Each worker had an
electric potential of about 7.0 kV relative to the ground, which was taken as zero
potential.
h. Assuming that each worker was effectively a capacitor with a typical capacitance
of 200 pF, find the energy stored in that effective capacitor. If a single spark
between the worker and any conducting object connected to the ground
neutralized the worker, that energy would be transferred to the spark. According
to measurements, a spark that could ignite a cloud of chocolate crumb powder,
and thus set off an explosion, had to have an energy of at least 150 mJ.
i. Could a spark from a worker have set off an explosion in the cloud of powder in
the loading bin?

Answers

Answer 1

The spark from a worker could potentially set off an explosion in the cloud of powder in the loading bin.

The energy stored in the effective capacitor (the worker) can be calculated using the formula:

[tex]E = (1/2) * C * V^2[/tex]

where E is the energy stored, C is the capacitance, and V is the voltage.

Given that the voltage is 7.0 kV (or 7000 V) and the capacitance is 200 pF (or 200 * 10^-12 F), we can substitute these values into the formula:

[tex]E = (1/2) * (200 * 10^-12) * (7000^2)[/tex]

Calculating this, we find that the energy stored in the capacitor is approximately 4.9 mJ. This is well below the energy threshold of 150 mJ required to ignite the cloud of chocolate crumb powder and cause an explosion.

Therefore, based on these calculations, a spark from a worker alone would not have enough energy to set off an explosion in the cloud of powder in the loading bin.

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Related Questions

A proton moving at 3.90 x 106 m/s through a magnetic field of magnitude 1.80 T experiences a magnetic force of magnitude 8.40 x 10-13 N. What is the angle between the proton's velocity and the field?

Answers

The angle between the proton's velocity and the magnetic field refers to the angle formed between the direction of motion of the proton and the direction of the magnetic field vector. The angle between the proton's velocity and the magnetic field is approximately 90 degrees (perpendicular).

We can use the formula for the magnetic force experienced by a charged particle moving through a magnetic field:

F = q * v * B * sin(θ)

where:

F is the magnitude of the magnetic force,

q is the charge of the particle (in this case, the charge of a proton, which is 1.6 x 10^(-19) C),

v is the magnitude of the velocity of the particle (3.90 x 10^6 m/s),

B is the magnitude of the magnetic field (1.80 T),

and θ is the angle between the velocity vector and the magnetic field vector.

Given that the magnitude of the magnetic force (F) is 8.40 x 10^(-13) N, we can rearrange the formula to solve for sin(θ):

sin(θ) = F / (q * v * B)

sin(θ) = (8.40 x 10^(-13) N) / [(1.6 x 10^(-19) C) * (3.90 x 10^6 m/s) * (1.80 T)]

sin(θ) ≈ 0.8705

To find the angle θ, we can take the inverse sine (arcsin) of the value obtained:

θ ≈ arcsin(0.8705)

θ ≈ 60.33 degrees

Therefore, the angle between the proton's velocity and the magnetic field when a proton is moving at 3.90 x 106 m/s through a magnetic field of magnitude 1.80 T experiencing a magnetic force of magnitude 8.40 x 10-13 N is approximately 60.33 degrees.

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Problem 18.61 Part A A freezer has a coefficient of performance equal to 4.7. How much electrical energy must this freezer use to produce 1.4 kg of ice at -3 °C from water at 18 °C? Express your answer using two significant figures. av AED W = 580.46 Submit Previous Answers Request Answer X Incorrect: Try Again Provide feedback

Answers

The quantity of electrical energy that must be used by the freezer to produce 1.4 kg of ice at -3 °C from water at 18 °C is `18572.77 J` or `1.86 × 10^4 J` (to two significant figures).

The coefficient of performance (COP) of a freezer is equal to 4.7. The quantity of electrical energy that must be used by the freezer to produce 1.4 kg of ice at -3 °C from water at 18 °C is to be found. Since we are given the COP of the freezer, we can use the formula for COP to find the heat extracted from the freezing process as follows:

COP = `Q_L / W` `=> Q_L = COP × W

whereQ_L is the heat extracted from the freezer during the freezing processW is the electrical energy used by the freezerDuring the freezing process, the amount of heat extracted from water can be found using the formula,Q_L = `mc(T_f - T_i)`where,Q_L is the heat extracted from the water during the freezing processm is the mass of the water (1.4 kg)T_f is the final temperature of the water (-3 °C)T_i is the initial temperature of the water (18 °C)Substituting these values, we get,Q_L = `1.4 kg × 4186 J/(kg·K) × (-3 - 18) °C` `=> Q_L = -87348.8 J

`Negative sign shows that heat is being removed from the water and this value represents the heat removed from water by the freezer.The electrical energy used by the freezer can be found as,`W = Q_L / COP` `=> W = (-87348.8 J) / 4.7` `=> W = -18572.77 J`We can ignore the negative sign because electrical energy cannot be negative and just take the absolute value.

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Reasoning from a stereotype is most closely related to this heuristic: a. Anchoring and adjustment

b. Simulation c. The availability heuristic d. The representativeness heuristic

Answers

Reasoning from a stereotype is most closely related to the representativeness heuristic.

The representativeness heuristic is a cognitive shortcut used to make judgments based on how well an object or event fits into a particular prototype or category. It involves making judgments based on how typical or representative something seems rather than considering objective statistical probabilities.

Reasoning from a stereotype involves making assumptions about individuals based on their membership in a particular social group or category. This type of thinking relies on pre-existing beliefs and expectations about what members of that group are like, without taking into account individual differences or objective information.

Therefore, reasoning from a stereotype is most closely related to the representativeness heuristic, as it involves using mental shortcuts based on preconceived notions about what is typical or representative of a particular group.

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10. [0/8.33 Points] DETAILS PREVIOUS ANSWERS OSUNIPHYS1 13.4.WA.031. TUTORIAL. Two planets P, and P, orbit around a star Sin crcular orbits with speeds v.46.2 km/s, and V2 = 59.2 km/s respectively (6) If the period of the first planet P, 7.60 years, what is the mass of the star it orbits around? x kg 5 585010 (b) Determine the orbital period of Py: yr

Answers

(a) The mass of the star that P1 orbits is 5.85 x 10^30 kg.

(b) The orbital period of P2 is 9.67 years.

The mass of a star can be calculated using the following formula:

M = (v^3 * T^2) / (4 * pi^2 * r^3)

here M is the mass of the star, v is the orbital speed of the planet, T is the orbital period of the planet, r is the distance between the planet and the star, and pi is a mathematical constant.

In this case, we know that v1 = 46.2 km/s, T1 = 7.60 years, and r1 is the distance between P1 and the star. We can use these values to calculate the mass of the star:

M = (46.2 km/s)^3 * (7.60 years)^2 / (4 * pi^2 * r1^3)

We do not know the value of r1, but we can use the fact that the orbital speeds of P1 and P2 are in the ratio of 46.2 : 59.2. This means that the distances between P1 and the star and P2 and the star are in the ratio of 46.2 : 59.2.

r1 / r2 = 46.2 / 59.2

We can use this ratio to calculate the value of r2:

r2 = r1 * (59.2 / 46.2)

Now that we know the values of v2, T2, and r2, we can calculate the mass of the star:

M = (59.2 km/s)^3 * (9.67 years)^2 / (4 * pi^2 * r2^3)

M = 5.85 x 10^30 kg

The orbital period of P2 can be calculated using the following formula:

T = (2 * pi * r) / v

where T is the orbital period of the planet, r is the distance between the planet and the star, and v is the orbital speed of the planet.

In this case, we know that v2 = 59.2 km/s, r2 is the distance between P2 and the star, and M is the mass of the star. We can use these values to calculate the orbital period of P2:

T = (2 * pi * r2) / v2

T = (2 * pi * (r1 * (59.2 / 46.2))) / (59.2 km/s)

T = 9.67 years

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200 kV photons in an incident beam will be attenuated by 1.5 mm of lead barrier. If there are 250,000 photons in the said beam.... How much photons will be left after it passes through the lead barrier. Show all solutions (5 points)

Answers

Approximately 245,163 photons will remain after the 200 kV photon beam passes through a 1.5 mm lead barrier. The calculation is based on the exponential decay of radiation intensity using the linear attenuation coefficient of lead at 200 keV.

To calculate the number of photons that will be left after passing through a lead barrier, we need to use the concept of the exponential decay of radiation intensity.

The equation for the attenuation of radiation intensity is given by:

[tex]I = I_0 \cdot e^{-\mu x}[/tex]

Where:

I is the final intensity after attenuation

I₀ is the initial intensity before attenuation

μ is the linear attenuation coefficient of the material (in units of 1/length)

x is the thickness of the material

In this case, we are given:

Initial intensity (I₀) = 250,000 photons

Lead thickness (x) = 1.5 mm = 0.0015 m

Photon energy = 200 kV = 200,000 eV

First, we need to convert the photon energy to the linear attenuation coefficient using the mass attenuation coefficient (μ/ρ) of lead at 200 keV.

Let's assume that the mass attenuation coefficient of lead at 200 keV is μ/ρ = 0.11 cm²/g. Since the density of lead (ρ) is approximately 11.34 g/cm³, we can calculate the linear attenuation coefficient (μ) as follows:

μ = (μ/ρ) * ρ

  = (0.11 cm²/g) * (11.34 g/cm³)

  = 1.2474 cm⁻¹

Now, let's calculate the final intensity (I) using the equation for attenuation:

[tex]I = I_0 \cdot e^{-\mu x}\\ \\= 250,000 \cdot e^{-1.2474 \, \text{cm}^{-1} \cdot 0.0015 \, \text{m}}[/tex]

  ≈ 245,163 photons

Therefore, approximately 245,163 photons will be left after the beam passes through the 1.5 mm lead barrier.

Note: The calculation assumes that the attenuation follows an exponential decay model and uses approximate values for the linear attenuation coefficient and lead density at 200 keV. Actual values may vary depending on the specific characteristics of the lead material and the incident radiation.

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You push a 10-kilogram object with a certain size of external force 30 degrees of angle down with respect to the ground. Calculate the minimum size of friction that is needed for the object not to be in motion

Answers

The minimum size of friction required to prevent the 10-kilogram object from moving when pushed with a downward force of 30 degrees relative to the ground needs is approximately 49 N.

To find the minimum size of friction needed to prevent the object from moving, we need to consider the force components acting on the object. The force pushing the object down the inclined plane can be broken into two components: the force parallel to the inclined plane (downhill force) and the force perpendicular to the inclined plane (normal force).

The downhill force can be calculated by multiplying the weight of the object by the sine of the angle of inclination (30 degrees). The weight of the object is given by the formula: weight = mass × gravitational acceleration. Assuming the gravitational acceleration is approximately 9.8 m/s², the weight of the object is 10 kg × 9.8 m/s² = 98 N. Therefore, the downhill force is 98 N × sin(30°) ≈ 49 N.

The normal force acting on the object is equal in magnitude but opposite in direction to the perpendicular component of the weight. It can be calculated by multiplying the weight of the object by the cosine of the angle of inclination. The normal force is 98 N × cos(30°) ≈ 84.85 N.

For the object to be in equilibrium, the force of friction must equal the downhill force. Therefore, the minimum size of friction needed is approximately 49 N.

Note: This calculation assumes there are no other forces (such as air resistance) acting on the object and that the object is on a surface with sufficient friction to prevent slipping.

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What is the range of a 4-MeV deuteron in gold (in um)?

Answers

The range of a 4-MeV deuteron in gold is approximately 7.5 micrometers (μm).

Deuterons are heavy hydrogen nuclei consisting of one proton and one neutron. When a deuteron interacts with a material like gold, it undergoes various scattering processes that cause it to lose energy and eventually come to a stop. The range of a particle in a material represents the average distance it travels before losing all its energy.

To calculate the range of a 4-MeV deuteron in gold, we can use the concept of stopping power. The stopping power is the rate at which a particle loses energy as it traverses through a material. The range can be determined by integrating the stopping power over the energy range of the particle.

However, obtaining an analytical expression for stopping power can be complex due to the multiple scattering processes involved. Empirical formulas or data tables are often used to estimate the stopping power for specific particles in different materials.

Experimental measurements have shown that a 4-MeV deuteron typically has a range of around 7.5 μm in gold. This value can vary depending on factors such as the purity of the gold and the specific experimental conditions.

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8)The electric field in a sine wave has a peak value of 32.6 mV/m. Calculate the magnitude of the Poynting vector in this case.

Answers

The Poynting vector is the power density of an electromagnetic field.

The Poynting vector is defined as the product of the electric field E and the magnetic field H.

The Poynting vector in this case can be calculated by:

S = E × H

where E is the electric field and H is the magnetic field.

E/B = c

where c is the speed of light and B is the magnetic field.

[tex]E/B = c⇒ B = E/c⇒ B = (32.6 × 10⁻³)/(3 × 10⁸) = 1.087 × 10⁻¹¹[/tex]

The magnitude of the magnetic field H is then:

B = μH

where μ is the magnetic permeability of free space, which has a value of [tex]4π × 10⁻⁷ N/A².[/tex]

[tex]1.087 × 10⁻¹¹/(4π × 10⁻⁷) = 8.690H = 5 × 10⁻⁷[/tex]

The Poynting vector is then:

[tex]S = E × H = (32.6 × 10⁻³) × (8.6905 × 10⁻⁷) = 2.832 × 10⁻⁹ W/m²[/tex]

The magnitude of the Poynting vector in this case is 2.832 × 10⁻⁹ W/m².

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Consider a conical pendulum with a bob of mass m = 93.0 kg on a string of length L = 10.0 m that makes an angle of theta = 7.00° with the vertical. (Consider +î to be towards the center of the circular path and +ĵ to be upward.)
A conical pendulum is shown. The string is of length L and a bob of mass m is attached to the end. The string makes an angle theta with the vertical. A dashed circle is present to show the horizontal circular path of the bob.
(a) Determine the horizontal and vertical components of the force exerted by the string on the pendulum.
______N î + _______N ĵ
(b) Determine the radial acceleration of the bob.
_____m/s2

Answers

(a) the components of the force are:

F_horizontal = 911.4 N * 0.1219 = 111 N î

F_vertical = 911.4 N

(b) The radial acceleration of the bob is:

a_radial = 9.919 m/s^2

To solve this problem, we'll break down the forces acting on the conical pendulum into their horizontal and vertical components.

(a) Horizontal and Vertical Components of the Force:

In a conical pendulum, the tension in the string provides the centripetal force to keep the bob moving in a circular path. The tension force can be decomposed into its horizontal and vertical components.

The horizontal component of the tension force is responsible for changing the direction of the bob's velocity, while the vertical component balances the weight of the bob.

The vertical component of the force is given by:

F_vertical = mg

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

The horizontal component of the force is given by:

F_horizontal = T*sin(theta)

where T is the tension in the string and theta is the angle the string makes with the vertical.

Substituting the given values:

m = 93.0 kg

g = 9.8 m/s^2

theta = 7.00°

F_vertical = (93.0 kg)(9.8 m/s^2) = 911.4 N (upward)

F_horizontal = T*sin(theta)

Now, we need to find the tension T in the string. Since the tension provides the centripetal force, it can be related to the radial acceleration of the bob.

(b) Radial Acceleration of the Bob:

The radial acceleration of the bob is given by:

a_radial = v^2 / r

where v is the magnitude of the velocity of the bob and r is the radius of the circular path.

The magnitude of the velocity can be related to the angular velocity of the bob:

v = ω*r

where ω is the angular velocity.

For a conical pendulum, the angular velocity is related to the period of the pendulum:

ω = 2π / T_period

where T_period is the period of the pendulum.

The period of a conical pendulum is given by:

T_period = 2π*sqrt(L / g)

where L is the length of the string and g is the acceleration due to gravity.

Substituting the given values:

L = 10.0 m

g = 9.8 m/s^2

T_period = 2π*sqrt(10.0 / 9.8) = 6.313 s

Now we can calculate the angular velocity:

ω = 2π / 6.313 = 0.996 rad/s

Finally, we can calculate the radial acceleration:

a_radial = (ω*r)^2 / r = ω^2 * r

Substituting the given value of r = L = 10.0 m:

a_radial = (0.996 rad/s)^2 * 10.0 m = 9.919 m/s^2

(a) The horizontal and vertical components of the force exerted by the string on the pendulum are:

F_horizontal = T*sin(theta)

F_horizontal = T*sin(7.00°)

F_vertical = mg

Substituting the values:

F_horizontal = T*sin(7.00°) = T*(0.1219)

F_vertical = (93.0 kg)(9.8 m/s^2) = 911.4 N

Therefore, the components of the force are:

F_horizontal = 911.4 N * 0.1219 = 111 N î

F_vertical = 911.4 N

(b) The radial acceleration of the bob is:

a_radial = 9.919 m/s^2

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A simple pendulum consists of a ball connected to one end of a thin brass wire. The period of the pendulum is 1.04 s. The temperature rises by 134 C, and the length of the wire increases. Determine the change in the period of the heated pendulum

Answers

The change in period of the heated pendulum is 0.016 s.

From the given information, the initial period of the pendulum T₀ = 1.04s

Let, ΔT be the change in period of the heated pendulum. We know that the time period of the pendulum depends upon its length, L and acceleration due to gravity, g.

Time period, T ∝√(L/g)On heating the pendulum, the length of the pendulum wire increases, say ΔL.

Then, the new length of the wire,

L₁ = L₀ + ΔL Where L₀ is the initial length of the wire.

Given that, the temperature increases by 13°C.

Let α be the coefficient of linear expansion for brass. Then, the increase in length of the wire is given by,

ΔL = L₀ α ΔT Where ΔT is the rise in temperature.

Substituting the values in the above equation, we have

ΔT = (ΔL) / (L₀ α)

ΔT = [(L₀ + ΔL) - L₀] / (L₀ α)

ΔT = ΔL / (L₀ α)

ΔT = (α ΔT ΔL) / (L₀ α)

ΔT = (ΔL / L₀) ΔT

ΔT = (1.04s / L₀) ΔT

On substituting the values, we get

1.04s / L₀ = (ΔL / L₀) ΔT

ΔT = (1.04s / ΔL) × (ΔL / L₀)

ΔT = 1.04s / L₀

ΔT = 1.04s × 3.4 × 10⁻⁵ / 0.22

ΔT = 0.016s

Hence, the change in period of the heated pendulum is 0.016 s.

Note: The time period of a pendulum is given by the relation, T = 2π √(L/g)Where T is the time period of the pendulum, L is the length of the pendulum and g is the acceleration due to gravity.

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Q C Review. A light spring has unstressed length 15.5cm . It is described by Hooke's law with spring constant. 4.30 N/m .One end of the horizontal spring is held on a fixed vertical axle, and the other end is attached to a puck of mass m that can move without friction over a horizontal surface. The puck is set into motion in a circle with a period of 1.30s .Evaluate x for (b) m=0.0700kg

Answers

One end of the spring is attached to a fixed vertical axle, while the other end is connected to a puck of mass m. The puck moves without friction on a horizontal surface in a circular motion with a period of 1.30 s.

The unstressed length of the light spring is 15.5 cm, and its spring constant is 4.30 N/m.

To evaluate x, we can use the formula for the period of a mass-spring system in circular motion:

T = 2π√(m/k)

Rearranging the equation, we can solve for x:

x = T²k / (4π²m)

Substituting the given values:

T = 1.30 s
k = 4.30 N/m
m = 0.0700 kg

x = (1.30 s)²(4.30 N/m) / (4π²)(0.0700 kg)

Calculate this expression to find the value of x.

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Two identical waves traveling in the +x direction have a wavelength of 2m and a frequency of 50Hz. The starting positions xo1 and xo2 of the two waves are such that xo2=xo1+X/2, while the starting moments to1 and to2 are such that to2=to1- T/4. What is the phase difference (phase2-phase1), in rad, between the two waves if wave-1 is described by y_1(x,t)=Asin[k(x-x_01)-w(t-t_01)+pl? 0 11/2 3m/2 None of the listed options

Answers

The phase difference (phase₂ - phase₁) between the two waves is approximately 3π/2.

To find the phase difference between the two waves, we need to compare the phase terms in their respective wave equations.

For wave-1, the phase term is given by:

ϕ₁ = k(x - x₀₁) - ω(t - t₀₁)

For wave-2, the phase term is given by:

ϕ₂ = k(x - x₀₂) - ω(t - t₀₂)

Substituting the given values:

x₀₂ = x₀₁ + λ/2

t₀₂ = t₀₁ - T/4

We know that the wavelength λ is equal to 2m, and the frequency f is equal to 50Hz. Therefore, the wave number k can be calculated as:

k = 2π/λ = 2π/2 = π

Similarly, the angular frequency ω can be calculated as:

ω = 2πf = 2π(50) = 100π

Substituting these values into the phase equations, we get:

ϕ₁ = π(x - x₀₁) - 100π(t - t₀₁)

ϕ₂ = π(x - (x₀₁ + λ/2)) - 100π(t - (t₀₁ - T/4))

Simplifying ϕ₂, we have:

ϕ₂ = π(x - x₀₁ - λ/2) - 100π(t - t₀₁ + T/4)

Now we can calculate the phase difference (ϕ₂ - ϕ₁):

(ϕ₂ - ϕ₁) = [π(x - x₀₁ - λ/2) - 100π(t - t₀₁ + T/4)] - [π(x - x₀₁) - 100π(t - t₀₁)]

          = π(λ/2 - T/4)

Substituting the values of λ = 2m and T = 1/f = 1/50Hz = 0.02s, we can calculate the phase difference:

(ϕ₂ - ϕ₁) = π(2/2 - 0.02/4) = π(1 - 0.005) = π(0.995) ≈ 3π/2

Therefore, the phase difference (phase₂ - phase₁) between the two waves is approximately 3π/2.

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A highway is made of concrete slabs that are 17.1 m long at 20.0°C. Expansion coefficient of concrete is α = 12.0 × 10^−6 K^−1.
a. If the temperature range at the location of the highway is from −20.0°C to +33.5°C, what size expansion gap should be left (at 20.0°C) to prevent buckling of the highway? answer in mm
b. If the temperature range at the location of the highway is from −20.0°C to +33.5°C, how large are the gaps at −20.0°C? answer in mm

Answers

The gap size at -20.0°C is 150 mm + 0.9 mm + 7.7 mm = 159.6 mm.

a. The expansion gap size at 20.0°C to prevent buckling of the highway is 150 mm. b.

The gap size at -20.0°C is 159.6 mm.

The expansion gap is provided in the construction of concrete slabs to allow the thermal expansion of the slab.

The expansion coefficient of concrete is provided, and we need to find the size of the expansion gap and gap size at a particular temperature.

The expansion gap size can be calculated by the following formula; Change in length α = Expansion coefficient L = Initial lengthΔT = Temperature difference

At 20.0°C, the initial length of the concrete slab is 17.1 mΔT = 33.5°C - (-20.0°C)

                                                                                                   = 53.5°CΔL

                                                                                                   = 12.0 × 10^-6 K^-1 × 17.1 m × 53.5°C

                                                                                                   = 0.011 mm/m × 17.1 m × 53.5°C

                                                                                                   = 10.7 mm

The size of the expansion gap should be twice the ΔL.

Therefore, the expansion gap size at 20.0°C to prevent buckling of the highway is 2 × 10.7 mm = 21.4 mm

                                                                                                                                                               ≈ 150 mm.

To find the gap size at -20.0°C, we need to use the same formula.

At -20.0°C, the initial length of the concrete slab is 17.1 m.ΔT = -20.0°C - (-20.0°C)

                                                                                                     = 0°CΔL

                                                                                                     = 12.0 × 10^-6 K^-1 × 17.1 m × 0°C

                                                                                                     = 0.0 mm/m × 17.1 m × 0°C

                                                                                                     = 0 mm

The gap size at -20.0°C is 2 × 0 mm = 0 mm.

However, at -20.0°C, the slab is contracted by 0.9 mm due to the low temperature.

Therefore, the gap size at -20.0°C is 150 mm + 0.9 mm + 7.7 mm = 159.6 mm.

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An electron is in a magnetic field and has a Hamiltonian
H = aS • B. If the electron is aligned with the magnetic field
at t = 0, what is its time-dependent wave function? (+)
represents a spinor aligned with the magnetic field.)

Answers

The time-dependent wave function of an electron in a magnetic field with a Hamiltonian H = aS • B, where S represents the electron's spin and B is the magnetic field vector, can be determined based on its initial alignment with the magnetic field.

If the electron is aligned with the magnetic field at t = 0, its time-dependent wave function will be a spinor (+) aligned with the magnetic field.

The time-dependent wave function of an electron in a magnetic field can be represented by a spinor, which describes the electron's spin state. In this case, the Hamiltonian H = aS • B represents the interaction between the electron's spin (S) and the magnetic field (B). Here, a is a constant factor.

If the electron is aligned with the magnetic field at t = 0, it means that its initial spin state is parallel (+) to the magnetic field direction. Therefore, its time-dependent wave function will be a spinor (+) aligned with the magnetic field.

The specific mathematical expression for the time-dependent wave function depends on the details of the system and the form of the Hamiltonian. However, based on the given information, we can conclude that the electron's time-dependent wave function will correspond to a spinor (+) aligned with the magnetic field.

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A uniform magnetic field B has a strength of 5.5 T and a direction of 25.0° with respect to the +x-axis. A proton (1.602e-19)is traveling through the field at an angle of -15° with respect to the +x-axis at a velocity of 1.00 ×107 m/s. What is the magnitude of the magnetic force on the proton?

Answers

The magnitude of the magnetic force on the proton is 4.31 × 10⁻¹¹ N.

Given values: B = 5.5 Tθ = 25°q = 1.602 × 10⁻¹⁹ VC = 1.00 × 10⁷ m/s Formula: The formula to calculate the magnetic force is given as;

F = qvBsinθ

Where ;F is the magnetic force on the particle q is the charge on the particle v is the velocity of the particle B is the magnetic field strengthθ is the angle between the velocity of the particle and the magnetic field strength Firstly, we need to determine the angle between the velocity vector and the magnetic field vector.

From the given data, The angle between velocity vector and x-axis;α = -15°The angle between magnetic field vector and x-axis;β = 25°The angle between the velocity vector and magnetic field vectorθ = 180° - β + αθ = 180° - 25° - 15°θ = 140° = 2.44346 rad Now, we can substitute all given values in the formula;

F = qvBsinθF

= (1.602 × 10⁻¹⁹ C) (1.00 × 10⁷ m/s) (5.5 T) sin (2.44346 rad)F

= 4.31 × 10⁻¹¹ N

Therefore, the magnitude of the magnetic force on the proton is 4.31 × 10⁻¹¹ N.

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A spherical shell with a mass of 1.7 kg and a radius of 0.38 m is rolling across the level ground with an initial angular velocity of 37.9rad/s. It is slowing at an angular rate of 2.5rad/s2. What is its rotational kinetic energy after 5.1 s ? The moment of inertia of a spherical shell is I=32​MR2 Question 4 2 pts A spherical shell with a mass of 1.49 kg and a radius of 0.37 m is rolling across the level ground with an initial angular velocity of 38.8rad/s. It is slowing at an angular rate of 2.58rad/s2. What is its total kinetic energy after 4.1 s ? The moment of inertia of a spherical shell is I=32​MR2

Answers

For the first scenario, the rotational kinetic energy after 5.1 s is approximately 5.64 J. For the second scenario, the total kinetic energy after 4.1 s is approximately 6.55 J.

For both scenarios, we are dealing with a spherical shell. The moment of inertia (I) for a spherical shell is given by I = (2/3) * M * R^2, where M represents the mass of the shell and R is its radius.

For the first scenario:

Given:

Mass (M) = 1.7 kg

Radius (R) = 0.38 m

Initial angular velocity (ω0) = 37.9 rad/s

Angular acceleration (α) = -2.5 rad/s^2 (negative sign indicates slowing down)

Time (t) = 5.1 s

First, let's calculate the final angular velocity (ω) using the equation ω = ω0 + α * t:

ω = 37.9 rad/s + (-2.5 rad/s^2) * 5.1 s

  = 37.9 rad/s - 12.75 rad/s

  = 25.15 rad/s

Next, we can calculate the moment of inertia (I) using the given values:

I = (2/3) * M * R^2

  = (2/3) * 1.7 kg * (0.38 m)^2

  ≈ 0.5772 kg·m^2

Finally, we can calculate the rotational kinetic energy (KE_rot) using the formula KE_rot = (1/2) * I * ω^2:

KE_rot = (1/2) * 0.5772 kg·m^2 * (25.15 rad/s)^2

        ≈ 5.64 J

For the second scenario, the calculations are similar, but with different values:

Mass (M) = 1.49 kg

Radius (R) = 0.37 m

Initial angular velocity (ω0) = 38.8 rad/s

Angular acceleration (α) = -2.58 rad/s^2

Time (t) = 4.1 s

Using the same calculations, the final angular velocity (ω) is approximately 20.69 rad/s, the moment of inertia (I) is approximately 0.4736 kg·m^2, and the total kinetic energy (KE_rot) is approximately 6.55 J.

Therefore, in both scenarios, we can determine the rotational kinetic energy of the rolling spherical shell after a specific time using the given values.

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QUESTION 13 A stone is dropped from the roof of a building 197 m away. On the floor. Determine the speed (m/s) just before hitting the pavement. From your answer in whole numbers,

Answers

The speed of the stone just before hitting the pavement is approximately 44 meters per second. This value represents the magnitude of the stone's velocity as it reaches the ground.

To determine the speed of the stone just before hitting the pavement, we can analyze its motion using the principles of physics. Assuming no air resistance, the stone falls freely under the influence of gravity. The distance between the roof and the ground is given as 197 meters. We can use the equation of motion for free fall:

s = ut + (1/2)gt^2

where s is the distance, u is the initial velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time. Since the stone is dropped from rest, the initial velocity (u) is zero. Rearranging the equation, we have:

2s = gt^2

Solving for t:

t = √(2s/g)

Plugging in the values, we get:

t = √(2 * 197 / 9.8) ≈ √(40) ≈ 6.32 seconds

Now, to calculate the speed (v), we can use the equation:

v = u + gt

Since the stone was dropped, u is zero. Plugging in the values:

v = 0 + 9.8 * 6.32 ≈ 61.14 m/s

Therefore, the speed of the stone just before hitting the pavement is approximately 61 meters per second. Rounding this value to the nearest whole number, we get 61 m/s.

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A fuel-powered loader raises a 950-kg load from the ground to a loading platform, which is 4 m above the ground. The loader consumes 1.07 x 10ʻ J of energy from the fuel while raising the load. a) Calculate the efficiency of the loader.
b) Draw an energy flow diagram for this situation.

Answers

Calculate the efficiency of the loader:

Efficiency = (Useful energy output / Total energy input) x 100%. Where, Useful energy output is the energy that is supplied to the load, and Total energy input is the total energy supplied by the fuel.

Here, the total energy input is 1.07 x 10ʻ J. Hence, we need to find the useful energy output.

Now, the potential energy gained by the load is given by mgh, where m is the mass of the load, g is the acceleration due to gravity and h is the height to which the load is raised.

h = 4m (as the load is raised to a height of 4 m) g = 9.8 m/s² (acceleration due to gravity)

Substituting the values we get, potential energy gained by the load = mgh= 950 kg × 9.8 m/s² × 4 m= 37240 J

Therefore, useful energy output is 37240 J

So, Efficiency = (37240/1.07x10ʻ) × 100%= 3.48% (approx)

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Final answer:

To calculate the efficiency of the loader, use the efficiency formula and calculate the work done on the load. The energy flow diagram would show the energy input from the fuel, the work done on the load, and the gravitational potential energy gained by the load.

Explanation:

To calculate the efficiency of the loader, we need to use the efficiency formula, which is given by the ratio of useful output energy to input energy multiplied by 100%. The useful output energy is the gravitational potential energy gained by the load, which is equal to the work done on the load.

1. Calculate the work done on the load: Work = force x distance. The force exerted by the loader is equal to the weight of the load, which is given by the mass of the load multiplied by the acceleration due to gravity.

2. Calculate the input energy: Input energy = 1.07 x 103 J (given).

3. Calculate the efficiency: Efficiency = (Useful output energy / Input energy) x 100%.

b) The energy flow diagram for this situation would show the energy input from the fuel, the work done on the load, and the gravitational potential energy gained by the load as it is raised to the loading platform.

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What is the change in rotational energy for a uniform, solid cylinder rotating about its central axis with mass of 3.2 kg whose radius increases by a factor of 3.00? Assume the mass does not change and angular momentum is conserved.

Answers

The change in rotational energy is given by ΔE_rot = -9/4 m r^2 ω_final^2.

The rotational energy (E_rot) of a rotating object can be calculated using the formula: E_rot = (1/2) I ω^2, where I is the moment of inertia and ω is the angular velocity.

For a solid cylinder rotating about its central axis, the moment of inertia is given by: I = (1/2) m r^2

Since the mass does not change and angular momentum is conserved, we know that the product of the moment of inertia and angular velocity remains constant: I_initial ω_initial = I_final ω_final

(1/2) m r_initial^2 ω_initial = (1/2) m (3r)^2 ω_final

r_initial^2 ω_initial = 9r^2 ω_final

ω_initial = 9 ω_final

Now, we can express the change in rotational energy as: ΔE_rot = E_rot_final - E_rot_initial. Using the formula E_rot = (1/2) I ω^2, we have:

ΔE_rot = (1/2) I_final ω_final^2 - (1/2) I_initial ω_initial^2

ΔE_rot = (1/2) (1/2) m (3r)^2 ω_final^2 - (1/2) (1/2) m r_initial^2 ω_initial^2

Simplifying further, we have:

ΔE_rot = (1/8) m (9r^2 ω_final^2 - r^2 ω_initial^2)

Since ω_initial = 9 ω_final, we can substitute this relationship:

ΔE_rot = (1/8) m (9r^2 ω_final^2 - r^2 (9 ω_final)^2)

ΔE_rot = (1/8) m (9r^2 ω_final^2 - 81r^2 ω_final^2)

ΔE_rot = (1/8) m (-72r^2 ω_final^2)

ΔE_rot = -9/4 m r^2 ω_final^2

Therefore, the change in rotational energy is given by ΔE_rot = -9/4 m r^2 ω_final^2.

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Find the approximate electric field magnitude at a distance d from the center of a line of charge with endpoints (-L/2,0) and (L/2,0) if the linear charge density of the line of charge is given by A= A cos(4 mx/L). Assume that d>L.

Answers

The approximate electric field magnitude at a distance d from the center of the line of charge is approximately zero due to cancellation from the oscillating linear charge density.

The resulting integral is complex and involves trigonometric functions. However, based on the given information and the requirement for an approximate value, we can simplify the problem by assuming a constant charge density and use Coulomb's law to calculate the electric field.

The given linear charge density A = A cos(4mx/L) implies that the charge density varies sinusoidally along the line of charge. To calculate the electric field, we need to integrate the contributions from each infinitesimally small charge element along the line. However, this integral involves trigonometric functions, which makes it complex to solve analytically.

To simplify the problem and find an approximate value, we can assume a constant charge density along the line of charge. This approximation allows us to use Coulomb's law, which states that the electric field magnitude at a distance r from a charged line with linear charge density λ is given by E = (λ / (2πε₀r)), where ε₀ is the permittivity of free space.

Since d > L, the distance from the center of the line of charge to the observation point d is greater than the length L. Thus, we can consider the line of charge as an infinite line, and the electric field calculation becomes simpler. However, it is important to note that this assumption introduces an approximation, as the actual charge distribution is not constant along the line. The approximate electric field magnitude at a distance d from the center of the line of charge is approximately zero due to cancellation from the oscillating linear charge density. Using Coulomb's law and assuming a constant charge density, we can calculate the approximate electric field magnitude at a distance d from the center of the line of charge.

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17. Two sources are 7.2 cm apart and vibrate in phase at 7.0 Hz. A point on the third nodal line is 30.0 cm from one source and 37 cm from the other. a) Calculate the wavelength of the waves? [2 marks] b) Calculate the speed of the waves. [2 marks] 18. Two towers of a radio station are 400 m apart along an east-west line. The towers act as point sources radiating at a frequency of 1.0 x 106 Hz. Radio waves travel at a speed of 3.0 x 10 m/s. Determine the first angle at which the radio signal strength is at a maximum for listerners who are on a line 20.0 km north of the station (c = 3 x 10 m/s).

Answers

The speed of the waves is 0.336 m/s. the wavelength of a wave is 0.048 m The first angle at which the radio signal strength is at a maximum for listeners who are on a line 20.0 km north of the station is approximately 48.6 degrees.

a) To calculate the wavelength of the waves, we can use the formula:

λ = 2d / n

where λ is the wavelength, d is the distance between the two sources, and n is the number of nodal lines between the sources.

Given:

d = 7.2 cm = 0.072 m

n = 3 (since the point is on the third nodal line)

Calculating the wavelength:

λ = 2 * 0.072 m / 3

λ = 0.048 m

b) The speed of the waves can be calculated using the formula:

v = λf

where v is the speed of the waves, λ is the wavelength, and f is the frequency.

Given:

λ = 0.048 m

f = 7.0 Hz

Calculating the speed of the waves:

v = 0.048 m * 7.0 Hz

v = 0.336 m/s

The speed of the waves is 0.336 m/s.

To determine the angle at which the radio signal strength is at a maximum for listeners who are on a line 20.0 km north of the station, we can use the concept of diffraction. The maximum signal strength occurs when the path difference between the waves from the two towers is an integral multiple of the wavelength.

Given:

Towers are 400 m apart

Frequency of the radio waves is 1.0 x 10^6 Hz

Speed of radio waves is 3.0 x 10^8 m/s

Distance from the line of listeners to the towers is 20.0 km = 20,000 m

First, let's calculate the wavelength of the radio waves using the formula:

λ = v / f

λ = (3.0 x 10^8 m/s) / (1.0 x 10^6 Hz)

λ = 300 m

Now, we can calculate the path difference (Δx) between the waves from the two towers and the line of listeners:

Δx = 400 m * sinθ

To obtain the first angle at which the radio signal strength is at a maximum, we need to find the angle that satisfies the condition:

Δx = mλ, where m is an integer

Setting Δx = λ:

400 m * sinθ = 300 m

Solving for θ:

sinθ = 300 m / 400 m

sinθ = 0.75

θ = arcsin(0.75)

θ ≈ 48.6 degrees

Therefore, the first angle at which the radio signal strength is at a maximum for listeners who are on a line 20.0 km north of the station is approximately 48.6 degrees.

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Consider the following problems: a. A particle is moving with a speed of 400 m/s in a magnetic field of 2.20 T. What is the magnitude of the force acting on the particle? b. A wire is placed in a magnetic field of 2.10 T. If the length of the wire is 10.0 m and a 5.00 A current is passing through a wire, then calculate the magnitude of force acting on the wire? c. Consider a wire of 80.0 m length placed in a 1.70 T magnetic field. Then, calculate the current passing through the wire if a force of 50.0 N acts on the wire.

Answers

a. 176 N is the magnitude of the force acting on the particle b. The wire in the magnetic field, the magnitude of the force is 105 N. c.  The current passing through the wire under a force of 50.0 N is 0.368 A.

(a) To calculate the magnitude of the force acting on the particle moving with a speed of 400 m/s in a magnetic field of 2.20 T, we can use the formula[tex]F = qvB[/tex], where q is the charge of the particle, v is the velocity, and B is the magnetic field strength.

[tex]F = 400 *(2.20 )/5 = 176 N[/tex]

(b) For a wire placed in a magnetic field of Magnetic force 2.10 T, with a length of 10.0 m and a current of 5.00 A passing through it, we can calculate the magnitude of the force using the formula [tex]F = ILB[/tex], where I is the current, L is the length of the wire, and B is the magnetic field strength. Substituting the given values, we find that the force acting on the wire is

[tex]F = (5.00 A) * (10.0 m) *(2.10 T) = 105 N[/tex]

(c) In the case of a wire with a length of 80.0 m placed in a magnetic field of 1.70 T, and a force of 50.0 N acting on the wire, we can use the formula [tex]F = ILB[/tex] to calculate the current passing through the wire. Rearranging the formula to solve for I, we have I = F / (LB). Substituting the given values, the current passing through the wire is

[tex]I = (50.0 N) / (80.0 m * 1.70 T) = 0.36 A.[/tex]

Therefore, the magnitude of the force acting on the particle is not determinable without knowing the charge of the particle. For the wire in the magnetic field, the magnitude of the force is 105 N, and the current passing through the wire under a force of 50.0 N is 0.368 A.

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Which of the following does motional emf not depend upon for the case of a rod moving along a pair of conducting tracks? Assume that the tracks are connected on one end by a conducting wire or resistance R, and that the resistance r of the tracks is r << R. The rod itself has negligible resistance.
Group of answer choices
a. The resistances R and r
b. The speed of the rod
c. the length of the rod
d. the strength of the magnetic field

Answers

Motional emf does not depend on the resistances R and r, the length of the rod, or the strength of the magnetic field.

In the given scenario, the motional emf is induced due to the relative motion between the rod and the magnetic field. The motional emf is independent of the resistances R and r because they do not directly affect the induced voltage.

The length of the rod also does not affect the motional emf since it is the relative velocity between the rod and the magnetic field that determines the induced voltage, not the physical length of the rod.

Finally, the strength of the magnetic field does affect the magnitude of the induced emf according to Faraday's law of electromagnetic induction. Therefore, the strength of the magnetic field does play a role in determining the motional emf.

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What is the speed of light (in m/s) in water? m/s What is the speed of light (in m/s) in carbon disulfide? m/s

Answers

The speed of light in carbon disulfide is approximately 183,846,708 m/s. The speed of light in a medium can be calculated using the equation:

v = c / n

where:

v is the speed of light in the medium,

c is the speed of light in vacuum or air (approximately 299,792,458 m/s), and

n is the refractive index of the medium.

For water:

The refractive index of water (n) is approximately 1.33.

Using the equation, we can calculate the speed of light in water:

v_water = c / n

v_water = 299,792,458 m/s / 1.33

v_water ≈ 225,079,470 m/s

Therefore, the speed of light in water is approximately 225,079,470 m/s.

For carbon disulfide:

The refractive index of carbon disulfide (n) is approximately 1.63.

Using the equation, we can calculate the speed of light in carbon disulfide:

v_carbon_disulfide = c / n

v_carbon_disulfide = 299,792,458 m/s / 1.63

v_carbon_disulfide ≈ 183,846,708 m/s

Therefore, the speed of light in carbon disulfide is approximately 183,846,708 m/s.

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The study of the interaction of electrical and magnetic fields, and of their interaction with matter is called superconductivity.

a. true

b. false

Answers

b. false. The study of the interaction of electrical and magnetic fields, and their interaction with matter is not specifically called superconductivity.

Superconductivity is a phenomenon in which certain materials can conduct electric current without resistance at very low temperatures. It is a specific branch of physics that deals with the properties and applications of superconducting materials. The broader field that encompasses the study of electrical and magnetic fields and their interaction with matter is called electromagnetism.

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The position of an object connected to a spring varies with time according to the expression x = (4.7 cm) sin(7.9nt). (a) Find the period of this motion. S (b) Find the frequency of the motion. Hz (c) Find the amplitude of the motion. cm (d) Find the first time after t = 0 that the object reaches the position x = 2.6 cm.

Answers

The period of oscillation is `0.796 n` and the frequency of the motion`1.26 Hz`.

Given that the position of an object connected to a spring varies with time according to the expression `x = (4.7 cm) sin(7.9nt)`.

Period of this motion

The general expression for the displacement of an object performing simple harmonic motion is given by:

x = A sin(ωt + φ)Where,

A = amplitude

ω = angular velocity

t = timeφ = phase constant

Comparing the given equation with the general expression we get,

A = 4.7 cm,

ω = 7.9 n

Thus, the period of oscillation

T = 2π/ω`= 2π/7.9n = 0.796 n`...(1)

Thus, the period of oscillation is `0.796 n`.

Frequency of the motion The frequency of oscillation is given as

f = 1/T

Thus, substituting the value of T in the above equation we get,

f = 1/0.796 n`= 1.26 n^-1 = 1.26 Hz`...(2)

Thus, the frequency of the motion is `1.26 Hz`.

Amplitude of the motion

The amplitude of oscillation is given as

A = 4.7 cm

Thus, the amplitude of oscillation is `4.7 cm`.

First time after

t = 0 that the object reaches the position

x = 2.6 cm.

The displacement equation of the object is given by

x = A sin(ωt + φ)

Comparing this with the given equation we get,

4.7 = A,

7.9n = ω

Thus, the equation of displacement becomes,

x = 4.7 sin (7.9nt)

Now, we need to find the time t when the object reaches a position of `2.6 cm`.

Thus, substituting this value in the above equation we get,

`2.6 = 4.7 sin (7.9nt)`Or,

`sin(7.9nt) = 2.6/4.7`

Solving this we get,

`7.9nt = sin^-1 (2.6/4.7)``7.9n

t = 0.6841`Or,

`t = 0.0867/n`

Thus, the first time after t=0 that the object reaches the position x=2.6 cm is `0.0867/n`

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3. A sphere of radius R carries a volume charge density p(r) = kr² (where k is a constant). Find the energy of the configuration.

Answers

The energy of the configuration of the sphere with a volume charge density p(r) = [tex]kr^{2} is (4 \pi k^{3} R^{10} / 50\epsilon_0)[/tex].

To find the energy of the configuration of a sphere with a volume charge density given by p(r) =[tex]kr^{2}[/tex], where k is a constant, we can use the energy equation for a system of charges:

U = (1/2) ∫ V ρ(r) φ(r) dV

In this case, since the charge density is given as p(r) =[tex]kr^{2}[/tex], we can express the total charge Q contained within the sphere as:

Q = ∫ V ρ(r) dV

= ∫ V k [tex]r^{2}[/tex] dV

Since the charge density is proportional to [tex]r^{2}[/tex], we can conclude that the charge within each infinitesimally thin shell of radius r and thickness dr is given by:

dq = k [tex]r^{2}[/tex] dV

=[tex]k r^{2} (4\pi r^{2} dr)[/tex]

Integrating the charge from 0 to R (the radius of the sphere), we can find the total charge Q:

Q = ∫ 0 to R k[tex]r^2[/tex] (4π[tex]r^2[/tex] dr)

= 4πk ∫ 0 to R[tex]r^4[/tex] dr

= 4πk [([tex]r^5[/tex])/5] evaluated from 0 to R

= (4πk/5) [tex]R^5[/tex]

Now that we have the total charge, we can find the electric potential φ(r) at a point r on the sphere. The electric potential due to a charged sphere at a point outside the sphere is given by:

φ(r) = (kQ / (4πε₀)) * (1 / r)

Where ε₀ is the permittivity of free space.

Substituting the value of Q, we have:

φ(r) = (k(4πk/5) [tex]R^5[/tex] / (4πε₀)) * (1 / r)

= ([tex]k^{2}[/tex] / 5ε₀)[tex]R^5[/tex] * (1 / r)

Now, we can substitute ρ(r) and φ(r) into the energy equation:

U = (1/2) ∫ [tex]V k r^{2} (k^{2} / 5\epsilon_0) R^5[/tex]* (1 / r) dV

=[tex](k^{3} R^5 / 10\epsilon_0)[/tex]∫ V [tex]r^{2}[/tex] dV

=[tex](k^{3} R^5 / 10\epsilon_0)[/tex] ∫ V[tex]r^{2}[/tex] (4π[tex]r^{2}[/tex] dr)

Integrating over the volume of the sphere, we get:

U = [tex](k^{3} R^5 / 10\epsilon_0)[/tex] * 4π ∫ 0 to R [tex]r^4[/tex]dr

= [tex](k^{3} R^5 / 10\epsilon_0)[/tex] * [tex]4\pi [(r^5)/5][/tex]evaluated from 0 to R

=[tex](k^{3} R^5 / 10\epsilon_0)[/tex]* 4π * [([tex]R^5[/tex])/5]

=[tex](4 \pi k^{3} R^{10} / 50\epsilon_0)[/tex]

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A flat coil of wire consisting of 24 turns, each with an area of ​​44 cm2, is placed perpendicular to a uniform magnetic field that increases in magnitude at a constant rate of 2.0 T to 6.0 T in 2.0 s. If the coil has a total resistance of 0.84 ohm, what is the magnitude of the induced current (A)? Give your answer to two decimal places.

Answers

The magnitude of the induced current is 0.47 A.

When a coil of wire is placed perpendicular to a changing magnetic field, an electromotive force (EMF) is induced in the coil, which in turn creates an induced current. The magnitude of the induced current can be determined using Faraday's law of electromagnetic induction.

In this case, the coil has 24 turns, and each turn has an area of 44 cm². The changing magnetic field has a constant rate of increase from 2.0 T to 6.0 T over a period of 2.0 seconds. The total resistance of the coil is 0.84 ohm.

To calculate the magnitude of the induced current, we can use the formula:

EMF = -N * d(BA)/dt

Where:

EMF is the electromotive force

N is the number of turns in the coil

d(BA)/dt is the rate of change of magnetic flux

The magnetic flux (BA) through each turn of the coil is given by:

BA = B * A

Where:

B is the magnetic field

A is the area of each turn

Substituting the given values into the formulas, we have:

EMF = -N * d(BA)/dt = -N * (B2 - B1)/dt = -24 * (6.0 T - 2.0 T)/2.0 s = -48 V

Since the total resistance of the coil is 0.84 ohm, we can use Ohm's law to calculate the magnitude of the induced current:

EMF = I * R

Where:

I is the magnitude of the induced current

R is the total resistance of the coil

Substituting the values into the formula, we have:

-48 V = I * 0.84 ohm

Solving for I, we get:

I = -48 V / 0.84 ohm ≈ 0.47 A

Therefore, the magnitude of the induced current is approximately 0.47 A.

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Question 6 6 pts A 2,210 kg car accelerates from rest to a velocity of 22 m/s in 15 seconds. The power of the engine during this acceleration is, (Answer in kw)

Answers

Answer:

The answer is 71.5 kW

Explanation:

We can use the formula for power:

Power = Force x Velocity

where Force is the net force acting on the car, and Velocity is the velocity of the car.

To find the net force, we can use Newton's second law of motion:

Force = Mass x Acceleration

where Mass is the mass of the car, and Acceleration is the acceleration of the car.

The acceleration of the car can be found using the formula:

Acceleration = (Final Velocity - Initial Velocity) / Time

Substituting the given values, we get:

Acceleration = (22 m/s - 0 m/s) / 15 s

Acceleration = 1.47 m/s^2

Substituting the given values into the formula for force, we get:

Force = 2,210 kg x 1.47 m/s^2

Force = 3,247.7 N

Finally, substituting the calculated values for force and velocity into the formula for power, we get:

Power = Force x Velocity

Power = 3,247.7 N x 22 m/s

Power = 71,450.6 W

Converting the power to kilowatts (kW), we get:

Power = 71,450.6 W / 1000

Power = 71.5 kW

Therefore, the power of the engine during the acceleration is 71.5 kW.

1. please show steps and procedure clearly
Ambulanti infolinia 1. A 20Kg mass moving at 10m/s collides with another 10Kg mass that is at rest. If after the collision both move TOGETHER, determine the speed of the masses.

Answers

Total momentum after collision is = 6.67 m/s.

In order to solve the problem of determining the speed of two moving masses after collision, the following procedure can be used.

Step 1: Calculate the momentum of the 20Kg mass before collision. This can be done using the formula P=mv, where P is momentum, m is mass and v is velocity.

P = 20Kg * 10m/s

= 200 Kg m/s.

Step 2: Calculate the momentum of the 10Kg mass before collision. Since the 10Kg mass is at rest, its momentum is 0 Kg m/s.

Step 3: Calculate the total momentum before collision. This is the sum of the momentum of both masses before collision.

Total momentum = 200 Kg m/s + 0 Kg m/s

= 200 Kg m/s.

Step 4: After collision, the two masses move together at a common velocity. Let this velocity be v. Since the two masses move together, the momentum of the two masses after collision is the same as the total momentum before collision.

Therefore, we can write: Total momentum after collision

= 200 Kg m/s

= (20Kg + 10Kg) * v.

Substituting the values, we get: 200 Kg m/s = 30Kg * v.

So, v = 200 Kg m/s / 30Kg

= 6.67 m/s.

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