Heat is produced within a cylindrical cable with a radius of 0.60 m and a length of 3 m with a heat conductivity of 85 W/m K. The amount of heat produced per unit volume and per unit time is given as Q (W/m3.s) = 4x10-3 T0.5 where T is the temperature (K). The surface temperature of the sphere is 120 °C. a) Construct an energy balance within the cylindrical cable. b) Solve the energy balance with MATLAB to obtain the temperature profile within the cylindrical cable by appropriate assumptions

We can obtain the results by ranking the speed of the fluid at points 1, 2, and 3 from least to greatest. 1 < 3 < 2

Point 1 : The fluid velocity is the least at point 1 because the pipe diameter is largest at this point. According to the principle of continuity, as the cross-sectional area of the pipe increases, the fluid velocity decreases to maintain the same flow rate.

Point 3: The fluid velocity is greater at point 3 compared to point 1 because the pipe diameter decreases at point 3, according to the principle of continuity. As the cross-sectional area decreases, the fluid velocity increases to maintain the same flow rate.

Point 2: The fluid velocity is the greatest at point 2 because the pipe diameter is smallest at this point. Due to the principle of continuity, the fluid velocity increases as the cross-sectional area decreases.

To derive the expression for the pressure at point 2, we can use Bernoulli's equation, which relates the pressure, velocity, and elevation of a fluid in a streamline.

Bernoulli's equation:

P1 + (1/2) * ρ * v1^2 + ρ * g * h1 = P2 + (1/2) * ρ * v2^2 + ρ * g * h2

Assumptions:

The fluid is incompressible.

The fluid is flowing along a streamline.

There is no change in elevation (h1 = h2).

Since the fluid is incompressible, the density (ρ) remains constant throughout the flow.

Given:

Pressure at point 1: P1

Velocity at point 1: v1

Cross-sectional area at point 1: A

Cross-sectional area at point 2: A/2

Simplifying Bernoulli's equation:

P2 = P1 + (1/2) * ρ * (v1^2 - v2^2)

Since the fluid is incompressible, the density (ρ) can be factored out:

P2 = P1 + (1/2) * ρ * (v1^2 - v2^2)

To determine the relationship between v1 and v2, we can use the principle of continuity:

A1 * v1 = A2 * v2

Substituting the relationship between v1 and v2 into the expression for P2:

P2 = P1 + (1/2) * ρ * (v1^2 - (A1^2 / A2^2) * v1^2)

Simplifying further:

P2 = P1 + (1/2) * ρ * v1^2 * (1 - (A1^2 / A2^2))

The final expression for the pressure at point 2 in terms of the given variables is:

P2 = P1 + (1/2) * ρ * v1^2 * (1 - (A1^2 / (A/2)^2))

Simplifying the expression:

P2 = P1 + (1/2) * ρ * v1^2 * (1 - 4)P2 = P1 - (3/2) * ρ * v1^2

This is the derived expression for the pressure in the pipe at point 2 in terms of the given variables: P2 = P1 - (3/2) * ρ * v1^2.

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For the given conditions, the frequency of the sound wave in optical waves is 4.3 × 1013 Hz.

Given that the velocity of sound in a solid is of the order 103 m/s, and the frequency of the sound wave is λ = 20 Å.

We have to compare the frequency of the sound wave for (a) a monoatomic system and (b) acoustic waves and optical waves in a diatomic system containing two identical atoms (M=m) per unit cell of interatomic spacing 2.2 Å.

(a) Monoatomic system

The relation between the frequency, wavelength, and velocity of sound wave in a solid is given by:

f = v / λ

Where,

f is frequency,

λ is wavelength, and

v is velocity of sound.

The frequency of the sound wave in monoatomic system is

f = 103 / 20 × 10^-10f = 5 × 10^12 Hz

(b) Diatomic system

The diatomic system contains two identical atoms (M=m) per unit cell of interatomic spacing 2.2 Å.

For diatomic system, there are two modes of vibration in a solid:

Acoustic mode and Optical mode.

Acoustic mode

For acoustic waves in a diatomic system, the angular frequency of the wave is given by:

ω = 2Vs × √(sin²(πn/Na)+(1 - sin²(πn/Na)) / 4) / a

Where,

ω is the angular frequency,

Vs is the velocity of sound in a solid,

n is the mode of vibration,

a is the interatomic spacing, and

Na is the number of atoms per unit cell of a crystal.

The frequency of the sound wave in acoustic mode is

f = ω / 2π

The frequency of the sound wave in acoustic mode for diatomic system is

f = Vs × √(sin²(πn/Na)+(1 - sin²(πn/Na)) / 4) / a × (1 / 2π)f

 = 103 × √(sin²(πn/2)+(1 - sin²(πn/2)) / 4) / 2.2 × (1 / 2π)

For n = 1, the frequency of the sound wave in acoustic mode is

f = 0.73 × 10^13 Hz

For n = 2, the frequency of the sound wave in acoustic mode is

f = 1.6 × 10^13 Hz

For n = 3, the frequency of the sound wave in acoustic mode is

f = 2.5 × 10^13 Hz

For n = 4, the frequency of the sound wave in acoustic mode is

f = 3.3 × 10^13 Hz

Optical mode

For optical waves in a diatomic system, the angular frequency of the wave is given by:

ω = 2Vs × √(sin²(πn/Na)-(1 - sin²(πn/Na)) / 4) / a

Where,

ω is the angular frequency,

Vs is the velocity of sound in a solid,

n is the mode of vibration,

a is the interatomic spacing, and

Na is the number of atoms per unit cell of a crystal.

The frequency of the sound wave in optical mode is

f = ω / 2π

The frequency of the sound wave in optical mode for diatomic system is

f = Vs × √(sin²(πn/Na)-(1 - sin²(πn/Na)) / 4) / a × (1 / 2π)

f = 103 × √(sin²(πn/2)-(1 - sin²(πn/2)) / 4) / 2.2 × (1 / 2π)

For n = 1, the frequency of the sound wave in optical mode is

f = 2.2 × 10^13 Hz

For n = 2, the frequency of the sound wave in optical mode is

f = 2.6 × 10^13 Hz

For n = 3, the frequency of the sound wave in optical mode is

f = 3.4 × 10^13 Hz

For n = 4, the frequency of the sound wave in optical mode is

f = 4.3 × 10^13 Hz

Therefore, the frequency of the sound wave for (a) a monoatomic system is 5 × 10^12 Hz and the frequency of the sound wave for (b) acoustic waves and optical waves in a diatomic system containing two identical atoms (M=m) per unit cell of interatomic spacing 2.2 Å are given in the table below:

Optical waves

Acoustic waves

11.3 × 10^13 Hz0.73 × 10^13 Hz22.6 × 10^13 Hz1.6 × 10^13 Hz33.4 × 10^13 Hz2.5 × 10^13 Hz44.3 × 10^13 Hz3.3 × 10^13 Hz

Therefore, for the given conditions, the frequency of the sound wave in optical waves is 4.3 × 1013 Hz.

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The frequency of sound waves in a monoatomic and diatomic system can be calculated using the velocity and wavelength of sound waves.

Frequency refers to the number of occurrences of a repeating event, such as a wave crest passing a fixed point, within a given unit of time, typically measured in Hertz (Hz). To compare the frequency of sound waves in different systems, we need to use the equation v = fλ, where v is the velocity of sound and λ is the wavelength.

In a monoatomic system, the frequency will be the same as in the given sound wave: f = v/λ = 103/20 = 5.15 x 10^3 Hz. In a diatomic system, where there are two identical atoms per unit cell, the effective mass is doubled. Therefore, the frequency will be half of that in the monoatomic system: f = v/λ = 103/20 = 2.58 x 10^3 Hz.

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The acceleration of the block as it moves along the surface is approximately 0.880 m/s².

To determine the acceleration of the block, we need to consider the forces acting on it.

The horizontal force of 40 lbs (pounds) is acting on the block in the direction of motion.

The frictional force opposes the motion of the block. There are two cases we need to consider:

The maximum static friction force can be calculated using the formula:

F_static_max = μ_static * N

where μ_static is the coefficient of static friction and N is the normal force.

The normal force is equal to the weight of the block, which can be calculated as:

N = m * g

where m is the mass of the block and g is the acceleration due to gravity.

The kinetic friction force can be calculated using the formula:

F_kinetic = μ_kinetic * N

where μ_kinetic is the coefficient of kinetic friction and N is the normal force.

Once we have the forces, we can use Newton's second law to determine the acceleration:

ΣF = m * a

where ΣF is the net force acting on the block, m is the mass of the block, and a is the acceleration.

Applied force = 40 lbs

Mass of the block (m) = 12 lbs

Coefficient of static friction (μ_static) = 0.40

Coefficient of kinetic friction (μ_kinetic) = 0.25

Acceleration due to gravity (g) = 32.2 ft/s²

First, let's convert the values to SI units (kilograms and meters):

1 lb ≈ 0.454 kg

1 ft ≈ 0.305 m

Applied force = 40 lbs ≈ 40 * 0.454 kg ≈ 18.16 kg

Mass of the block (m) = 12 lbs ≈ 12 * 0.454 kg ≈ 5.448 kg

Acceleration due to gravity (g) = 32.2 ft/s² ≈ 32.2 * 0.305 m/s² ≈ 9.817 m/s²

Now, let's calculate the forces:

Static friction force:

N = m * g = 5.448 kg * 9.817 m/s² ≈ 53.467 N

F_static_max = μ_static * N = 0.40 * 53.467 N ≈ 21.387 N

F_kinetic = μ_kinetic * N = 0.25 * 53.467 N ≈ 13.367 N

Since the applied force (40 lbs or 18.16 kg) exceeds the maximum static friction force (21.387 N), the block will start moving, and the kinetic friction force will be in effect. Therefore, the net force acting on the block is the difference between the applied force and the kinetic friction force:

ΣF = Applied force - F_kinetic = 18.16 kg - 13.367 N ≈ 4.793 N

Finally, we can use Newton's second law to calculate the acceleration:

ΣF = m * a

4.793 N = 5.448 kg * a

a ≈ 4.793 N / 5.448 kg ≈ 0.880 m/s²

Therefore, the acceleration of the block as it moves along the surface is approximately 0.880 m/s².

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The total resistance of a gold wire can be calculated using its resistivity, length, and radius. In this case, with a resistivity of 4.14x10^8 Ωm, a length of 6.57 m, and a radius of 0.080 m, we can determine the total resistance.

The resistance of a wire can be calculated using the formula R = (ρ * L) / A, where R is the resistance, ρ is the resistivity, L is the length of the wire, and A is the cross-sectional area of the wire. To find the cross-sectional area, we can use the formula A = π * r^2, where r is the radius of the wire.

Plugging in the given values, we have A = π * (0.080 m)^2 = 0.0201 m^2. Now, we can calculate the resistance using the formula R = (4.14x10^8 Ωm * 6.57 m) / 0.0201 m^2.

Simplifying this expression, we get R ≈ 1.34 Ω. Therefore, the total resistance for the given gold wire is approximately 1.34 ohms.

Note: It's worth mentioning that the resistivity value provided (4.14x10^8 Ωm) seems unusually high for gold. The resistivity of gold is typically around 2.44x10^-8 Ωm. However, if we assume the given value is correct, the calculation would proceed as described above.

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The wave produced on the vibrating guitar string is transverse and progressive.

When a guitar string is plucked, it produces a wave that travels along the string. This wave is transverse in nature, meaning that the particles of the medium (the string) vibrate perpendicular to the direction of wave propagation. As the string oscillates up and down, it creates peaks and troughs in the wave pattern, forming a characteristic waveform.

The wave is also progressive, which means it propagates through space. As the plucked string vibrates, the disturbance travels along the length of the string, carrying the energy of the wave with it. This progressive motion allows the sound wave to reach our ears, where we perceive it as a sound of constant frequency.

In summary, when a guitar string is plucked, it generates a transverse and progressive wave. The transverse nature of the wave arises from the perpendicular vibrations of the string's particles, while its progressiveness refers to the propagation of the wave through space, enabling us to hear a sound of constant frequency.

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The objective is to design an absorption packed tower to reduce NH3 concentration in air, and the parameters to be determined are the mole fraction of pollutant in the gas phase at the tower inlet (y), the equilibrium mole fraction of pollutant in the liquid phase (x), the slope of the equilibrium curve (m), the absorption factor (A), and the height of the tower.

The given problem involves the design of an absorption packed tower for removing NH3 from air. The tower should reduce the NH3 concentration from 0.10 kg/m3 to 0.0005 kg/m3.

The operating conditions include a column diameter of 3.00 m, operating temperature of 20.0°C, air density at 20.0°C of 1.205 kg/m3, and operating pressure of 101.325 kPa. The relevant data includes the measured partial pressure of NH3, the flow rate of H2O, and the molecular weights of NH3, air, and H2O.

The objectives are to determine the mole fraction of the pollutant in the gas phase at the inlet of the tower (y), the equilibrium mole fraction of the pollutant in the liquid phase (x), the slope of the equilibrium curve (m), the absorption factor (A), and the height of the absorption packed tower.

These parameters will help in designing an effective tower for NH3 removal.

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The approximate gravitational red shift in 500-nm light emitted by a white dwarf star whose mass is that of the sun but whose radius is that of the earth, 6.4 x 105 m is 5.96 x 10-6. The gravitational redshift is defined as the decrease in frequency and energy of a photon as it moves from a higher gravitational potential to a lower one. Gravitational redshift happens because of the effect of gravity on light.

Explanation:

The gravitational red shift is given by

Δλ/λ = GM/(Rc²)

where

Δλ/λ = fractional shift of the wavelength of light.

G = gravitational constant (6.67 × 10-11 Nm²/kg²)

M = mass of the object (1 M☉ = 1.99 × 10³⁰ kg)

R = radius of the object (earth radius, 6.4 × 10⁶ m)

c = speed of light (3 × 10⁸ m/s)

Substitute the values in the above formula

Δλ/λ = (6.67 × 10-11 Nm²/kg²) × (1.99 × 1030 kg) / [(6.4 × 106 m) × (3 × 108 m/s)²]

Δλ/λ = 5.96 × 10-6

Therefore, the approximate gravitational red shift in 500-nm light emitted by a white dwarf star whose mass is that of the sun but whose radius is that of the earth, 6.4 × 105 m is 5.96 × 10-6.

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The final temperature of the gas after compression is approximately 150.38°C.

To determine the final temperature of the gas after compression, using the combined gas law:

(P₁ ×V₁) / T₁= (P₂ × V₂) / T₂

Where:

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Given:

P₁ = 2.00 atm (constant pressure)

V₁ = Initial volume

T₁ = 178°C + 273.15

P₂ = 2.00 atm (constant pressure)

V₂ = (1/3) × V₁

T₂ = Final temperature

Substituting the values and solving for T₂

(2.00 atm × V₁) / (178°C + 273.15) = (2.00 atm × (1/3) × V₁) / T₂

V₁ / (178°C + 273.15) = (1/3) × V₁ / T₂

T₂ = (178°C + 273.15) × (1/3)

T₂ ≈ 150.38°C

Therefore, the final temperature of the gas after compression is approximately 150.38°C.

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The electric potential function in a volume of space is given by V(x,y,z) = x2 + xy2 + 2yz.The electric field in the region at the coordinate (3, 4, 5) is -22i - 24j - 8k.

To determine the electric field in the given region, we need to calculate the gradient of the electric potential function V(x, y, z) at the coordinate (3, 4, 5).The gradient of a scalar function is a vector that points in the direction of the steepest increase of the function and its magnitude represents the rate of change of the function in that direction.

The electric potential function is given as V(x, y, z) = x^2 + xy^2 + 2yz.

To find the gradient, we need to calculate the partial derivatives of V with respect to each coordinate (x, y, z):

∂V/∂x = 2x + y^2

∂V/∂y = 2xy

∂V/∂z = 2y

Now, we can evaluate these partial derivatives at the coordinate (3, 4, 5):

∂V/∂x = 2(3) + (4)^2 = 6 + 16 = 22

∂V/∂y = 2(3)(4) = 24

∂V/∂z = 2(4) = 8

Therefore, the electric field at the coordinate (3, 4, 5) is given by the vector E = -(∂V/∂x)i - (∂V/∂y)j - (∂V/∂z)k:

E = -22i - 24j - 8k

So, the electric field in the region at the coordinate (3, 4, 5) is -22i - 24j - 8k.

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Given that a police officer in a police car finds that a vehicle is travelling beyond the speed limit in a low-velocity zone with a constant speed of 24 m/s. As soon as the vehicle passes the police car, the police officer begins pursuing the vehicle with a constant acceleration of 6 m/s² until the police office catches up with and stops the speeding vehicle. Here, the distance covered and the time elapsed are the same for both the POLICE CAR and the SPEEDING VEHICLE, from the time the police car begins pursuing the vehicle to the time the police car catches up and stops the vehicle.

The time taken by the police car to catch up with and stop the speeding vehicle is 4 seconds.

We need to find the time taken by the police car to catch up with and stop the speeding vehicle.

Solution:

Let the time taken to catch up with and stop the vehicle be t.

So, the distance covered by the police car during the time t = distance covered by the speeding vehicle during the time Distance = speed × time.

Distance covered by the speeding vehicle during the time t is 24t.

Distance covered by the police car during the time t is 1/2 × 6t², since it starts from rest and its acceleration is 6 m/s².

We know that both distances are the same.

Therefore, 24t = 1/2 × 6t²

⇒ 4t = t²

⇒ t = 4 s.

Therefore, the time taken by the police car to catch up with and stop the speeding vehicle is 4 seconds.

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The problem involves a radio station broadcasting at a frequency of 92.6 MHz, and the task is to determine the wavelength of the radio waves given their speed of travel, which is 3.00 x 10^8 m/s.

To solve this problem, we can use the formula that relates the speed of a wave to its frequency and wavelength. The key parameters involved are frequency, wavelength, and speed.

The formula is: speed = frequency * wavelength. Rearranging the formula, we get: wavelength = speed / frequency. By substituting the given values of the speed (3.00 x 10^8 m/s) and the frequency (92.6 MHz, which is equivalent to 92.6 x 10^6 Hz), we can calculate the wavelength of the radio waves.

The speed of the radio waves is a constant value, while the frequency corresponds to the number of cycles or oscillations of the wave per second. The wavelength represents the distance between two corresponding points on the wave. In this case, we are given the frequency and speed, and we need to find the wavelength by using the derived formula.

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The mass rate of change of the space ship is 190,000 kg/s and the required displacement is 8.88 m (upwards).

Question 1A The space probe lands on an alien planet with a gravitational acceleration of 5.00 m/s².

Now, the upward acceleration required is 2.50 m/s². Hence, the required acceleration can be calculated as:

∆v/∆t = a Where,

∆v = change in velocity = 40,000 m/s

a = acceleration = 2.50 m/s²

∆t can be calculated as:

∆t = ∆v/a

= 40,000/2.5

= 16,000 seconds

Therefore, the mass rate of change of the space ship is calculated as:

∆m/∆t = (F/a)

Where, F = force

= m × a

F = (190,000 kg) × (2.5 m/s²)

F = 475,000 N

∆m/∆t = (F/a)∆m/∆t

= (475,000 N) / (2.5 m/s²)

∆m/∆t = 190,000 kg/s

Question 2 Mass of the roller coaster, m = 114 kg

Spring constant, k = 550 N/m

Compression, x = 11.0 m

Initial velocity of the roller coaster, u = 0

Final velocity of the roller coaster, v = 7.0 m/s

At point A (Start)

Potential Energy + Kinetic Energy = Total Energy

[tex]1/2 kx^2+ 0 = 1/2 mv^2 + mgh[/tex]

[tex]0 + 0 = 1/2 \times 114 \times 7^2 + 114 \times g \times h[/tex]

[tex]1/2 \times 114 \times 7^2 + 0 = 114 \times 9.8 \times h[/tex]

h = 16.43 m

At point B (End)

Potential Energy + Kinetic Energy = Total Energy

[tex]0 + 1/2 \ mv^2 = 1/2 \ mv^2 + mgh[/tex]

[tex]0 + 1/2 \times 114 \times 7^2= 0 + 114 \times 9.8 \times h[/tex]

h = -7.55 m

So, the vertical displacement is 16.43 m - 7.55 m = 8.88 m (upwards)

Therefore, the required displacement is 8.88 m (upwards).

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The vertical displacement from the starting level to the second (flat) level.

To determine the mass rate of change of the space ship (Δm/Δt) needed to achieve an upward acceleration of 2.50 m/s², we can use the rocket equation, which states:

Δv = (ve * ln(m0 / mf))

Where:

Δv is the desired change in velocity (2.50 m/s² in the upward direction),

ve is the exhaust velocity of the fuel (40,000 m/s),

m0 is the initial mass of the space probe (190,000 kg + fuel mass),

mf is the final mass of the space probe (190,000 kg).

Rearranging the equation, we get:

Δm = m0 - mf = m0 * (1 - e^(Δv / ve))

To find the mass rate of change, we divide Δm by the time it takes to achieve the desired acceleration:

(Δm / Δt) = (m0 * (1 - e^(Δv / ve))) / t

To determine the vertical displacement of the roller coaster from its starting level when it reaches the second (flat) level with a velocity of 7.0 m/s, we can use the conservation of mechanical energy. At the starting level, the only form of energy is the potential energy stored in the compressed spring, which is then converted into kinetic energy at the second level.

Potential energy at the starting level = Kinetic energy at the second level

0.5 * k * x^2 = 0.5 * m * v^2

where:

k is the spring constant (550.0 N/m),

x is the compression of the spring (11.0 m),

m is the mass of the roller coaster cart (114.0 kg),

v is the velocity at the second level (7.0 m/s).

Plugging in the values:

0.5 * (550.0 N/m) * (11.0 m)^2 = 0.5 * (114.0 kg) * (7.0 m/s)^2

Solving this equation will give us the vertical displacement from the starting level to the second (flat) level.

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Two transverse waves y1 = 2 sin(2ttt - itx) and y2 = 2 sin(2nt - TeX + Tt/3) are moving in the same direction. The resultant amplitude of the interference between the two waves is √(8 + 8cos(nx)).

To find the resultant amplitude of the interference between the two waves, we need to add their individual amplitudes. The given waves are:

y1 = 2 sin(2ωt - k1x)

y2 = 2 sin(2ωt - k2x + φ)

where ω is the angular frequency, t is the time, k1 and k2 are the wave numbers, x is the position, and φ is the phase difference.

Comparing the equations, we can see that the angular frequency ω is the same for both waves (2ωt term). However, the wave numbers and phase differences are different.

k1 = ω, which implies k1 = 2t

k2 = ω, which implies k2 = n

Using the formula for the resultant amplitude of two interfering waves, we have:

Resultant amplitude = √(A1^2 + A2^2 + 2A1A2cos(φ))

In this case, A1 = 2 and A2 = 2 (both waves have the same amplitude).

To find the phase difference φ, we equate the phase terms in the given wave equations:

-itx = -k2x + φ

-itx = -nx + φ

Since the waves are moving in the same direction, we can assume that the phase difference φ is constant and does not depend on x. Therefore, we can rewrite the equation as:

φ = -itx + nx

Since we don't have specific values for t and n, we cannot determine the exact value of the phase difference φ.

However, if we assume that t = 0, then the equation becomes:

φ = 0 + nx = nx

In this case, the phase difference φ is directly proportional to x.

Now we can calculate the resultant amplitude:

Resultant amplitude = √(A1^2 + A2^2 + 2A1A2cos(φ))

= √(2^2 + 2^2 + 2(2)(2)cos(nx))

= √(4 + 4 + 8cos(nx))

= √(8 + 8cos(nx))

Therefore, the resultant amplitude of the interference between the two waves is √(8 + 8cos(nx)).

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The reading on the ammeter would be approximately 2.14 Amperes.

To calculate the reading on the ammeter, we need to determine the resistance of the 14 AWG iron wire. The resistance can be calculated using the formula

[tex]R = ρ * (L / A)[/tex]

where:

R is the resistance,

ρ is the resistivity of the material (in this case, iron),

L is the length of the wire, and

A is the cross-sectional area of the wire.

First, let's calculate the cross-sectional area of the 14 AWG wire. The diameter of the wire can be obtained from the wire gauge size. For 14 AWG, the diameter is approximately 1.628 mm.

The radius (r) can be calculated by dividing the diameter by 2:

r = 1.628 mm / 2 = 0.814 mm = 0.000814 m

The cross-sectional area (A) can be calculated using the formula:

[tex]R = ρ * (L / A)[/tex]

[tex]A = 3.14159 * (0.000814 m)^2 ≈ 2.07678 × 10^(-6) m^2[/tex]

Next, we need to find the resistivity of iron. The resistivity of iron (ρ) is approximately 9.71 × 10^(-8) Ω·m.

Now, we can calculate the resistance (R) using the formula mentioned earlier:

[tex]R = (9.71 × 10^(-8) Ω·m) * (100 m / 2.07678 × 10^(-6) m^2)[/tex]

[tex]R ≈ 4.675 Ω[/tex]

Therefore, with a 10 V potential difference across the 14 AWG iron wire resistor, the reading on the ammeter would be:

[tex]I = V / R[/tex]

[tex]I = 10 V / 4.675 Ω[/tex]

[tex]I ≈ 2.14 A[/tex]

So, the reading on the ammeter would be approximately 2.14 Amperes.

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The ratio of high tension to low tension is `1.22`.Hence, option D is correct.

Given data: Frequency of both the string,

`f = 440 Hz`

Diameter of high tension string, `d_1 = 0.432 mm

`Diameter of low tension string, `d_2 = 0.381 mm`

The density of both strings is the same.

Let the tension in high tension string and low tension string be `T_1` and `T_2` respectively.

Ratio of tension in both strings:

`T_1/T_2= [(π/4)d_1²p(f₁)²]/[(π/4)d_2²p(f₂)²]`

Here, `f₁ = f₂ = f =

440 Hz`.

So,

`T_1/T_2 = d_1²/d_2² = (0.432)²/(0.381)²

≈ 1.22`

Therefore, the ratio of high tension to low tension is `1.22`.

Hence, option D is correct.

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The ratio of the high-tension to the low-tension is 1.3616:1.Given Data: Diameter of high tension string: d₁ = 0.432 mm Diameter of low tension string:

d₂ = 0.381 mm

The strings are made of the same material, so they have the same density p.

Frequency of both the strings: f = 440 Hz Formula Used:

The tension (T) in a string is given by, T = μf²d²π² Where, μ is the linear density of the string (mass per unit length)d is the diameter of the string f is the frequency of vibration of the stringπ = 3.14 Calculation:

Let the tension in the high-tension string be T₁ and the tension in the low-tension string be T₂ We know that,μ = pA where, p is the density of the string

A = πd²/4 is the cross-sectional area of the string As the strings are made of the same material, they have the same density.

Therefore,μ₁ = μ₂

⇒ pA₁ = pA₂

⇒ A₁ = A₂d₁²

= d₂²

= (0.432 mm)²

= 0.186624 mm²

= A₁A₂

= (0.381 mm)²

= 0.144961 mm²

Therefore, A₁/A₂ = (0.432 mm)²/(0.381 mm)²

= 1.3616/1T₁ = μf²d₁²π²and,T₂ = μf²d₂²π²Dividing these two equations,  

T₁/T₂ = μ₁f²d₁²π²/μ₂f²d₂²π²

⇒ T₁/T₂ = d₁²/d₂²

⇒ T₁/T₂ = (0.432 mm)²/(0.381 mm)²

⇒ T₁/T₂ = 1.3616/1

⇒ T₁/T₂ = 1.3616:1

Therefore, the ratio of the high-tension to the low-tension is 1.3616:1.

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The image is formed at a distance of 425.77 cm to the left of the mirror. The absolute value of the image distance will be 425.77 cm.

It is given that ,Height of object h1 = 0.858 cm, Distance of object from mirror u = -15.0 cm, Radius of curvature R = -20.6 cm

Since the mirror is concave in shape, its radius of curvature will be negative. By applying the mirror formula, we have the ability to determine the distance at which the image is positioned relative to the mirror.

That is, 1/f = 1/v + 1/u where,

the focal length of the mirror is denoted by f, and

v is the distance of the image from the mirror.

Rearranging the equation, we get,

1/v = 1/f - 1/u

1/f = 1/R

Therefore, substituting the values in the above equation, we get,

1/v = 1/R - 1/u = 1/-20.6 - 1/-15 = -0.0485v = -20.6/-0.0485v = 425.77 cm

As the image is formed on the same side of the object, the image distance v is negative. Thus, the image is formed at a distance of 425.77 cm to the left of the mirror. The absolute value of the image distance will be 425.77 cm.

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In the absorption and emission of electromagnetic radiation by atoms, energy transfer occurs in discrete amounts of energy.

When atoms absorb or emit electromagnetic radiation, such as photons, the energy transfer occurs in discrete amounts called quanta. This phenomenon is explained by quantum theory and is commonly known as the quantization of energy. According to this theory, atoms can only absorb or emit energy in specific discrete packets, corresponding to the energy difference between their energy levels.

The statement that atoms are able to absorb and emit energy for a continuous range of wavelengths is not correct. While there is a continuous spectrum of electromagnetic radiation, the energy transfer at the atomic level occurs in quantized steps.

The other two statements regarding the transmission of energy in the photoelectric effect, cell phone transmission, chemical reactions, and collisions are not relevant to the question and do not accurately describe energy transmission in the context of electromagnetic radiation and atoms.

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The question involves deriving the First Law equation, determining specific enthalpy of a wet steam mixture, finding the final state of the system, calculating volumes and pV work, assessing thermodynamic principles and properties in a closed system.

The given question focuses on the application of the First Law of Thermodynamics for a closed system and involves various calculations related to enthalpy, heat energy, work, specific enthalpy, system states, and volume changes.

(a) In part (a), the derivation of the First Law of Thermodynamics at constant pressure is requested, showing the relationship AH = QH - W₂, where AH represents the enthalpy change, QH is the supplied heat energy, and W₂ is the non-pV work done by the system.

(b) In part (b), the specific enthalpy of a wet steam mixture is to be determined based on the provided information from steam tables.

(c) Part (c) involves determining the final state of the system, including the final temperature and the phases present, when a specific amount of heat is added while maintaining constant pressure.

(d) The volumes occupied by the initial steam/water mixture described in part (b) and the final volume of the system after the heat addition are requested in part (d).

(e) Part (e) requires the calculation of the pV work done by the system using two different approaches: the volume change in the system and the change in internal energy for the system.

Overall, the question assesses the understanding and application of thermodynamic principles and properties to analyze and solve problems related to energy, heat transfer, work, and system states in a closed system.

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The solution we get is V = 10V * (1 - e-0.01s/29.4μs) = 2.93V.

The step-by-step answer for Part A of the RC circuit problem:

The time constant of the circuit is τ = RC = 7.2kΩ * 4.0μF = 29.4μs.

The voltage across the capacitor at time t = 0.01s is given by the equation

V = V0(1 - e-t/τ) = 10V * (1 - e-0.01s/29.4μs) = 2.93V.

Therefore, the voltage across the capacitor at time t = 0.01s is 2.93V.

Here is a more detailed explanation of each step:

The time constant of an RC circuit is the time it takes for the voltage across the capacitor to reach 63.2% of its final value. The time constant is calculated by multiplying the resistance of the circuit by the capacitance of the circuit.

The voltage across the capacitor at time t is given by the equation V = V0(1 - e-t/τ), where V0 is the initial voltage across the capacitor, t is the time in seconds, and τ is the time constant of the circuit.

In this problem, V0 = 10V, t = 0.01s, and τ = 29.4μs. Substituting these values into the equation, we get V = 10V * (1 - e-0.01s/29.4μs) = 2.93V.

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The wavelength of an electron with a kinetic energy of 2.4 eV can be calculated using the de Broglie wavelength equation. The wavelength, expressed in nanometers (nm) to two decimal places, can be determined numerically.

The de Broglie wavelength equation relates the wavelength (λ) of a particle to its momentum (p). For an electron, the equation is given by:

λ = h / p

Where:

λ is the wavelength,

h is the Planck's constant (approximately 6.626 x 10^-34 J·s), and

p is the momentum.

The momentum of an electron can be calculated using its kinetic energy (KE) and mass (m) through the equation:

p = sqrt(2 * m * KE)

To find the wavelength, we first need to convert the kinetic energy from electron volts (eV) to joules (J) using the conversion factor: 1 eV = 1.602 x 10^-19 J. Then, we can calculate the momentum and substitute it into the de Broglie wavelength equation.

By plugging in the appropriate values and performing the calculations, we can find the wavelength of the electron in nanometers to two decimal places.

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The time elapsed since the original amount of tritium is one-sixty-fourth of its original amount can be determined by using the concept of half-life.

Tritium has a half-life of 12.3 years, which means that in every 12.3-year period, half of the tritium atoms decay.

To find the time elapsed, we can determine the number of half-lives that have occurred. Since the sample is one-sixty-fourth of its original amount, it has undergone 6 half-lives because 2^6 = 64.

Each half-life corresponds to a time period of 12.3 years, so the total time elapsed is 6 times the half-life, which is 6 * 12.3 = 73.8 years.

Therefore, the time elapsed since the original amount of tritium is one-sixty-fourth of its original amount is 73.8 years.

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The Millikan experiment was carried out to determine the value of the electric charge carried by an electron.'

The method was to suspend oil droplets in a uniform electric field between two metal plates by adjusting the voltage applied to the plates such that the force on the droplet was balanced by the force of gravity. The excess or deficit charge on the droplet could then be calculated and from this,

The charge carried by an electron could be determined.What is an oil drop?An oil drop is a charged droplet of oil that is formed in a high voltage field. An oil droplet carries an electric charge because when it comes into contact with an ion.

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For the time-dependent Schrödinger equation with zero potential and given wave function

V (x, t) = A sin (k x – w t),

its form is as follows:

Hψ(x, t) = I ħ (∂ψ/∂t)

And,

ψ(x, t) = A sin (k x – w t) ∂ψ/∂

t = -A kw cos (k x-w t)

Hence,

Hψ(x, t) = iħ∂ψ/∂

t = -iħAk^2w cos (k x-w t)

Thus, the above wave function

V (x, t) = A sin (k x – w t)

cannot be a solution of the time- dependent Schrödinger equation with zero potential.

For the time-dependent Schrödinger equation with zero potential and given wave function

V (x, t) = cos (k r - w t) + i sin (k x – w t)),

its form is as follows:

Hψ(x, t) = i ħ (∂ψ/∂t) And,

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When a ray of light strikes a surface at a 90-degree angle, which means it is parallel to the normal, the angle of refraction is 90 degrees (Option B).

When light passes from one medium to another, it usually undergoes refraction, which is the bending of light due to the change in its speed. The angle of refraction is determined by Snell's law, which states that the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the speeds of light in the two media.
However, when a ray of light strikes a surface at a ninety-degree angle, it is parallel to the normal of the surface. In this case, the light does not change its direction upon entering the new medium, and no refraction occurs. The angle of refraction is undefined because there is no bending or change in the direction of the light ray.
Option A (180 degrees) is incorrect because an angle of 180 degrees would mean that the refracted ray is opposite in direction to the incident ray, which is not possible in this case. Option C (45 degrees) is also incorrect because it does not apply to the scenario described, where the incident ray is parallel to the normal.
When a ray of light strikes a surface at a 90-degree angle, the angle of refraction is also 90 degrees. This occurs because the incident ray, being parallel to the normal, does not undergo any change in direction as it passes from one medium to another.

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The correct option is : The time constant is 2.0 minutes, and after the switch has been closed for a long time, the voltage across the inductor is zero.

To determine the time constant and the voltage across the inductor after a long time, we can use the formula for the time constant of an RL circuit:

τ = L/R

where τ is the time constant, L is the inductance, and R is the resistance.

In this case, the inductance (L) is given as 6.0 H and the resistance (R) is given as 0.050 Ω.

Using the formula, we can calculate the time constant:

τ = 6.0 H / 0.050 Ω = 120 seconds

Since the time constant is given in seconds, we need to convert it to minutes:

τ = 120 seconds * (1 minute / 60 seconds) = 2.0 minutes

So, the correct option is:

The time constant is 2.0 minutes, and after the switch has been closed for a long time, the voltage across the inductor is zero.

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a) The energy transferred to the electron for the quantum jump from the first excited state to the state with n = 9 is 1.52 eV.

b) The shortest wavelength emitted when the electron de-excites back to its ground state is approximately 410 nm.

c) The second shortest wavelength emitted is approximately 821 nm.

d) The longest wavelength emitted is approximately 4100 nm.

e) The second longest wavelength emitted is approximately 8210 nm.

a) The energy transferred to the electron for the quantum jump can be calculated using the formula for the energy levels of a particle in an infinite well. The energy of the nth level is given by Eₙ = (n²h²)/(8mL²), where h is the Planck's constant, m is the mass of the electron, and L is the width of the well. By calculating the energy difference between the first excited state (n = 2) and the state with n = 9, we can determine the energy transferred, which is approximately 1.52 eV.

b), c), d), e) When the electron de-excites back to its ground state, it emits light with various wavelengths. The wavelength can be determined using the formula λ = 2L/n, where λ is the wavelength, L is the width of the well, and n is the quantum number of the state.

The shortest wavelength corresponds to the highest energy transition, which occurs when n = 2. Substituting the values, we find the shortest wavelength to be approximately 410 nm.

Similarly, we can calculate the wavelengths for the second shortest, longest, and second longest emitted light, which are approximately 821 nm, 4100 nm, and 8210 nm, respectively. These values correspond to the different possible transitions the electron can undergo during de-excitation.

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The energy required for the electron to transition from its first excited state to the state with n = 9 can be calculated using a formula. The shortest, second shortest, longest, and second longest wavelengths that can be emitted when the electron de-excites can also be calculated using a formula.

(a) The energy required for the electron to transition from its first excited state to the state with n = 9 can be calculated using the formula:

E = ((n^2)π^2ħ^2) / (2mL^2)

where n is the quantum number, ħ is the reduced Planck's constant, m is the mass of the electron, and L is the width of the infinite well.

(b) The shortest wavelength that can be emitted corresponds to the transition from the excited state with n = 9 to the ground state with n = 1. This can be calculated using the formula:

λ = 2L / n

(c), (d), and (e) The second shortest, longest, and second longest wavelengths that can be emitted correspond to other possible transitions from the excited state with n = 9 to lower energy states. These can be calculated using the same formula.

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The net torque acting on the cylinder is approximately 0.031 N·m.

To find the net torque acting on the cylinder, we can use the rotational motion equation:

Torque (τ) = Moment of inertia (I) × Angular acceleration (α).

Given that the cylinder starts from rest and reaches 200 rpm (revolutions per minute) in half a minute, we can calculate the angular acceleration. First, we convert the angular velocity from rpm to radians per second (rad/s):

ω = (200 rpm) × (2π rad/1 min) × (1 min/60 s) = 20π rad/s.

The angular acceleration (α) can be calculated by dividing the change in angular velocity (Δω) by the time taken (Δt):

α = Δω/Δt = (20π rad/s - 0 rad/s)/(30 s - 0 s) = (20π/30) rad/s².

Next, we need to calculate the moment of inertia (I) for the cylinder. The moment of inertia of a solid cylinder rotating about its central axis is given by:

I = (1/2)mr²,

where m is the mass of the cylinder and r is its radius.

Converting the mass of the cylinder from grams to kilograms, we have:

m = 20 g = 0.02 kg.

Substituting the values of m and r into the moment of inertia equation, we get:

I = (1/2)(0.02 kg)(0.05 m)² = 2.5 × 10⁻⁵ kg·m².

Now, we can calculate the net torque by multiplying the moment of inertia (I) by the angular acceleration (α):

τ = I × α = (2.5 × 10⁻⁵ kg·m²) × (20π/30) rad/s² ≈ 0.031 N·m.

Therefore, the net torque acting on the cylinder is approximately 0.031 N·m.

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i) The worker must apply a horizontal force of 39.2 N to maintain the constant motion. ii) The box slides a distance of 8.75 m before coming to a rest.

i) To maintain a constant speed, the applied force must balance the frictional force acting on the box. The frictional force can be calculated using the formula F_friction = μ_s * N, where μ_s is the coefficient of static friction and N is the normal force.

Given that the coefficient of static friction is 0.40, we can find the normal force N using the equation N = mg, where m is the mass of the box and g is the acceleration due to gravity.

N = (11.2 kg)(9.8 m/s2) = 109.76 N

The frictional force is then F_friction = (0.40)(109.76 N) = 43.904 N.

ii) When the force is removed, the box experiences a deceleration due to the kinetic friction. The deceleration can be calculated using the formula a = F_friction / m, where F_friction is the kinetic frictional force and m is the mass of the box.

Using the kinematic equation [tex]v^{2}[/tex] = [tex]u^{2}[/tex] + 2as, where v is the final velocity, u is the initial velocity (3.50 m/s), a is the acceleration, and s is the distance traveled, we can solve for s.

0 = (3.50 m/s)2 + 2(-1.964 m/s2) * s

Simplifying the equation, we find s = 8.75 m.

Therefore, the box slides a distance of approximately 8.75 m before coming to a rest.

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When the light beam from the underwater spotlight exits the water at an angle of 64.8 degrees to the vertical, it hits the air-water interface from below the surface with an angle of incidence of 25.2 degrees.

The problem involves a light beam coming from an underwater spotlight and exiting the water at an angle of 64.8 degrees to the vertical. We need to determine the angle of incidence at which the light beam hits the air-water interface from below the surface.

By applying the laws of reflection and refraction, we can calculate the angle of incidence. In this case, the angle of incidence is found to be 25.2 degrees.

When light passes from one medium to another, such as from water to air, it undergoes both reflection and refraction. The angle of incidence (θ₁) is the angle between the incident ray and the normal to the interface, and the angle of refraction (θ₂) is the angle between the refracted ray and the normal.

In this problem, the light beam exits the water at an angle of 64.8 degrees to the vertical. The vertical direction is perpendicular to the surface of the water. Therefore, the angle of incidence is given by:

θ₁ = 90° - 64.8° = 25.2°

This means that the light beam, upon hitting the air-water interface from below the surface, makes an angle of incidence of 25.2 degrees with the normal to the interface.

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A function or collection of functions will be the solution to a differential equation, which is made up of a function and one or more of its derivatives.

Thus, These equations can be used to represent movement, growth, oscillations, waves, and any other phenomenon with a rate of change.

In some differential equations, the variables must be separated since there may be multiple variables at play and a solution may exist for one or more of them. In a different example, the (y) needs to be isolated on one side of the equation if there are two variables in the equation.

It is necessary to move the second variable (x) to the opposing side of the equation.

Thus, A function or collection of functions will be the solution to a differential equation, which is made up of a function and one or more of its derivatives.

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