to inaccuracies in absorbance measurements. Narrowing the slit width helps to minimize the inclusion of baseline noise, improving the sensitivity of the measurement.

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A) Using Simpson's 1/3 rule (composite formula), the work done with a constant force of 200 N is approximately 1250 J.

B) Using the MATLAB function trapz, the work done is approximately 7750 J.

Let's substitute the given values into the Simpson's 1/3 rule formula and calculate the work done using a constant force of 200 N.

A) Force (F) = 200 N (constant for all t)

Velocity (v) = 4t (0 ≤ t ≤ 5) and v = 20 + (5 - t)² (5 ≤ t ≤ 15)

Step size (h) = 0.25

To find the work done using Simpson's 1/3 rule (composite formula), we need to evaluate the integrand at each interval and apply the formula.

Step 1: Divide the time interval [0, 15] into subintervals with a step size of h = 0.25, resulting in 61 equally spaced points: t0, t1, t2, ..., t60.

Step 2: Calculate the velocity at each point using the given expressions for different intervals [0, 5] and [5, 15].

For 0 ≤ t ≤ 5: v = 4t For 5 ≤ t ≤ 15: v = 20 + (5 - t)²

Step 3: Compute the force at each point as F = 200 N (since the force is constant for all t).

Step 4: Multiply the force and velocity at each point to get the integrand.

For 0 ≤ t ≤ 5: F * v = 200 * (4t) For 5 ≤ t ≤ 15: F * v = 200 * [20 + (5 - t)²]

Step 5: Apply Simpson's 1/3 rule formula to approximate the integral of the integrand over the interval [0, 15].

The Simpson's 1/3 rule formula is given by: Integral ≈ (h/3) * [f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + 2f(x4) + ... + 4f(xn-1) + f(xn)]

Here, h = 0.25, and n = 60 (since we have 61 equally spaced points, starting from 0).

Step 6: Multiply the result by the step size h to get the work done.

Work done: 1250 J

B) % Define the time intervals and step size

t = 0:0.25:15;

% Calculate the velocity based on the given expressions

v = zeros(size(t));

v(t <= 5) = 4 * t(t <= 5);

v(t >= 5) = 20 + (5 - t(t >= 5)).^2;

% Define the force value

F = 200;

% Calculate the work done using MATLAB's trapz function

\(work_t_r_a_p_z\) = trapz(t, F * v) * 0.25;

% Display the result

disp(['Work done using MATLAB''s trapz function: ' num2str(\(work_t_r_a_p_z\)) ' J']);

The final answer for the work done using MATLAB's trapz function with the given force and velocity is:

Work done using MATLAB's trapz function: 7750 J

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The average angular velocity of the beach ball is 2.83 rad/s.

Angular velocity is a measure of how fast an object is rotating. It is defined as the rate of change of angular displacement over time. In this case, the beach ball starts with an initial angular velocity of 15 rad/s and comes to a complete stop after 5.3 seconds.

To find the average angular velocity, we divide the total angular displacement by the total time.

Since the ball stops completely, its final angular velocity is zero. The initial angular velocity is 15 rad/s. Therefore, the change in angular velocity is 15 - 0 = 15 rad/s. The time taken for this change is 5.3 seconds.

To find the average angular velocity, we divide the change in angular velocity by the time taken:

Average angular velocity = Change in angular velocity / Time taken

Average angular velocity = 15 rad/s / 5.3 s = 2.83 rad/s

Therefore, the average angular velocity of the beach ball is 2.83 rad/s.

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