A thin coil has 17 rectangular turns of wire. When a current of 4 A runs through the coil, there is a total flux of 5 ✕ 10−3 T · m2 enclosed by one turn of the coil (note that

Φ = kI,

and you can calculate the proportionality constant k). Determine the inductance in henries.

We can find the distance the sled travels before it comes to rest by equating the work done by the frictional force with the initial kinetic energy of the sled. We first calculate the initial kinetic energy with (1/2)mv². The force of friction is given by μmg. The work done by friction is equal to this force times the distance traveled, and we set this equal to the negative initial kinetic energy to account for the sled stopping.

The subject of this question is about the concept of work and energy in a physics problem involving a sled on a frozen pond. To solve this, we would use the principle of work and energy where the work done on the sled is equal to the change in its kinetic energy according to the work-energy theorem. Since the sled comes to rest, the final kinetic energy is zero.

First, we calculate the initial kinetic energy of the sled using the formula (1/2)mv², where m is the mass (8.54 kg), and v is the velocity (1.87 m/s). Next, we know that the work done by the friction on the sled is equal to the force of friction times the distance traveled which is equal to the change in kinetic energy. The force of friction can be calculated by multiplying the mass of the sled, the acceleration due to gravity (9.8 m/s²), and the coefficient of kinetic friction (0.087). Set this work equal to the negative initial kinetic energy (as it acts to bring the sled to a stop) and solve for the distance.

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To find the distance the sled moves before coming to rest, we can use the work and energy concepts. Work done by the friction force is equal to the change in kinetic energy. By setting the work done by friction equal to the change in kinetic energy and using the given values, we can solve for the distance the sled moves.

To find the distance the sled moves before coming to rest, we will use the work and energy concepts. The work done on an object is equal to the change in its kinetic energy, so we can use the work-energy principle to solve the problem. The work done by the friction force is equal to the force of friction multiplied by the distance the sled moves. Setting the work done by friction equal to the change in kinetic energy, we can solve for the distance the sled moves.

The work done by the friction force is given by the equation:
W = F * d * cos(180°)

where W is the work done, F is the force of friction, d is the distance, and cos(180°) is the angle between the force of friction and the displacement. The work done by the friction force is equal to the negative change in kinetic energy of the sled, which can be calculated as:
ΔKE = -(1/2) * m * v²

where ΔKE is the change in kinetic energy, m is the mass of the sled, and v is the final velocity of the sled.

Setting the work done by friction equal to the negative change in kinetic energy and solving for the distance d:

W = -(1/2) * m * v²
F * d * cos(180°) = -(1/2) * m * v²

Using the given values for the mass of the sled (m = 8.54 kg), the initial velocity (v = 1.87 m/s), and the coefficient of kinetic friction (µ = 0.087), we can solve for the distance d.

First, we need to calculate the magnitude of the friction force using the equation:
F = µ * N

where µ is the coefficient of kinetic friction and N is the normal force. The normal force is equal to the weight of the sled, which can be calculated as:
N = m * g

where g is the acceleration due to gravity (approximately 9.8 m/s²).

Substituting the values into the equations:

F = 0.087 * (8.54 kg * 9.8 m/s²) = 7.94 N

Now, we can substitute the values for the force of friction and the change in kinetic energy into the equation:

F * d * cos(180°) = -(1/2) * (8.54 kg) * (1.87 m/s)²

Using the cosine of 180° as -1:

7.94 N * d * (-1) = -(1/2) * (8.54 kg) * (1.87 m/s)²

Simplifying the equation:

-7.94 N * d = -(1/2) * (8.54 kg) * (1.87 m/s)²

Dividing both sides of the equation by -7.94 N:

d = [(1/2) * (8.54 kg) * (1.87 m/s)²] / 7.94 N

Calculating the distance d:

d = 2.008 m

Therefore, the sled moves a distance of 2.008 meters before coming to rest.

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Answer:

[tex]V_p = 267.258 m/s[/tex]

[tex]V_n = 38.375 m/s[/tex]      

Explanation:

using the law of the conservation of the linear momentum:

[tex]P_i = P_f[/tex]

where [tex]P_i[/tex] is the inicial momemtum and [tex]P_f[/tex] is the final momentum

the linear momentum is calculated by the next equation

P = MV

where M is the mass and V is the velocity.

so:

[tex]P_i = m(270 m/s)[/tex]

[tex]P_f = mV_P + M_nV_n[/tex]

where m is the mass of the proton and [tex]V_p[/tex] is the velocity of the proton after the collision, [tex]M_n[/tex] is the mass of the nucleus and [tex]V_n[/tex] is the velocity of the nucleus after the collision.

therefore, we can formulate the following equation:

m(270 m/s) = m[tex]V_p[/tex] + 14m[tex]V_n[/tex]

then, m is cancelated and we have:

270 = [tex]V_p[/tex] + [tex]14V_n[/tex]

This is a elastic collision, so the kinetic energy K is conservated. Then:

[tex]K_i = \frac{1}{2}MV^2 = \frac{1}{2}m(270)^2[/tex]

and

Kf = [tex]\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2[/tex]

then,

[tex]\frac{1}{2}m(270)^2[/tex] =  [tex]\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2[/tex]

here we can cancel the m and get:

[tex]\frac{1}{2}(270)^2[/tex] =  [tex]\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2[/tex]

now, we have two equations and two incognites:

270 = [tex]V_p[/tex] + [tex]14V_n[/tex]  (eq. 1)

[tex]\frac{1}{2}(270)^2[/tex] =  [tex]\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2[/tex]

in the second equation, we have:

36450 =  [tex]\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2[/tex]  (eq. 2)

from this last equation we solve for [tex]V_n[/tex] as:

[tex]V_n = \sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }[/tex]

and replace in the other equation as:

270 = [tex]V_p +[/tex] 14[tex]\sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }[/tex]

so,

[tex]V_p = -267.258 m/s[/tex]

Vp is negative because the proton go in the -i hat direction.

Finally, replacing this value on eq. 1 we get:

[tex]V_n = \frac{270+267.258}{14}[/tex]

[tex]V_n = 38.375 m/s[/tex]  

Answer:

[tex]1\times 10^{8}\ Pa[/tex]

Explanation:

k = Boltzmann constant = [tex]1.38\times 10^{-23}\ J/K[/tex]

r = Radius of gas molecule = [tex]2\times 10^{-10}\ m[/tex]

t = Temperature = 300 K

P = Pressure

Volume of gas per molecule is given by

[tex]V=\frac{4}{3}\pi r^3\\\Rightarrow V=\frac{4}{3}\pi (2\times 10^{-10})^3\\\Rightarrow V=3.35103\times 10^{-29}\ m^3[/tex]

From the ideal gas law we have

[tex]PV=kt\\\Rightarrow P=\frac{kt}{V}\\\Rightarrow P=\frac{1.38\times 10^{-23}\times 300}{3.35103\times 10^{-29}}\\\Rightarrow P=123544104.35\ Pa=1\times 10^{8}\ Pa[/tex]

The pressure at which the finite volume of the molecules should cause noticeable deviations from ideal-gas behavior is [tex]1\times 10^{8}\ Pa[/tex]

The pressure at which the finite volume of oxygen molecules causes deviations from ideal-gas behavior can be estimated using the ideal gas law and considering the volume of a molecule. At ordinary temperatures, the approximate pressure is 2.04 × 10^11 Pa.

To estimate the pressure at which the finite volume of oxygen molecules causes noticeable deviations from ideal-gas behavior, we need to equate the volume of gas per molecule to the volume of a molecule. The volume of the gas per molecule is equivalent to the volume of a sphere with a radius of 2.0 × 10-10 m, which is approximately 4.19 × 10-30 m3. Using the ideal gas law, we can calculate the pressure using the formula PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. Assuming 1 mole of oxygen gas (which contains 6.022 × 1023 molecules), the number of moles can be calculated by dividing the volume of the gas per molecule by Avogadro's number. The approximate pressure at which deviations from ideal-gas behavior become noticeable is around 2.04 × 1011 Pa (Pascals).

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Answer:

Explanation:

If the acceleration due to gravity on the Moon's surface is [tex]a_M[/tex] and the acceleration due to gravity on the Earth is [tex]a_E[/tex], we can write that:

[tex]a_M=\frac{1}{6}a_E[/tex]

[tex]\frac{a_M}{a_E}=\frac{1}{6}[/tex]

If the radius of the Moon is [tex]r_M[/tex] and the radius of the Earth is [tex]r_E[/tex], we can write that:

[tex]r_M=\frac{1}{4}r_E[/tex]

[tex]\frac{r_M}{r_E}=\frac{1}{4}[/tex]

By Newton's 2nd Law we know that F=ma and using Newton's law of universal gravitation  we can calculate the gravitational force an object with mass m experiments from a planet with mass M being at a distance r from it. We will assume our object is on the surface so this distance will be the radius of the planet.

Since the force the object experiments is the force of gravitation we can write, for Earth:

[tex]F=ma_E=\frac{GM_Em}{r_E^2}[/tex]

which means:

[tex]a_E=\frac{GM_E}{r_E^2}[/tex]

[tex]M_E=\frac{a_Er_E^2}{G}[/tex]

And for the Moon:

[tex]F=ma_M=\frac{GM_Mm}{r_M^2}[/tex]

which means:

[tex]a_M=\frac{GM_M}{r_M^2}[/tex]

[tex]M_M=\frac{a_Mr_M^2}{G}[/tex]

We can then write the fraction:

[tex]\frac{M_M}{M_E}=\frac{a_Mr_M^2}{G}\frac{G}{a_Er_E^2}=\frac{a_M}{a_E}(\frac{r_M}{r_E})^2=\frac{1}{6}(\frac{1}{4})^2=0.01[/tex]

Which means:

[tex]M_M=0.01M_E[/tex]

Final answer:

To find the mass of the Moon in terms of the mass of the Earth, we can use the formula for gravitational acceleration and the given ratios between the Moon and Earth's acceleration and radii. Using these values, we can solve for the mass of the Moon in terms of the mass of the Earth.

Explanation:

To find the mass of the Moon in terms of the mass of the Earth, we can use the formula for gravitational acceleration: g = GM/r². Given that the acceleration due to gravity on the Moon is one-sixth that of Earth and the radius of the Moon is one-quarter that of Earth, we can set up the following equation:

(1/6) * (9.8 m/s²) = GM/(1/4 * R)²

Simplifying, we get: 1/6 * 9.8 = GM/(1/16)

Now, we can solve for the mass of the Moon (M) in terms of the mass of the Earth (m):

M = (1/6 * 9.8 * R²)/(1/16 * G)

Substituting the values for R and G, we get:

M = (1/6 * 9.8 * (1/4)²)/(1/16 * 6.67 × 10⁻¹¹)

M ≈ 0.123m

Answer:

Width of central bright fringe will be 0.00228 m

Explanation:

We have given wavelength of light [tex]\lambda =632.8nm=632.8\times 10^{-9}m[/tex]

Distance between the slits [tex]d=0.5mm=0.5\times 10^{-3}m[/tex]

Distance between slit and screen D = 1.8 m

We have to find the width

We know that width is given by

[tex]width=\frac{\lambda D}{d}=\frac{632.8\times 10^{-9}\times 1.8}{0.5\times 10^{-3}}=0.00228m[/tex]

So width of central bright fringe will be 0.00228 m

Applying the given values to the formula for the width of the central bright fringe in a double-slit interference pattern yields a result of approximately 4.556 × 10^(-3) m.

The width of the central bright fringe in a double-slit interference pattern can be calculated using the formula:

w = λL / d

where:

w is the width of the central bright fringe,

λ is the wavelength of the light,

L is the distance from the slits to the screen, and

d is the distance between the slits.

In this case:

λ = 632.8 nm = 6.328 × 10^(-7) m,

L = 1.8 m, and

d = 0.5 mm = 5 × 10^(-4) m.

Plug these values into the formula:

w = (6.328 × 10^(-7) m × 1.8 m) / (5 × 10^(-4) m)

Calculating this expression gives:

w ≈ 4.556 × 10^(-3) m

So, the width of the central bright fringe is indeed approximately 4.556 × 10^(-3) m

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The value of all options are mathematically given as

a) a=1.5 rad/s^2

b)theta=248 rad

c) theta '=248 radians

d) angles in degree=14207.511

Question Parameter(s):

Generally, the equation for the final angular speed  is mathematically given as

w2=w+at

Therefore

37=25+a 8

a=1.5 rad/s^2

b)

Average angular speed

[tex]\theta =\omega _1t+\frac{1}{2}\alpha t^2[/tex]

[tex]\theta =25\times 8+\frac{1}{2}\times 1.5\times 8^2[/tex]

theta=248 rad

Hence

theta =200+48

theta=248 rad

In conclusion,

c) theta '=248 radians

d) angles in degree=14207.511

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Final answer:

The angular acceleration is 1.5 rad/s², the average angular speed is 31 rad/s, the angle through which the wheel rotates is 320 rad, and the angle in degrees is 18333 degrees.

Explanation:

(a) To find the angular acceleration, we can use the formula: angular acceleration = (final angular velocity - initial angular velocity) / time interval. Plugging in the given values, we get: angular acceleration = (37 rad/s - 25 rad/s) / 8 s = 1.5 rad/s².

(b) The average angular speed can be found by taking the average of the initial and final angular velocities. So, the average angular speed = (25 rad/s + 37 rad/s) / 2 = 31 rad/s.

(c) The angle through which the wheel rotates can be found using the formula: angle = initial angular velocity * time + (1/2) * angular acceleration * time². Plugging in the given values, we get: angle = 25 rad/s * 8 s + (1/2) * 1.5 rad/s² * (8 s)² = 320 rad.

(d) To convert radians to degrees, we use the conversion factor: 1 radian = 180 degrees / pi. So, the angle in degrees = 320 rad * 180 degrees / pi = 18333 degrees (rounded to the nearest whole number).

Answer

given,

Distance for decibel reading

 r₁ = 13 m

 r₂ = 24 m

When the engineer is at r₁ reading is β₁ = 101 dB

now, Calculating the Intensity at r₁

Using formula

[tex]\beta = 10 log(\dfrac{I}{I_0})[/tex]

I₀ = 10⁻¹² W/m²

now inserting the given values

[tex]101= 10 log(\dfrac{I}{10^{-12}})[/tex]

[tex]10.1= log(\dfrac{I}{10^{-12}})[/tex]

[tex]\dfrac{I}{10^{-12}}= 10^{10.1}[/tex]

[tex]I =10^{10.1} \times 10^{-12}[/tex]

I = 0.01258 W/m²

now, calculating power at r₁

P₁ = I₁ A₁

P₁ = 0.01258 x 4 π r²

P₁ = 0.01258 x 4 π x 13²

P₁ = 26.72 W

Final answer:

The question deals with acoustics in physics, where an audio engineer uses the decibel scale to measure sound pressure levels and requires an understanding of the inverse square law and logarithmic measurements of sound intensity.

Explanation:

The student's question involves the concept of sound intensity levels and decibels, which are part of the field of acoustics, a topic in physics. Specifically, the student is dealing with the measurement of sound pressure levels at various distances from a sound source, which is an application of the inverse square law in acoustics. The decibel (dB) scale is a logarithmic unit used to measure sound intensity, with 0 dB representing the threshold of human hearing at an intensity of 10-12 W/m².

The loudness experienced by the audio engineer at various distances can be computed using the formula for sound intensity level (SIL), which is SIL = 10 log(I/I0), where I is the sound intensity and I0 is the reference intensity of 10-12 W/m².

Answer:

λ = 547.96 nm

Explanation:

given,

thickness of soap bubble = 103 nm

refractive index of thin film = 1.33

using formula of constructive interference

[tex]2 n t = (m + \dfrac{1}{2})\times \lambda[/tex]

t is thickness of the medium

n is refractive index

m = 0,1,2...

now,

[tex]2 n t = (m + 0.5)\times \lambda[/tex]

at m = 0

[tex]2\times 1.33 \times 103 =0.5\times \lambda[/tex]

λ = 547.96 nm

at m = 1

[tex]2 n t = (m + 0.5)\times \lambda[/tex]

[tex]2\times 1.33 \times 103 =1.5\times \lambda[/tex]

λ = 182.65 nm

the only visible light enhanced by the thin film is of wavelength

λ = 547.96 nm

Answer:

The speed of the water is 14.68 m/s.

Explanation:

Given that,

Time = 30 minutes

Distance = 11.0 m

Pressure = 101.3 kPa

Density of water = 1000 kg/m³

We need to calculate the speed of the water

Using equation of motion

[tex]v^2=u^2+2gs[/tex]

Where, u = speed of water

g = acceleration due to gravity

h = height

Put the value into the formula

[tex]0=u^2-2\times9.8\times11.0[/tex]

[tex]u=\sqrt{2\times9.8\times11.0}[/tex]

[tex]u=14.68\ m/s[/tex]

Hence, The speed of the water is 14.68 m/s.

To develop this problem it is necessary to apply the concepts of sum of capacitors in a circuit, either in parallel or in series.

When capacitors are connected in series, they consecutively add their capacitance.

However, when they are connected in parallel, the sum that is made is that of the inverse of the capacitance, that is [tex]\frac {1} {C}[/tex] where, C is the capacitance.

For the given case, it is best to connect two of these capacitors in series and one in parallel, so

We have three 100 pF capacitors, then

[tex]C_1 = 100 pF\\C_2 = 100 pF\\C_3 = 100 pF\\[/tex]

Here we can see how two capacitors 1 and 2 are in series and the third in parallel.

[tex]\frac{1}{C_{serie}} = \frac{1}{100pF}+\frac{1}{100pF}[/tex]

[tex]\frac{1}{C_{serie}} = \frac{1}{50pF}[/tex]

Investing equality

[tex]C_{serie} = 50pF[/tex]

Adding it in parallel, then

[tex]C_{total} = C_{serie} +C_3[/tex]

[tex]C_{total} = 50pF+100pF[/tex]

[tex]C_{total} = 150pF[/tex]

Answer:3.51

Explanation:

Given

Coefficient of Friction [tex]\mu =0.4 [/tex]

Consider a small element at an angle \theta having an angle of [tex]d\theta [/tex]

Normal Force[tex]=T\times \frac{d\theta }{2}+(T+dT)\cdot \frac{d\theta }{2}[/tex]

[tex]N=T\cdot d\theta [/tex]

Friction [tex]f=\mu \times Normal\ Reaction[/tex]

[tex]f=\mu \cdot N[/tex]

and [tex]T+dT-T=f=\mu Td\theta [/tex]

[tex]dT=\mu Td\theta [/tex]

[tex]\frac{dT}{T}=\mu d\theta [/tex]

[tex]\int_{T_2}^{T_1}\frac{dT}{T}=\int_{0}^{\pi }\mu d\theta [/tex]

[tex]\frac{T_2}{T_1}=e^{\mu \pi}[/tex]

[tex]\frac{T_2}{T_1}=e^{0.4\times \pi }[/tex]

[tex]\frac{T_2}{T_1}==e^{1.256}[/tex]

[tex]\frac{T_2}{T_1}=3.51[/tex]

Answer:

It slows down from 0-31.25s

It speeds up from 31.25-62.507s

Explanation:

If we find the maximum of the equation, we will know the moment when it changes direction. It will slow down on the first interval, and speed up on the second one.

[tex]Y = -16t^2+1000t+7[/tex]

[tex]Y' = -32t+1000 = 0[/tex]  Solving for t:

t = 31.25s

The bullet will slow down in the interval 0-31.25s

Let's now find the moment when it hits ground:

[tex]Y = 0 = -16t^2+1000t+7[/tex]   Solving for t:

t1 = -0.0069s     t2 = 62.507s

The bullet will speed up in the interval 31.25-62.507s

Answer:

Zero

Explanation:

Total energy of the EARTH  is the sum total of the all the kinetic and the gravitation potential energy of the earth.

Which can be written as

[tex]E_{total} = KE+U[/tex]

[tex]= \frac{1}{2}mv^2+ \frac{GMm}{R}[/tex]

At very large distances the kinetic as well as the gravitational potential energies become zero.

therefore, E_total= 0

When an object is launched at its escape velocity, its total mechanical energy at infinite distance from the planet is zero. The object loses all its kinetic energy, and its potential energy also becomes zero as it reaches infinity.

Escape velocity is the minimum initial velocity required for an object to escape the gravitational pull of a planet or other large body and reach an infinite distance where gravitational force becomes zero. When an object is launched at this escape velocity, its total mechanical energy (Etotal) at an infinite distance is zero. This is because the object gives up all its kinetic energy and the potential energy approaches zero as the distance (r) approaches infinity.

Using conservation of energy, we can illustrate this with the following steps:

To find the angular acceleration of the cylinder, calculate the moment of inertia of the cylinder, determine the torque due to the falling bucket, and then use Newton's second law for rotation to find the angular acceleration.

The subject involves calculating the angular acceleration of a cylinder when the bucket of water tied to a rope around the cylinder is let go. We assume no friction or air resistance in this case.

First, we need to calculate the moment of inertia (I) of the solid cylinder, which can be calculated using the formula I = 0.5 * m * r^2. Here, m is the mass of the cylinder (50 kg) and r is the radius (half of the diameter, 0.125 m). After calculating I, use Newton's second law for rotation τ = I * α, where τ is torque and α is the angular acceleration.

The torque τ can be calculated as the product of the force exerted by the falling bucket (F) and the radius r of the cylinder. The force F is the weight of the falling bucket, m * g, where m is the mass of the bucket (20 kg) and g is acceleration due to gravity (about 9.8 m/s^2). Once you have calculated τ, you can solve for α by rearranging the equation to α = τ/I.

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The angular acceleration of the 50-kg cylinder is found to be approximately 62.7 rad/s².

To find the angular acceleration of the 50-kg cylinder when the bucket of water falls down to the bottom of the well, we will use the concept of rotational dynamics. The moment of inertia (I) of a solid cylinder is given by:

I = (1/2) [tex]\times[/tex] M[tex]\times[/tex] R²

where M is the mass and R is the radius. Given the mass M is 50 kg and the diameter is 25 cm, so the radius R is 0.125 m:

I = (1/2) * 50 kg[tex]\times[/tex] (0.125 m)² = 0.390625 kg m²

The torque (τ) caused by the falling bucket is:

The force exerted by the bucket of water is its weight, F = m [tex]\times[/tex] g, where m is the mass of the bucket and water (20 kg) and g is the acceleration due to gravity (9.8 m/s²):

So,

Using Newton's second law for rotation:

where α is the angular acceleration:

Solving for α:

Therefore, the angular acceleration of the 50-kg cylinder is approximately 62.7 rad/s².

Answer:

heat absorbed by water = 2.39 KJ

Explanation:

given,                                    

mass of molten NaCl = 5 g

mass of  water = 25 g                    

Temperature of water = 25.0 °C

Temperature increased to = 47.9 °C

heat absorbed by water = m S ΔT

 S is specific heat in J/gm °C                

 Specific heat of water = 4.184 J/gm °C

 m is mass of water                                            

heat absorbed by water = 25 x 4.184 x (47.9-25)

                                         = 25 x 4.184 x 22.9

                                         = 2395.34 J

heat absorbed by water = 2.39 KJ

The amount of heat transferred is 2.39 KJ.

This is the form of energy which is transferred from one body to another as the result of a difference in temperature.

mass of molten NaCl = 5 g

mass of  water = 25 g                    

Temperature of water = 25.0 °C

Final Temperature of water = 47.9 °C

Specific heat of water = 4.184 J/gm °C

Heat = m S ΔT

Heat absorbed by water = 25 x 4.184 x (47.9 -25)

= 25 x 4.184 x 22.9 = 2395.34 J

= 2.39 KJ.

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Explanation:

Part 1.

The force of gravity on the apple, [tex]F_1=9.39\ N[/tex] (downward)

The force of the wind on the apple, [tex]F_2=1.5\ N[/tex] (right)

Let F is the magnitude of the net force acting on it. The resultant of two vectors is given by :

[tex]F=\sqrt{F_1^2+F_2^2}[/tex]

[tex]F=\sqrt{9.39^2+1.5^2}[/tex]

F = 9.509 N

Part 2.

[tex]tan\theta=\dfrac{F_2}{F_1}[/tex]

[tex]tan\theta=\dfrac{1.5}{9.39}[/tex]

[tex]\theta=9.07^{\circ}[/tex]

So, the direction of the net external force is 9.07 degrees wrt vertical. Hence, this is the required solution.

Answer:

[tex]\frac{d\phi}{dt}=\pi r^2\frac{dB}{dt}[/tex]

Explanation:

the rate of the magnetic flux is given by,

[tex]\frac{d\phi}{dt}= \frac{d(BA)}{dt}[/tex]

where B= magnetic strength

A= area of cross section

r= radius of circle

\phi= flux

=[tex]A\frac{dB}{dt}[/tex]

=[tex]\pi r^2\frac{dB}{dt}[/tex]

Answer:

I₂ = 25.4 W

Explanation:

Polarization problems can be solved with the malus law

     I = I₀ cos² θ

Let's apply this formula to find the intendant intensity (Gone)

Second and third polarizer, at an angle between them is

    θ₂ = 68.0-22.2 = 45.8º

    I = I₂ cos² θ₂

    I₂ = I / cos₂ θ₂

    I₂ = 75.5 / cos² 45.8

    I₂ = 155.3 W

We repeat for First and second polarizer

   I₂ = I₁ cos² θ₁

   I₁ = I₂ / cos² θ₁

   I₁ = 155.3 / cos² 22.2

   I₁ = 181.2 W

Now we analyze the first polarizer with the incident light is not polarized only half of the light for the first polarized

    I₁ = I₀ / 2

   I₀ = 2 I₁

   I₀ = 2 181.2

   I₀ = 362.4 W

Now we remove the second polarizer the intensity that reaches the third polarizer is

    I₁ = 181.2 W

The intensity at the exit is

    I₂ = I₁ cos² θ₂

    I₂ = 181.2 cos² 68.0

   I₂ = 25.4 W

Final answer:

In Situation B, the speed of the block at x = x0e is 3.5 m/s.

Explanation:

In Situation A, a 20-kg block is pressed against a single spring and released. The speed of the block at x = x0e is va,f = 5.0 m/s. In Situation B, the same 20-kg block is pressed against two identical springs in series. The entire spring assembly is compressed by Δx = 10 cm. The question asks for the speed of the block at x = x0e, which we can find using the principle of conservation of mechanical energy.

The speed of the block in Situation B, vb,f, can be found as follows:

Solving this equation gives us vb,f = 3.5 m/s.

Answer:

The speed of electron is [tex]3.2\times10^{6}\ m/s[/tex]

Explanation:

Given that,

Energy density = 0.1 J/m³

Separation = 0.2 mm

We need to calculate the potential difference

Using formula of energy density

[tex]J=\dfrac{1}{2}\epsilon_{0}E^2[/tex]

[tex]J=\dfrac{1}{2}\epsilon_{0}\dfrac{V^2}{d^2}[/tex]

[tex]V^2=\dfrac{0.1\times(0.2\times10^{-3})^2\times2}{8.85\times10^{-12}}[/tex]

[tex]V^2=\sqrt{903.95}[/tex]

[tex]V=30.06\ V[/tex]

We need to calculate the speed of electron

Using energy conservation

[tex]U=eV=\dfrac{1}{2}mv^2[/tex]

Put the value into the formula

[tex]1.6\times10^{-19}\times30.06=\dfrac{1}{2}\times9.1\times10^{-31}\times v^2[/tex]

[tex]v^2=\dfrac{1.6\times10^{-19}\times30.06\times2}{9.1\times10^{-31}}[/tex]

[tex]v=\sqrt{1.057\times10^{13}}[/tex]

[tex]v=3.2\times10^{6}\ m/s[/tex]

Hence, The speed of electron is [tex]3.2\times10^{6}\ m/s[/tex]

To solve this exercise it is necessary to take into account the concepts related to the magnetic field and its vector representation through the cross product or vector product.

According to the definition the direction of the electromagnetic wave propagation is given by

[tex]\hat{n} = \vec{E} \times \vec{B}[/tex]

Where,

E = Electric Field

B = Magnetic Field

According to the information provided, the direction of propagation of the electromagnetic wave is on the X axis, which for practical purposes we will denote as [tex]\hat {i}[/tex], on the other hand it is also indicated that the magnetic field is in the Y direction, that for practical purposes we will denote it as [tex]\hat {j}[/tex]. In this way using the previous equation we would have to,

[tex]\hat{n} = \vec{E} \times \vec{B}[/tex]

[tex]\hat{i} = \vec{E} \times \hat{j}[/tex]

The cross product identity is

[tex]\hat{i}=-\hat{k} \times \hat{j}[/tex]

From the equation we can notice that the electric field would be given by,

[tex]\vec{E} = -\hat{k}[/tex]

Therefore the direction of electric field is negative z-axis.

Answer:

Velocity of the helium nuleus  = 1.44x10⁴m/s

Velocity of the proton = 2.16x10⁴m/s

Explanation:

From the conservation of linear momentum of the proton collision with the He nucleus:

[tex] P_{1i} + P_{2i} = P_{1f} + P_{2f] [/tex] (1)

where [tex]P_{1i}[/tex]: is the proton linear momentum initial, [tex]P_{2i}[/tex]: is the helium nucleus linear momentum initial, [tex]P_{1f}[/tex]: is the proton linear momentum final, [tex]P_{2f}[/tex]: is the helium nucleus linear momentum final

From (1):

[tex] m_{1}v_{1i} + 0 = m_{1}v_{1f} + m_{2}v_{2f} [/tex] (2)

where m₁ and m₂: are the proton and helium mass, respectively, [tex]v_{1i}[/tex] and [tex]v_{2i}[/tex]: are the proton and helium nucleus velocities, respectively, before the collision, and [tex]v_{1f}[/tex] and [tex]v_{2f}[/tex]: are the proton and helium nucleus velocities, respectively, after the collision

By conservation of energy, we have:

[tex] K_{1i} + K_{2i} = K_{1f} + K_{2f} [/tex] (3)

where [tex]K_{1i}[/tex] and  [tex]K_{2i}[/tex]: are the kinetic energy for the proton and helium, respectively, before the colission, and [tex]K_{1f}[/tex] and  [tex]K_{2f}[/tex]: are the kinetic energy for the proton and helium, respectively, after the colission

From (3):

[tex] \frac{1}{2}m_{1}v_{1i}^{2} + 0 = \frac{1}{2}m_{1}v_{1f}^{2} + \frac{1}{2}m_{2}v_{2f}^{2} [/tex] (4)  

Now we have two equations: (2) ad (4), and two incognits: [tex]v_{1f}[/tex] and [tex]v_{2f}[/tex].

Solving equation (2) for [tex]v_{1f}[/tex], we have:

[tex] v_{1f} = v_{1i} -\frac{m_{2}}{m_{1}} v_{2f} [/tex] (5)

From getting (5) into (4) we can obtain the [tex]v_{2f}[/tex]:

[tex] v_{2f}^{2} \cdot (\frac{m_{2}^{2}}{m_{1}} + m_{2}) - 2v_{2f}v_{1i}m_{2} = 0 [/tex]

[tex] v_{2f}^{2} \cdot (\frac{(4u)^{2}}{1u} + 4u) - v_{2f}\cdot 2 \cdot 3.6 \cdot 10^{4} \cdot 4u = 0 [/tex]

From solving the quadratic equation, we can calculate the velocity of the helium nucleus after the collision:

[tex] v_{2f} = 1.44 \cdot 10^{4} \frac{m}{s} [/tex] (6)

Now, by introducing (6) into (5) we get the proton velocity after the collision:

[tex] v_{1f} = 3.6 \cdot 10^{4} -\frac{4u}{1u} \cdot 1.44 \cdot 10^{4} [/tex]

[tex] v_{1f} = -2.16 \cdot 10^{4} \frac{m}{s} [/tex]

The negative sign means that the proton is moving in the opposite direction after the collision.

I hope it helps you!

Final answer:

In an elastic collision between a proton and a helium nucleus, their final velocities can be found using the principles of conservation of momentum and kinetic energy.

Explanation:

In an elastic collision, both momentum and kinetic energy are conserved. To find the final velocities of the proton and helium nucleus after the collision, we can use the principles of conservation of momentum and kinetic energy.

Since the helium nucleus is initially at rest, its momentum before the collision is zero. The momentum before the collision can be calculated as:

Initial momentum proton = proton mass × proton velocity

Final momentum proton + Final momentum helium nucleus = proton mass × proton final velocity + helium nucleus mass × helium nucleus final velocity

Using the conservation of momentum, we can solve for the final velocities of the proton and helium nucleus. The final kinetic energy of the helium nucleus can also be calculated using the equation:

Final kinetic energy helium nucleus = 0.5 × helium nucleus mass × helium nucleus final velocity^2

Answer:

68.6 m/s

Explanation:

v = Speed of sound in air = 343 m/s

u = Speed of train

[tex]f_1[/tex] = Actual frequency

From the Doppler effect we have the observed frequency as

When the train is approaching

[tex]f=f_1\frac{v+u}{v}[/tex]

When the train is receeding

[tex]\frac{2f}{3}=f_1\frac{v-u}{v}[/tex]

Dividing the above equations we have

[tex]\frac{f}{\frac{2f}{3}}=\frac{f_1\frac{v+u}{v}}{f_1\frac{v-u}{v}}\\\Rightarrow \frac{3}{2}=\frac{v+u}{v-u}\\\Rightarrow 3v-3u=2v+2u\\\Rightarrow v=5u\\\Rightarrow u=\frac{v}{5}\\\Rightarrow u=\frac{343}{5}\\\Rightarrow u=68.6\ m/s[/tex]

The speed of the train is 68.6 m/s

Answer:

[tex]k= \frac{8\mu^2mg^2}{v^2}[/tex]

Explanation:

The mass of object is m

The coefficient of friction is μ

the speed of object at x= 0 is v

when the object compress the springs stops so,

The kinetic energy of the spring is equal to and friction work is done

[tex]\frac{1}{2}mv^2 = \frac{1}{2}kx^2+\mu(mg)x[/tex]...........

after that it returns from the rest and finally comes to rest after coming to x=0

position.

So, now the potential energy of the spring is equal to the work done by the friction force.

[tex]\frac{1}{2}kx^2= \mu mgx[/tex]

[tex]\frac{1}{2}kx= \mu mg[/tex]

[tex]x= \frac{2\mu mg}{x}[/tex]

substitute the x value in equation 1

mv^2 = kx^2+2\mu (mg)x

[tex]mv^2= k\frac{2\mu mg}{k}+ 2\mu mg \frac{2\mu mg }{k}[/tex]

solving this we get

[tex]mv^2= \frac{8m^2\mu^2g^2}{k}[/tex]

therefore the force constant

[tex]k= \frac{8\mu^2mg^2}{v^2}[/tex]

Spring Constant is the measure of a spring's stiffness. The value of the spring constant k in terms of µ, m, g, and v is [tex]\dfrac{8m \mu ^2 g^2}{v^2 }[/tex].

Spring Constant is the measure of a spring's stiffness.

Given to us

Mass of object = m

The coefficient of friction = μ

The speed of the object at (x= 0) = v

When the object compresses the spring after a distance the spring and the object both stop, therefore, the Kinetic energy of the object transforms to the kinetic energy of the spring during compression and the potential energy when fully compressed, therefore,

The kinetic energy of the object during the compression,

[tex]\dfrac{1}{2} mv^2 = \dfrac{1}{2}kx^2 + \mu (mg)x[/tex]

The kinetic energy of the object after complete compression,

[tex]\dfrac{1}{2}kx^2 = \mu m g x\\\\\dfrac{1}{2}kx = \mu m g \\\\x = \dfrac{2 \mu m g}{k}[/tex]

Substitute the value of x in the kinetic energy of the object during the compression,

[tex]\dfrac{1}{2} mv^2 = \dfrac{1}{2}kx^2 + \mu (mg)x\\\\\dfrac{1}{2} mv^2 = \dfrac{1}{2}k(\dfrac{2 \mu m g}{k})^2 + \mu (mg)(\dfrac{2 \mu m g}{k})\\\\\\mv^2 = \dfrac{8m^2 \mu ^2 g^2}{k}\\\\k= \dfrac{8m^2 \mu ^2 g^2}{mv^2 }\\\\k= \dfrac{8m \mu ^2 g^2}{v^2 }[/tex]

Hence, the value of the spring constant k in terms of µ, m, g, and v is [tex]\dfrac{8m \mu ^2 g^2}{v^2 }[/tex].

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According to the information presented, it is necessary to take into account the concepts related to mass flow, specific potential energy and the power that will determine the total work done in the system.

By definition we know that the change in mass flow is given by

[tex]\dot{m} = \rho AV[/tex]

[tex]\dot {m} = \rho Q[/tex]

Remember that the Discharge is defined as Q = AV, where A is the Area and V is the speed.

Substituting with the values we have we know that the mass flow is defined by

[tex]\dot{m} = 1000*0.03[/tex]

[tex]\dot{m} = 30kg/s[/tex]

To calculate the power we need to obtain the specific potential energy, which is given by

[tex]\Delta pe = gh[/tex]

[tex]\Delta pe = 9.8*45[/tex]

[tex]\Delta pe = 441m^2/s^2[/tex]

So the power needed to deliver the water into the storage tank would be

[tex]\dot {E} = \dot{m}\Delta pe[/tex]

[tex]\dot {E} = 30*441[/tex]

[tex]\dot{E} = 13230W = 13.23kW[/tex]

Finally the mechanical power that is converted to thermal energy due to friction effects is:

[tex]\dot{W}_f = \dot{W}_s - \dot{E}[/tex]

[tex]\dot{W}_f 20-13.23[/tex]

[tex]\dot{W} = 6.77kW[/tex]

Therefore the mechanical power due to friction effect is 6.77kW

The irreversible head loss in this pump system is 23 meters and the lost mechanical power during the pumping process is 6.8 kW.

In this system, the pump supplies energy to the water and drives it up to the higher reservoir. Out of this energy provided by the pump, part of it is used for lifting the water while some is lost as head loss due to friction and other losses in the system. Using the principle of conservation of energy and the Bernoulli's equation, one can calculate the irreversible head loss and the mechanical power loss.

Given the mechanical power provided by the pump, P = 20 kW, the height difference h = 45 m, the flow rate Q = 0.03 m3/s, and the density of water ρ = 1000 kg/m3, the useful power for lifting the water equals ρghQ=1000x 9.81x 45 x 0.03=13.2 kW. Thus, the lost power due to irreversible head loss equals the total power minus the useful power, or 20 kW - 13.2 kW = 6.8 kW. The corresponding head loss equals the lost power divided by the power used for lifting the water, or h_loss = (6.8 kW)/(ρgQ) = 23 m.

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Answer:

(a) Angular acceleration will be [tex]100.53rad/sec^2[/tex]

(b) 50 revolution

Explanation:

We have given time t = 2.5 sec

Initial speed of the drill [tex]\omega _0=0rad/sec[/tex]

Speed after 2.5 sec [tex]=2400rpm=\frac{2\pi \times 2400}{60}=251.327rad/sec[/tex]

From first equation of motion we know that

[tex]\omega =\omega _0+\alpha t[/tex]

[tex]251.327 =0+\alpha \times 2.5[/tex]

[tex]\alpha =100.53rad/sec^2[/tex]

(b) From second equation of motion we know that

[tex]\Theta =\omega _0t+\frac{1}{2}\alpha t^2[/tex]

[tex]\Theta =0\times 2.5+\frac{1}{2}\times 100.53\times  2.5^2=314.16rad=\frac{314.16}{2\pi }=50revolution[/tex]

Answer:

I= 6.292  kg.m²

Explanation:

Given that

m = 1.3 kg

Side of square a= 1.1 m

The distance r

[tex]r=\sqrt{{a^2}+{a^2}}[/tex]

[tex]r={a}{\sqrt 2}[/tex]

[tex]r={1.1}{\sqrt 2}[/tex]

The moment of inertia I

The axis passes through one of the mass then the distance of the that mass from the axis will be zero.

I = m a² + m a² + m r²

By putting the values

I = m a² + m a² + m r²

I =m( 2 a² +  r²)

I =1.3( 2 x 1.1² +  2 x 1.1²)

I = 1.3 x 4  x 1.1² kg.m²

I= 6.292  kg.m²

Explanation:

It is given that,

Force exerted by a boy, F = 11.8 N

Angle above the horizontal, [tex]\theta=28^{\circ}[/tex]

Mass of the sled, m = 6.15 kg

Distance moved, d = 2.75 m

Initial speed, u = 0.37 m/s

Let W is the work done by the boy. Using the expression for the work done to find it as :

[tex]W=Fd\ cos\theta[/tex]

[tex]W=11.8\times 2.75\ cos(28)[/tex]

W = 28.65 joules

Let v is the final speed of the sled. Using the work energy theorem to find it. It states that the work done is equal to the change in kinetic energy of an object. It is given by :

[tex]W=\dfrac{1}{2}m(v^2-u^2)[/tex]

[tex]v=\sqrt{\dfrac{2W}{m}+u^2}[/tex]

[tex]v=\sqrt{\dfrac{2\times 28.65}{6.15}+(0.37)^2}[/tex]

v = 3.07 m/s

So, the final speed of the sled after it moves 2.75 m is 3.07 m/s. Hence, this is the required solution.

The final speed of the sled is approximately 3.073 m/s and the the work done by the boy on the sled is 28.64 J.

To solve this problem, we need to calculate the work done by the boy on the sled and then use that to find the final speed of the sled.

First, let's calculate the work done by the boy. Work (W) is defined as the product of the force (F) applied in the direction of motion and the distance (d) over which the force is applied. The force applied in the direction of motion is the horizontal component of the force exerted by the boy. We can calculate this using trigonometry:

[tex]\[ F_{\text{horizontal}} = F \cdot \cos(\theta) \][/tex]

[tex]\[ F_{\text{horizontal}} = 11.8 \, \text{N} \cdot \cos(28.0^\circ) \][/tex]

[tex]\[ F_{\text{horizontal}} \approx 11.8 \, \text{N} \cdot 0.8829 \][/tex]

[tex]\[ F_{\text{horizontal}} \approx 10.42 \, \text{N} \][/tex]

Now, we can calculate the work done:

[tex]\[ W = F_{\text{horizontal}} \cdot d \][/tex]

[tex]\[ W = 10.42 \, \text{N} \cdot 2.75 \, \text{m} \][/tex]

\[ W \approx 28.64 \, \text{J} \]

Next, we need to find the final speed of the sled. The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy (KE):

[tex]\[ W = \Delta KE \][/tex]

[tex]\[ W = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \][/tex]

Given that the initial speed [tex]\( v_i \)[/tex] is 0.370 m/s and the mass [tex]\( m \)[/tex] of the sled is 6.15 kg, we can solve for the final speed [tex]\( v_f \)[/tex]:

[tex]\[ 28.64 \, \text{J} = \frac{1}{2} \cdot 6.15 \, \text{kg} \cdot v_f^2 - \frac{1}{2} \cdot 6.15 \, \text{kg} \cdot (0.370 \, \text{m/s})^2 \][/tex]

[tex]\[ 28.64 \, \text{J} = 3.075 \, \text{kg} \cdot v_f^2 - 0.405975 \, \text{kg} \cdot \text{m}^2/\text{s}^2 \][/tex]

[tex]\[ 28.64 \, \text{J} + 0.405975 \, \text{kg} \cdot \text{m}^2/\text{s}^2 = 3.075 \, \text{kg} \cdot v_f^2 \][/tex]

[tex]\[ 29.046 \, \text{kg} \cdot \text{m}^2/\text{s}^2 = 3.075 \, \text{kg} \cdot v_f^2 \][/tex]

[tex]\[ v_f^2 = \frac{29.046 \, \text{kg} \cdot \text{m}^2/\text{s}^2}{3.075 \, \text{kg}} \][/tex]

[tex]\[ v_f^2 \approx 9.444 \, \text{m}^2/\text{s}^2 \][/tex]

[tex]\[ v_f \approx \sqrt{9.444 \, \text{m}^2/\text{s}^2} \][/tex]

[tex]\[ v_f \approx 3.073 \, \text{m/s} \][/tex]

Answer:

24.616 mm

12.384 mm

Explanation:

[tex]D_b[/tex] = Blanking die diameter = 25 mm

[tex]D_h[/tex] = Punch diameter = 12 mm

a = Allowance = 0.06 for aluminium

t = Thickness = 3.2 mm

Clearance

[tex]c=at\\\Rightarrow c=0.06\times 3.2\\\Rightarrow c=0.192[/tex]

Diameter of punch is

[tex]d_p=D_b-2c\\\Rightarrow d_p=25-2\times 0.192\\\Rightarrow d_p=24.616\ mm[/tex]

Diameter of blanking punch is 24.616 m

For punching operation

[tex]d_d=D_h+2c\\\Rightarrow d_d=12+2\times 0.192\\\Rightarrow d_d=12.384\ mm[/tex]

The diameter of punching die is 12.384 mm

For the blanking operation, the punch diameter should be 12.0 mm and the die diameter should be 25.0 mm. For the punching operation, the punch diameter should be 12.0 mm and the die diameter should be slightly larger than 12.0 mm, typically by a clearance of 0.05 mm to 0.15 mm per side.

(a) Blanking Operation:

The blanking operation involves cutting the outside profile of the washer from the aluminum sheet stock. The punch and die sizes for this operation are determined by the outside and inside dimensions of the washer.

- The die diameter for the blanking operation should match the outside diameter of the washer, which is 25.0 mm. This is the size of the hole in the die through which the punch will pass to cut the washer's outer edge.

- The punch diameter for the blanking operation should match the inside diameter of the washer, which is 12.0 mm. This is the size of the punch that will push through the sheet stock to create the inner hole of the washer.

(b) Punching Operation:

The punching operation involves cutting the inside hole of the washer after the blanking operation has been completed. The punch and die sizes for this operation are determined by the inside diameter of the washer and the required clearance.

 - The punch diameter for the punching operation should also be 12.0 mm, the same as the inside diameter of the washer. This ensures that the punch cuts the material exactly at the desired inner diameter.

- The die diameter for the punching operation should be slightly larger than the punch diameter to allow for clearance. Clearance is necessary to prevent the punch and die from binding together and to allow for the removal of the slug (the piece of material that is punched out). The typical clearance ranges from 0.05 mm to 0.15 mm per side. Therefore, the die diameter would be 12.0 mm plus two times the clearance. If we take the middle value of the clearance range, which is 0.10 mm per side, the die diameter would be 12.0 mm + 2 * 0.10 mm = 12.20 mm.

Answer:

b = 1.2 m

L = 4.8 m

d= 1.2 m

Explanation:

Lets take

Width of the tank = b

Depth = d

length = L

Given that L = 4 b

v= 2 cm/s

Flow rate Q= 10,000 m³/d

We know that 1 d = 24 hr = 24 x 3600 s

1 d= 86400 s

Q= 0.115  m³/s

Flow rate Q

Q= Area x velocity

Q= L . b .v

0.115 = 4 b . b . 0.02

4 b² = 5.78  

b = 1.2 m

L = 4.8 m

d= 1.2 m

This is the dimension of the tank.

To design the settling tank with the given water flow and settling velocity, we use conservation of mass and settling speed to calculate the dimensions. We convert the flow rate to m3/s, relate it to the tank's cross-sectional area, and given settling characteristics to find the tank's width, length, and depth.

To design a rectangular settling tank for settling sand particles with a settling velocity of 2.0 cm/s and a water flow of 10,000 m3/d, we will apply the principles of conservation of mass and the given settling velocity to determine the dimensions of the tank. The problem states that the tank length (L) should be four times the width (W), and the width should be equal to the depth (D), therefore L = 4W and W = D.

To find the dimensions, we first convert the water flow rate into m3/s: flow rate = 10,000 m3/d × (1 day/86400 s) ≈ 0.1157 m3/s. Since the sand particles settle at 2 cm/s, for a particle to settle before reaching the end of the tank, the residence time should be at least the depth divided by the settling velocity. From here, we use the flow rate (Q), which is also equal to the cross-sectional area (A) times the water velocity (V), and we know that A = W² for this tank because W = D.

Therefore, Q = W²V. Using V = L / (residence time) and substituting for L = 4W, we find that Q = W² × (4W / (W/0.02 m/s)) which simplifies to Q = 0.08 m/s × W³. Solving for W gives us W = ∛(Q / 0.08 m/s) and subsequently we find L = 4W and D = W. Finally, we can plug in the flow rate and solve for the dimensions in meters.

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