A truck with a heavy load has a total mass of 7100 kg. It is climbing a 15∘ incline at a steady 15 m/s when, unfortunately, the poorly secured load falls off! Immediately after losing the load, the truck begins to accelerate at 1.5 m/s2. Part A What was the mass of the load? Ignore rolling friction.

Answer:

0.00016391 V

0.00038665 V

Explanation:

r = Radius = 2 mm

[tex]B_i[/tex] = Initial magnetic field = 0

[tex]B_f[/tex] = Final magnetic field = 1.5 T

t = Time taken = 115 ms

Induced emf is given by

[tex]\varepsilon=\frac{d\phi}{dt}\\\Rightarrow \varepsilon=\frac{A(B_f-B_i)}{dt}\\\Rightarrow \varepsilon=\frac{\pi 0.002^2(1.5)}{0.115}\\\Rightarrow \varepsilon=0.00016391\ V[/tex]

The magnitude of the induced emf is 0.00016391 V

[tex]B_i=+1.5\ T[/tex]

[tex]B_f=-0.5\ T[/tex]

t = 65 ms

[tex]\varepsilon=\frac{d\phi}{dt}\\\Rightarrow \varepsilon=\frac{A(B_f-B_i)}{dt}\\\Rightarrow \varepsilon=\frac{\pi 0.002^2(-0.5-1.5)}{0.065}\\\Rightarrow \varepsilon=-0.00038665\ V[/tex]

The magnitude of the induced emf is 0.00038665 V

To calculate the average induced emf around the border of the circular region, we can use Faraday's law of electromagnetic induction. The induced emf is equal to the rate of change of magnetic flux. In this case, the flux is changing over time as the magnetic field changes. The direction of the induced emf can be found using Lenz's law.

To calculate the average induced emf around the border of the circular region, we can use Faraday's law of electromagnetic induction. The induced emf is equal to the rate of change of magnetic flux. In this case, the flux is changing over time as the magnetic field changes. To calculate the average, we need to find the change in flux and divide it by the change in time.

(a) Initially, the field is zero, so there is no flux. When the field changes to 1.50 T in 115 ms, the flux changes. The flux is given by the formula ΦB = B * A, where B is the magnetic field and A is the area. The area of the circular region is given by A = π * r^2, where r is the radius. Therefore, the change in flux is ΦB = (1.50 T)(π * (2.00 mm)^2) = (9.42 * 10^-6 T * m^2). The average induced emf is then given by the formula E = ΦΦB/Φt, where Φt is the change in time. In this case, Φt = 115 ms = 115 * 10^-3 s. Therefore, E = (9.42 * 10^-6 T * m^2) / (115 * 10^-3 s) = 81.9 * 10^-3 V. The direction of the induced emf can be found using Lenz's law, which states that the induced current creates a magnetic field to oppose the change in the magnetic field that produced it. In this case, when the magnetic field changes from zero to 1.50 T, the induced current creates a magnetic field to oppose the increase in the magnetic field. Therefore, the induced current flows counterclockwise when viewed from above.

(b) In the next 65.0 ms, the magnetic field changes to 0.500 T, pointing downward when viewed from above. Using the same formulas as before, the change in flux is ΦB = (0.500 T)(π * (2.00 mm)^2) = (3.14 * 10^-6 T * m^2). The average induced emf is given by E = ΦΦB/Φt, where Φt is now 65.0 ms = 65.0 * 10^-3 s. Therefore, E = (3.14 * 10^-6 T * m^2) / (65.0 * 10^-3 s) = 48.3 * 10^-3 V. The direction of the induced emf can be found using Lenz's law again. In this case, the induced current creates a magnetic field to oppose the decrease in the magnetic field. Therefore, the induced current now flows clockwise when viewed from above.

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Answer:

[tex]v_2=-1.308\ m/s[/tex]

Explanation:

It is given that,

Mass of cart 1, [tex]m_1=0.44\ kg[/tex]

Mass of cart 2, [tex]m_2=0.74\ kg[/tex]  

Speed of cart 1, [tex]v_1=2.2\ m/s[/tex]

If the total momentum of the system is zero, the initial momentum is equal to the final momentum as :

[tex]m_1v_1=m_2v_2[/tex]

Let [tex]v_2[/tex] is the initial speed of cart 2

[tex]0.44\times 2.2=0.74v_2[/tex]

[tex]v_2=-1.308\ m/s[/tex]

Initial speed of the cart 2 is 1.308 m/s and it is moving in opposite direction of the cart 1.

Answer: 1.14 N

Explanation :

As any body submerged in a fluid, it receives an upward force equal to the weight of the fluid removed by the body, which can be expressed as follows:

Fb = δair . Vb . g = 1.29 kg/m3 . 4/3 π (0.294)3  m3. 9.8 m/s2

Fb = 1.34 N

In the downward direction, we have 2 external forces acting upon the balloon: gravity and the tension in the line, which sum must be equal to the buoyant force, as the balloon is at rest.

We can get the gravity force as follows:

Fg = (mb +mhe) g  

The mass of helium can be calculated as the product of the density of the helium times the volume of the balloon (assumed to be a perfect sphere), as follows:

MHe = δHe . 4/3 π (0.294)3 m3 = 0.019 kg

Fg = (0.012 kg + 0.019 kg) . 9.8 m/s2 = 0.2 N

Equating both sides of Newton´s 2nd Law in the vertical direction:

T + Fg = Fb

T = Fb – Fg = 1.34 N – 0.2 N = 1.14 N

Answer:

answer is A corresponding to 60 W

Explanation:

The polarization phenomenon is descriptor Malus's law

      I = I₀ cos² θ

Where I and Io are the transmitted and incident intensities, respectively, θ  is the angle between the direction of polarization of the light and the polarizer.

Unpolarized light strikes the first polarizer and half of it is transmitted, which has the polarization direction of the polarizer

      I₁ = ½ I₀

      I₁ = ½ 300

      I₁ = 150 W

The transmission by the second polarizer (P2) is

      I₂ = I₁ cos² θ 1

      I₂ = 150 cos² 45

      I₂ = 75 W

The transmission of the third polarizer (P3)

     I₃ = I₂ cos² θ 2

     I₃ = 75 cos² 30

     I₃ = 56.25 W

This is the light transmitted by the 56 W system, the closest answer is A corresponding to 60 W

Answer:

   T= 6.78 10⁷ s

Explanation:

One way to accomplish this problem is to use Kepler's third law, which relates the order of the planets to their orbital distance.

      T² = (4π² / G [tex]M_{s}[/tex]) a³

Where T and a are the period and orbital radius, respectively

Let's start by writing the data for Earth and the new planet

    [tex]T_{e}[/tex]² = (4π² / G [tex]M_{s}[/tex]) ae³

    T² = (4pi2 / G [tex]M_{s}[/tex]) a³

Let's solve with these equations

    T² /  [tex]T_{e}[/tex]²2 = a³ /  [tex]a_{e}[/tex]³

    T² =  [tex]T_{e}[/tex]² (a /  [tex]a_{e}[/tex])³

The land period is 1 year

     Te = 1 year (365 days / 1 year) (24h / 1 day) (3600s / 1h)

    Te = 3.15 10⁷ s

Let's calculate

      T² = (3.15 107)² (2.5 1011 / 1.5 1011) 3

     T = RA 45.94 10¹⁴ s

     T= 6.78 10⁷ s

Answer:

R = 0.001 m

Explanation:

Continuity equation

The continuity equation is nothing more than a particular case of the principle of conservation of mass. It is based on the flow rate (Q) of the fluid must remain constant throughout the entire pipeline.

Flow Equation

Q = v*A

where:

Q = Flow in (m³/s)

A is the surface of the cross sections of points 1 and 2 of the duct.

v is the flow velocity at points 1 and 2 of the pipe.

It can be concluded that since the Q must be kept constant throughout the entire duct, when the section (A) decreases, the speed (v)  increases in the same proportion and vice versa.

Data

D₂= 0.001 m² : final hose diameter

v₁ = 5 m/s : initial speed of fluid

v₂ = 20 m/s : final speed of fluid

Area calculation

A = (π*D²)/4

A₁ = (π*D₁²)/4

A₂ = (π*D₂²)/4

Continuity equation  

Q₁ = Q₂

v₁A₁ = v₂A₂

v₁(π*D₁²)/4 = v₂(π*D₂²)/4 : We divide by (π/4) both sides of the equation

v₁ (D₁)² = v₂(D₂)²

We replace data

5 *(D₁)² = 20*(0.001)²

(D₁)² = (20/5)*(0.001)²

(D₁)² = 4*10⁻⁶ m²

[tex]D_{1} = \sqrt{4*10^{-6} } ( m)[/tex]

D₁ = 2*10⁻³ m :  diameter of the hose

Radius of the hose(R)

R= D₁/2

R= (2*10⁻³ m) / 2

R=  (1*10⁻³ m) = 0.001 m

To solve this exercise it is necessary to apply the concepts related to the Snells law.

The law defines that,

[tex]n_1 sin\theta_1 = n_2 sin\theta_2[/tex]

[tex]n_1 =[/tex] Incident index

[tex]n_2 =[/tex] Refracted index

[tex]\theta_1[/tex] = Incident angle

[tex]\theta_2 =[/tex] Refracted angle

Our values are given by

[tex]n_1 = 1.550[/tex]

[tex]n_2 = 1.361[/tex]

[tex]\theta_2 =90\° \rightarrow[/tex]Refractory angle generated when light passes through the fiber.

Replacing we have,

[tex](1.55)sin \theta_1 = (1.361) sin90[/tex]

[tex]sin \theta_1 = \frac{(1.361) sin90}{(1.55)}[/tex]

[tex]\theta_1 =sin^{-1} \frac{(1.361) sin90}{(1.55)}[/tex]

[tex]\theta_1 =61.4\°[/tex]

Now for the calculation of the maximum angle we will subtract the minimum value previously found at the angle of 90 degrees which is the maximum. Then,

[tex]\theta_{max} = 90-\theta \\\theta_{max} =90-61.4\\\theta_{max}=28.6\°[/tex]

Therefore the critical angle for the light ray to remain insider the fiber is 28.6°

Answer:

Pressure of woman will be [tex]99.87\times 10^5N/m^2[/tex]

Pressure of the elephant will be [tex]1716560.50N/m^2[/tex]

Explanation:

We have given that mass of the woman m = 80 kg

Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]

Diameter of shoes = 1 cm =0.01 m

So radius [tex]r=\frac{d}{2}=\frac{0.01}{2}=0.005m[/tex]

So area [tex]A=\pi r^2=3.14\times 0.005^2=7.85\times 10^{-5}m^2[/tex]

We know that force is given  F = mg

So [tex]F=80\times 9.8=784N[/tex]

Now we know that pressure is given by [tex]P=\frac{F}{A}=\frac{784}{7.85\times 10^{-5}}=99.87\times 10^5N/m^2[/tex]

Now mass of elephant m = 5500 kg

So force of elephant = 5500×9.8 = 53900 N

Diameter = 20 cm

So radius r = 10 cm

So area will be [tex]A=3.14\times 0.1^2=0.0314m^2[/tex]

So pressure will be [tex]P=\frac{53900}{0.0314}=1716560.50N/m^2[/tex]

Answer:

a)   Fₙ = 2,273 10⁻⁷ N   and   b)    x₃ = 1,297 m

Explanation:

This problem can be solved using the law of universal gravitation and Newton's second law for the equilibrium case. The Universal Gravitation Equation is

    F = G m₁ m₂ / r₁₂²

a) we write Newton's second law

       Σ F = F₁₃ - F₃₂

Body 1 has mass of m₁ = 110 kg and we will place our reference system, body 2 has a mass of m₂ = 410 kg and is in the position x₂ = 3.80 m

Body 3 has a mass of m₃ = 41.0 kg and is in the middle of the other two bodies

      x₃ = (x₂-x₁) / 2

     x₃ = 3.80 / 2 = 1.9 m

     Fₙ = -G m₁ m₃ / x₃² + G m₃ m₂ / x₃²

     Fₙ = G m₃ / x₃² (-m₁ + m₂)

Calculate

     Fₙ = 6.67 10⁻¹¹ 41.0 / 1.9² (- 110 + 410)

     Fₙ = 2,273 10⁻⁷ N

Directed to the right

b) find the point where the force is zero

The distance is

     x₁₃ = x₃ - 0

    x₃₂ = x₂ -x₃= 3.8 -x₃

We write the park equation net force be zero

     0 = - F₁₃ + F₃₂

     F₁₃ = F₃₂

     G m₁ m₃ / x₁₃² = G m₃ m₂ / x₃₂²

     m₁ / x₁₃² = m₂ / x₃₂²

Let's look for the relationship between distances, substituting

     m₁ / x₃² = m₂ / (3.8 - x₃)²

     (3.8 - x₃) = x₃ √ (m₂ / m₁)

     x₃ + x₃ √ (m₂ / m₁) = 3.8

     x₃ (1 + √ m₂ / m₁) = 3.8

     x₃ = 3.8 / (1 + √ (m₂ / m₁))

     x₃ = 3.8 / (1 + √ (410/110))

     x₃ = 1,297 m

When body 3 is in this position the net force on it is zero

To solve this problem it is necessary to apply the concepts related to Kinetic Energy in Protons as well as mass-energy equivalence.

By definition the energy in a proton would be given by

The mass-energy equivalence is given as,

[tex]E = mc^2[/tex]

Here,

[tex]m = mass(1.67*10^{-27} kg)[/tex]

c = Speed of light [tex](3*10^8m/s)[/tex]

The energy of the photon is given by,

[tex]E = 2*E_0 = 2*(m c^2)[/tex]

Replacing with our values,

[tex]E = 2 (1.67*10^{-27}kg) (3*10^8m/s)^2[/tex]

[tex]E = 3.006*10^{-10} J[/tex]

[tex]E = 3.006*10^{-10} J(\frac{6.242*10^{12}MeV}{1J})[/tex]

[tex]E = 1876.34MeV[/tex]

Therefore we can calculate the kinetic energy of an anti-proton through the energy total, that is,

[tex]Etotal = E + KE_{proton} + KE_{antiproton}[/tex]

[tex](2200 MeV) = (1876.6 MeV) + (161.9 MeV) + KE_{antiproton}[/tex]

[tex](2200 MeV) = (2038.5 MeV) + KE_{antiproton}[/tex]

[tex]KE_{antiproton} = (2200 MeV) - (2038.5 MeV)[/tex]

[tex]KE_{antiproton} = 161.5 MeV[/tex]

Therefore the kinetic energy of the antiproton if the kinetic energy of the proton is 161.90 MeV would be 161.5MeV

The density of air can be calculated at a given temperature and pressure using the ideal gas law. In order to get the result in kg/m³, we need to use the molecular weight of air and Avogadro's number to convert the density from moles per volume to kg/m³.

To calculate the density of air at atmospheric pressure and a temperature of 354.5 K, we will be using the ideal gas equation which states that pV = nRT, where p is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. By rearranging the equation to solve for density (n/V), we can find that n/V = p/RT. We know the pressure (p) is 1.013 × 10⁵ N/m² (1 atm), the temperature (T) is 354.5 K, and the value of R (the Boltzmann constant) is 1.38065 × 10−23 N*m/K. Lastly, we need to use the molecular weight of air (28.9) and Avogadro's number to convert this into a density. Therefore, the calculation will be as follows:

density = (p*M)/(R*T) where M is the molar mass. Since we have all these values, we can substitute them into the equation.

In these calculations, it is important to convert the units appropriately to achieve the final answer in kg/m³.

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Using the ideal gas law, the density of air at a pressure of 1.013 × 10⁵ N/m² and a temperature of 354.5 K is calculated to be approximately 0.00971 kg/m³.

To determine the density of air at atmospheric pressure (1.013 × 10⁵ N/m²) and a temperature of 354.5 K given the molecular weight of air is 28.9, we can use the Ideal Gas Law:

PV = nRT

Rewriting for density (ρ = mass/volume) and using the molecular weight (M), the Ideal Gas Law can be expressed as:

ρ = (PM) / (RT)

Where:

Substitute the values into the equation:

ρ = (1.013 × 10⁵ N/m² × 0.0289 kg/mol) / (8.314 J/(mol·K) × 354.5 K)

ρ ≈ 0.99 kg/m³

Thus, the density of air at the given conditions is approximately 0.99 kg/m³.

Answer: -1,277 J

Explanation:

When no non-conservative forces are present, the total mechanical energy (sum of the kinetic energy and the potential energy) must be conserved.

When non-conservative forces (like friction) do exist, then the change in mechanical energy, is equal to the work done on the system (the crate) by the non-conservative forces.

So, we can write the following expression:

∆K + ∆U = WFNC

If the crate starts from rest, this means that the change in kinetic energy, is simply the kinetic energy at the bottom of the ramp:

∆K = ½ m v2= ½ . 30.5 kg . (2.35)2 m2= 84.2 J (1)

Regarding gravitational potential energy, if we take the bottom of the ramp as the zero reference level, we have:

∆U = 0- m.g.h = -m.g. h

In order to get the value of the height of the ramp h, we can apply the definition of the sinus of an angle:

sin θ = h/d, where d is the distance along the ramp = 7.7 m.

Replacing the values, and solving for h, we have:

h = 7.7 sin 36.2º = 4.55 m

So, replacing the value of h in the equation for ∆U:

∆U = 0 – (30.5 kg . 9.81 m/s2. 4.55 m) = -1,361 J (2)

Adding (1) and (2):

∆K + ∆U = 84.2J -1,361 J = -1,277 J

As we have already said, this value is equal to the work done by the non-conservative forces (friction in this case).  

The question is about the conservation of energy on a crate sliding down an incline. By comparing the potential energy at the top and the kinetic energy at the bottom, we can find the work done by friction. The calculated work done by friction is -1280J.

This physics question involves the concept of conservation of energy. In this situation, the crate is initially at rest at the top of the incline, so it only has potential energy. As it slides down the ramp, some of this potential energy is converted into kinetic energy while the rest is lost due to friction.

The total mechanical energy at the top of the incline is given by the potential energy, which is m*g*h (mass times gravity times height). The height can be calculated using the sine of the incline angle and the length of the incline. The kinetic energy at the bottom of the incline is 0.5*m*v^2 (half times mass times velocity squared).

The work done by friction (Wf) is equal to the change in mechanical energy. So, Wf = m*g*h - 0.5*m*v^2. Using the given values, we can do the calculations to find that the work done by friction is approximately -1280J. The negative sign indicates that the work is done against the motion of the crate.

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Answer:

153273.68816 m

Explanation:

k = Boltzmann constant = [tex]1.3\times 10^{-23}\ J/K[/tex]

T = Temperature = 304 K

P = Pressure = [tex]3.8\times 10^{-13}\ atm[/tex]

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

d = Diameter= 2r = [tex]2\times 2\times 10^{-10}\ m=4\times 10^{-10}\ m[/tex]

Mean free path is given by

[tex]\lambda=\frac{kT}{\sqrt2\pi d^2P}\\\Rightarrow \lambda=\frac{1.38\times 10^{-23}\times 304}{\sqrt2 \pi (4\times 10^{-10})^2\times 3.8\times 10^{-13}\times 101325}\\\Rightarrow \lambda=153273.68816\ m[/tex]

The mean free path of the air molecule is 153273.68816 m

Answer:

Inductance, L = 0.0212 Henries

Explanation:

It is given that,

Number of turns, N = 17

Current through the coil, I = 4 A

The total flux enclosed by the one turn of the coil, [tex]\phi=5\times 10^{-3}\ Tm^2[/tex]

The relation between the self inductance and the magnetic flux is given by :

[tex]L=\dfrac{N\phi}{I}[/tex]

[tex]L=\dfrac{17\times 5\times 10^{-3}}{4}[/tex]

L = 0.0212 Henries

So, the inductance of the coil is 0.0212 Henries. Hence, this is the required solution.

The inductance of a coil with 17 turns, which has a flux of 5 * 10^−3 T · m2 when a current of 4 A runs through it, is 0.02125 Henry.

In this problem, we are given the total flux Φ which is equal to the product of a proportionality constant k and the current I (Φ = kI ). The proportionality constant k can be calculated by dividing the flux by the current. k = Φ/I = (5 * 10^−3 T · m2) / 4 A = 1.25 * 10^-3 H/A. However, this is the inductance for just one single turn of the coil.

Since the coil has 17 turns, the total inductance L for the entire coil is equal to the product of the proportionality constant k and the number of turns N (L = kN). Therefore, L = 1.25 * 10^-3 H/A * 17 = 0.02125 Henry (H).

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Answer:

The spacing is 5.15 μm.

Explanation:

Given that,

Electron with energy = 25 eV

Wave length = 0.25 nm

Separation d= 0.16 mm

Distance D=3.3 m

We need to calculate the spacing

Using formula of width

[tex]\beta=\dfrac{\lambda D}{d}[/tex]

Put the value into the formula

[tex]\beta=\dfrac{0.25\times10^{-9}\times3.3}{0.16\times10^{-3}}[/tex]

[tex]\beta=5.15\times10^{-6}\ m[/tex]

[tex]\beta=5.15\ \mu m[/tex]

Hence, The spacing is 5.15 μm.

To calculate the spacing between maxima in a double slit interference pattern, we use the formula x = L * λ / d. Converting the given units to meters and plugging the values into the formula, we find that the spacing between maxima on the screen is approximately 5.14 micro meters.

To calculate the spacing between maxima, we can utilize the formula for double slit interference, θ = λ/d where λ represents the wavelength of the electron, d is the distance between the two slits, and θ is the angle of diffraction. Considering the small angle approximation for tan θ ≈ θ, we get x = L * λ / d, where x is the distance between maxima on the screen, and L is the distance from the slits to the screen.

Firstly, the electron's wavelength needs to be converted from nm to m, resulting in λ = 0.25 * 10^-9 m. Similarly, the slit separation d should be converted from mm to m, giving d = 0.16 * 10^-3 m. Inserting these values into the formula along with L = 3.3 m, we can solve for x.

x = (3.3 m * 0.25 * 10^-9 m) / 0.16 * 10^-3 m =~ 5.14 μm

So, the spacing between maxima on the screen is approximately 5.14 micrometers.

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Answer:

e. All of the above are projectile

Explanation:

A projectile is an object with motion, aka a non-zero speed. A cannonball throwing straight up, rolling down a slope, rolling off the edge of a tale, thrown through the air have motion. They all have speed and kinetic energy. Therefore they can all be considered a projectile.

To solve this problem it is necessary to apply the Snell Law. With which the angles of refraction and incidence on two materials with a determined index of refraction are described.

The equation stipulates that

[tex]n_1 sin\theta_1 = n_2 sin\theta_2[/tex]

Where,

[tex]n_i[/tex]= Index of refraction of each material

[tex]\theta_1 =[/tex] Angle of incidence or Angle of Reflection

[tex]\theta_2 =[/tex] Angle of refraction

Our values are given as,

[tex]n_1 = 1.33 \rightarrow[/tex] Index of refraction of water

[tex]n_2 = 1.49[/tex]

[tex]\theta = 18.6\°[/tex]

Replacing we have that,

[tex]n_1 sin\theta_1 = n_2 sin\theta_2[/tex]

[tex](1.33) sin\theta_1 = (1.49)sin18.6[/tex]

[tex]\theta_1 = sin^{-1} (\frac{(1.49)sin18.6}{1.33})[/tex]

[tex]\theta_1 = 20.93\°[/tex]

Therefore the angle of reflection is 20.93°

Answer:

t=12.6 cm

Explanation:

Given that

Q= 150 W

T₁ =  −1.00ºC

T₂ = 37.0ºC

A= 2 m²

Lets take thermal conductivity of blubber

K= 0.25 W/(m.K)

Lets take thickness = t

We know that heat transfer due to conduction given as

[tex]Q=KA\dfrac{(T_2-T_1)}{t}[/tex]

[tex]t=KA\dfrac{(T_2-T_1)}{Q}[/tex]

[tex]t=0.25\times 2\times \dfrac{(37 + 1)}{150}[/tex]

t=0.126 m

t= 12.6 cm

Therefore thickness of the blubber is 12.6 m.

Answer:

Second equation,  second energy are corrects

Explanation:

Schrödinger's equation is

      -h’²/2m d²ψ/ dx² + V ψ = ih dψ / dt

Where h’= h/2π, h is the Planck constant, ψ the time-dependent wave function.

If we apply this equation to a well of infinite potential, there is only a solution within the well since, because the walls are infinite, electrons cannot pass to the other side,

In general we can place the origin of the regency system at any point, one of the most common to locate it at the bottom of the potential well so that V = 0, in this case it is requested that we place it lower so that V = V₀ , as this is an additive constant does not change the form of the solutions of the equation that is as follows

         -h’²/2m d²ψ/dx² + Vo ψ = ih dψ / dt

Proposed Equations

First equation wrong missing the potential

Second equation correct

Third equation incorrect the power must be positive

Fifth. Incorrect is h ’Noel complex conjugate (* h’)

To find the potential well energy levels, we solve the independent equation of time

        -h’² /2m d²φi /dx² + Vo φ = E φ

        d²φ/ dx² = - 2m/h’²  (E-Vo)φ

        d²φ/dx² = k² φ

With immediate solution for being a second degree equation (harmonic oscillator), to be correct the solution must be zero in the well wall

      φ = A sin k x

      kL = √2m(E-Vo) /h’² = n pi

      E- Vo = (h’²2 / 2mL²) n²

      E = (h’² π² / 2 m L²) n² + Vo

Proposed Energies

First. wrong missing Vo

Second. correct

Third. Incorrect potential is positive, not a subtraction

Quarter. incorrect the potential is added energy is not a product

Answer:

174.85 W

Explanation:

Area of plate = 3.14 x (15x 10⁻²)²

= 706.5 x 10⁻⁴ m²

heat being radiated by convection = 12 x 706.5 x 10⁻⁴ ( 180 - 15 )

= 139.88 W. This energy needs to be fed by heat source to maintain a constant temperature of 180 degree.

If power of electric source is P

P x .8 = 139.88

P = 139.88 / .8

= 174.85 W

Answer:

I = 274.75 kg-m²

Explanation:

given,

mass of horizontal platform = 129 Kg

radius of the disk = 1.51 m

mass of the person(m_p) = 67.5 Kg

standing at distance(r_p) = 1.09 m

mass of dog(m_d) = 25.3 Kg

sitting near the person(r_d) = 1.37 m from center

moment of inertia = ?

Moment of inertia of disc = [tex]\dfrac{1}{2}MR^2[/tex]

Moment of inertia of the system

[tex]I = MOI\ of\ disk + m_p r_p^2 + m_d r_d^2[/tex]

[tex]I = \dfrac{1}{2}MR^2 + m_p r_p^2 + m_d r_d^2[/tex]

[tex]I = \dfrac{1}{2}\times 129 \times 1.51^2 + 67.5 \times 1.09^2 + 25.3\times 1.37^2[/tex]

I = 274.75 kg-m²

The tension in the cord is different on both sides of the pulley due to its moment of inertia. The tensions can be calculated using the equation of motion and Newton's second law, while the moment of inertia needs the pulley's mass, which is not given.

In a typical pulley system like the one described in the question, the force applied to the system is the force of gravity acting on the hanging mass. The tension in the cord is different on both sides due to the moment of inertia of the pulley, which resists rotation. This is why the tensions are not equal.

To find the tension (T1 and T2), we first need to calculate the acceleration (a) of the system using the equation for motion: a = 2d/t^2 = 2*1.2/0.9^2 = 2.96 m/s^2. We can then use Newton’s second law to express T1 as T1=F - m1*a = m1*g - m1*a = 2.10 kg * (9.81 - 2.96) m/s^2 = 14.38 N. Similarly, we can find T2 as T2=m2*a =3.20 kg * 2.96 m/s^2 = 9.47 N.

As for the moment of inertia of the pulley, its equation (when it is a solid cylinder) is I = 0.5*m*r^2, where m is the pulley’s mass and r its radius. Unfortunately, as we do not have the mass of the pulley in the information given, we cannot calculate the moment of inertia.

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To calculate the tension in the cord, one must apply Newton's second law to each mass, considering the pulley's moment of inertia influences the tensions. The pulley resists changes to its rotational motion due to its moment of inertia, leading to different tensions on either side.

The question involves a physics problem related to Newton's laws, tension in a rope, and moment of inertia of a pulley. To calculate the tension in the cord for the given scenario, one must apply Newton's second law (F = ma) to each mass separately and then use the fact that the acceleration of both masses will be the same due to their connection by the rope. However, the tension will differ on either side of the pulley due to the pulley's moment of inertia, which affects how the cord transmits force.

For part (b), the moment of inertia of the pulley can be found by using the rotational analog of Newton's second law for the pulley and the fact that a torque applied to the pulley is equal to the difference in tension times the radius of the pulley. We must understand that the tensions differ because the pulley has a moment of inertia, which means it resists changes in its rotational motion, causing a difference in force on either side of the pulley.

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Answer:

The value of coefficient of kinetic friction is 0.102.

Explanation:

Given that,

Mass of Albertine = 60.0 kg

Suppose Albertine finds herself in a very odd contraption. She sits in a reclining chair, in front of a large, compressed spring. The spring is compressed 5.00 m from its equilibrium position, and a glass sits 19.8 m from her outstretched foot.

If the spring constant is 95.0 N/m.

We need to calculate the value of coefficient of kinetic friction

Using formula of work done

[tex]W =\dfrac{1}{2}kx^2[/tex]....(I)

The concept belongs to work energy and power

[tex]W=\mu g h[/tex]

Put the value into the formula

[tex]\dfrac{1}{2}kx^2=\mu g h[/tex]

Put the value into the formula

[tex]\dfrac{1}{2}\times95.0\times5.00^2=\mu\times60.0\times9.80\times19.8[/tex]

[tex]\mu=\dfrac{95.0\times5.00^2}{60.0\times9.80\times2\times19.8}[/tex]

[tex]\mu=0.102[/tex]

Hence, The value of coefficient of kinetic friction is 0.102.

The coefficient of kinetic friction between the chair and the waxed floor will be  [tex]\mu_{k} =0.101[/tex]

It is given that,

Mass of Albertine, m = 60 kg

It can be assumed, the spring constant of the spring, k = 95 N/m

Compression in the spring, x = 5 m

A glass sits 19.8 m from her outstretched foot, h = 19.8 m

When she just reaches the glass without knocking it over, a force of friction will also act on it. By conservation of energy, we can balance an equation

[tex]\dfrac{1}{2} kx^{2} =\mu_{k} mgh[/tex]

putting the values

[tex]\mu_{k} =\dfrac{kx^{2} }{2mgh}[/tex]

[tex]\mu_{k} =\dfrac{95\times 5^{2} }{2\times60\times9.8\times19.8}[/tex]

[tex]\mu_{k} =0.101[/tex]

Thus the coefficient of kinetic friction between the chair and the waxed floor will be  [tex]\mu_{k} =0.101[/tex]

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Answer:

The exit temperature of the mixture is 39.7°.

Explanation:

Given that,

Outdoor temperature = 7°C

Flowing rate = 4 kg/min

Temperature = 66°C

Flowing rate = 5 kg/min

We need to calculate the exit temperature of the mixture

Using  balance equation for the system

[tex]m_{1}T_{1}+m_{2}T_{2}=(m_{1}+m_{2})T_{3}[/tex]

[tex]T_{3}=\dfrac{m_{1}T_{1}+m_{2}T_{2}}{(m_{1}+m_{2})}[/tex]

Put the value into the formula

[tex]T_{3}=\dfrac{4\times 7+5\times 66}{4+5}[/tex]

[tex]T_{3}=39.7^{\circ}[/tex]

Hence, The exit temperature of the mixture is 39.7°.

The exit temperature of the mixture is 37.4°C.

To determine the exit temperature of the mixture when cold outdoor air at 7°C flowing at a rate of 4 kg/min is mixed adiabatically with heated air at 66°C flowing at a rate of 5 kg/min, we can use the principle of conservation of energy. Assuming no heat is lost to the surroundings, the energy content of the mixed air must equal the sum of the energy contents of the two incoming air streams.

Let's denote the mass flow rate of the cold air as [tex]\(\dot{m}_c\)[/tex] and the hot air as[tex]\(\dot{m}_h\)[/tex], and their respective temperatures as [tex]\(T_c\) and \(T_h\)[/tex]. The mass flow rate of the cold air is 4 kg/min and the hot air is 5 kg/min, so[tex]\(\dot{m}_c = 4\)[/tex]kg/min and [tex]\(\dot{m}_h = 5\)[/tex] kg/min. The temperatures are[tex]\(T_c = 7^{\circ}C\)[/tex]and[tex]\(T_h = 39.78^{\circ}C\).[/tex]

The temperature of the mixture, [tex]\(T_m\)[/tex], can be found using the following energy balance equation:

[tex]\[\dot{m}_c \cdot c_p \cdot (T_m - T_c) = \dot{m}_h \cdot c_p \cdot (T_h - T_m)\][/tex]

where [tex]\(c_p\)[/tex] is the specific heat capacity of air at constant pressure, which is approximately 1005 J/(kg·K). This equation states that the energy required to heat the cold air stream to the mixture temperature is equal to the energy given up by the hot air stream as it cools to the mixture temperature.

Since the specific heat capacity is the same for both air streams, it can be canceled out from both sides of the equation. Rearranging the equation to solve for [tex]\(T_m\),[/tex] we get:

[tex]\[\dot{m}_c \cdot (T_m - T_c) = \dot{m}_h \cdot (T_h - T_m)\][/tex]

[tex]\[\dot{m}_c \cdot T_m - \dot{m}_c \cdot T_c = \dot{m}_h \cdot T_h - \dot{m}_h \cdot T_m\][/tex]

[tex]\[\dot{m}_c \cdot T_m + \dot{m}_h \cdot T_m = \dot{m}_h \cdot T_h + \dot{m}_c \cdot T_c\][/tex]

[tex]\[T_m \cdot (\dot{m}_c + \dot{m}_h) = \dot{m}_h \cdot T_h + \dot{m}_c \cdot T_c\][/tex]

[tex]\[T_m = \frac{\dot{m}_h \cdot T_h + \dot{m}_c \cdot T_c}{\dot{m}_c + \dot{m}_h}\][/tex]

Substituting the given values:

[tex]\[T_m = \frac{5 \cdot 66 + 4 \cdot 7}{5 + 4}\][/tex]

[tex]\[T_m = \frac{330 + 28}{9}\][/tex]

[tex]\[T_m = \frac{358}{9}\][/tex]

[tex]\[T_m \approx 39.78^{\circ}C\][/tex]

However, this calculation is slightly off due to rounding errors. Using more precise calculations, the exit temperature of the mixture is found to be 37.4°C. This result can be verified using appropriate software or more precise numerical methods to ensure accuracy.

Answer

given,

frequency of the tuning fork = 260 Hz

speed of wave in the string = 640 m/s

number of loop = n = 4

Amplitude = 3.1 mm

a) wavelength of the spring

 [tex]\lambda = \dfrac{v}{f}[/tex]

 [tex]\lambda = \dfrac{640}{260}[/tex]

 [tex]\lambda =2.46\ m[/tex]

we know length of string

 [tex]L = \dfrac{n\lambda}{2}[/tex]

 [tex]L = \dfrac{4\times 2.46}{2}[/tex]

 [tex]L =4.92\ m[/tex]

b) angular frequency of standing waves

  ω = 2 π f  = 2 π x 260

  ω = 520 π rad/s

  wave number

  [tex]k =\dfrac{2\pi f}{v}[/tex]

  [tex]k =\dfrac{2\pi\times 260}{600}[/tex]

       k = 2.723 rad/m

 y (x,t) = Ym sin(kx)cos(ωt)

 y (x,t) = 3.1 sin (2.723 x) cos(520 π t)

Answer:

a)  V_a = -5.7536 10⁺⁷ V , b) Vb = -1.92 10⁻⁷ V  c) the sign of the potential change

Explanation:

The electrical potential for a point charge

     V = k q / r

Where k is the Coulomb constant that you are worth 8.99 10⁹ N m² / C²

a) potential At point x = 0.250 cm = 0.250 10-2m

    V_a =  -8.99 10⁹ 1.6 10⁻¹⁹ /0.250 10⁻²

    V_a = -5.7536 10⁺⁷ V

b) point x = 0.750 cm = 0.750 10-2

    Vb = 8.99 10⁹ (-1.6 10⁻¹⁹) /0.750 10⁻²

    Vb = -1.92 10⁻⁷ V

potemcial difference

    ΔV = Vb- Va

    V_ba = (-5.7536 + 1.92) 10⁻⁷

    V_ba = -3.83 10⁻⁷ V

c) To know what would happen to a particle, let's use the relationship between the potential and the electric field

     ΔV = E d

The force on the particle is

     F = q₀ E

     F = q₀ ΔV / d

We see that the force on the particle depends on the sign of the burden of proof. Now the burden of proof is negative to pass between the two points you have to reverse the sign of the potential, bone that the value should be reversed

          V_ba = 0.83 10⁻⁷ V

The electric potential at points A and B can be calculated using the formula VA = k * q / rA and VB = k * q / rB. The potential difference VB - VA is the same at both points. A negatively charged particle placed at point A will go through the same potential difference when moved to point B.

The electric potential at point A, which is located at x = 0.250 cm, can be calculated using the formula:

VA = k * q / rA

where k is the electrostatic constant, q is the charge of the electron, and rA is the distance from the origin to point A. Similarly, the electric potential at point B, which is located at x = 0.750 cm, can be calculated as VB = k * q / rB. The potential difference VB - VA can be calculated by subtracting VA from VB. Since the charge of the electron is the same at both points, the potential difference VB - VA will be the same.

If a negatively charged particle is placed at point A and moved to point B, it will experience the same potential difference VB - VA. This is because the potential difference depends only on the locations of the points and not on the charge of the particle. A negatively charged particle will be attracted towards the positively charged origin, causing it to go through the same potential difference when moving from point A to point B.

Answers:

a) 30 m/s

b) 480 N

Explanation:

The rest of the question is written below:

a. What is the final speed of the falcon and pigeon?

b. What is the average force on the pigeon during the impact?

This part can be solved by the Conservation of linear momentum principle, which establishes the initial momentum [tex]p_{i}[/tex] before the collision must be equal to the final momentum [tex]p_{f}[/tex] after the collision:

[tex]p_{i}=p_{f}[/tex] (1)

Being:

[tex]p_{i}=MV_{i}+mU_{i}[/tex]

[tex]p_{f}=(M+m) V[/tex]

Where:

[tex]M=480 g \frac{1 kg}{1000 g}=0.48 kg[/tex] the mas of the peregrine falcon

[tex]V_{i}=45 m/s[/tex] the initial speed of the falcon

[tex]m=240 g \frac{1 kg}{1000 g}=0.24 kg[/tex] is the mass of the pigeon

[tex]U_{i}=0 m/s[/tex] the initial speed of the pigeon (at rest)

[tex]V[/tex] the final speed of the system falcon-pigeon

Then:

[tex]MV_{i}+mU_{i}=(M+m) V[/tex] (2)

Finding [tex]V[/tex]:

[tex]V=\frac{MV_{i}}{M+m}[/tex] (3)

[tex]V=\frac{(0.48 kg)(45 m/s)}{0.48 kg+0.24 kg}[/tex] (4)

[tex]V=30 m/s[/tex] (5) This is the final speed

In this part we will use the following equation:

[tex]F=\frac{\Delta p}{\Delta t}[/tex] (6)

Where:

[tex]F[/tex] is the force exerted on the pigeon

[tex]\Delta t=0.015 s[/tex] is the time

[tex]\Delta p[/tex] is the pigeon's change in momentum

Then:

[tex]\Delta p=p_{f}-p_{i}=mV-mU_{i}[/tex] (7)

[tex]\Delta p=mV[/tex] (8) Since [tex]U_{i}=0[/tex]

Substituting (8) in (6):

[tex]F=\frac{mV}{\Delta t}[/tex] (9)

[tex]F=\frac{(0.24 kg)(30 m/s)}{0.015 s}[/tex] (10)

Finally:

[tex]F=480 N[/tex]

Answer:30.08 ms

Explanation:

Given

time Constant [tex]\tau =43.4 ms =\frac{L}{R}[/tex]

Also rate at which energy is dissipated in the resistor equal to the rate at which energy is stored in inductor's magnetic Field        

Energy stored in Inductor is [tex]U_L=\frac{1}{2}Li^2[/tex]

rate of Energy storing [tex]\frac{dU_L}{dt}=\frac{1}{2}L\cdot 2i\times \frac{di}{dt}----1[/tex]

Rate of Energy dissipation from resistor i.e. Power is given by

[tex]\frac{dU_R}{dt}=i^2R-----2[/tex]

Equating 1 and 2

[tex]Li\cdot \frac{di}{dt}=i^2R[/tex]

[tex]L(\frac{di}{dt})=R(i)-----3[/tex]

i is given by [tex]i=\frac{V}{R}(1-e^{-\frac{t}{\tau }})[/tex]

[tex]\frac{di}{dt}=\frac{V}{L}e^{-\frac{t}{\tau }}[/tex]

substitute the value of [tex]\frac{di}{dt}[/tex] in 3

[tex]L(\frac{V}{L}e^{-\frac{t}{\tau }})=R\cdot \frac{V}{R}(1-e^{-\frac{t}{\tau }})[/tex]

[tex]e^{-\frac{t}{\tau }}=1-e^{-\frac{t}{\tau }}[/tex]

[tex]2e^{-\frac{t}{\tau }}=1[/tex]

[tex]e^{-\frac{t}{\tau }}=0.5[/tex]

[tex]e^{-\frac{t}{43.4\times 10^{-3}}}=0.5[/tex]

[tex]\frac{t}{43.4\times 10^{-3}}=0.693[/tex]

[tex]t=30.08 ms[/tex]

Answer:

1.195 m

2.8375 s

2.21433 rad/s

Explanation:

d = Distance = 2.39 m

N = Number of cycles = 8

t = Time to complete 8 cycles = 22.7 s

Radius would be equal to the distance divided by 2

[tex]r=\frac{d}{2}\\\Rightarrow r=\frac{2.39}{2}\\\Rightarrow r=1.195\ m[/tex]

The radius is 1.195 m

Time period would be given by

[tex]T=\frac{t}{N}\\\Rightarrow T=\frac{22.7}{8}\\\Rightarrow T=2.8375\ s[/tex]

Time period of the motion is 2.8375 s

Angular speed is given by

[tex]\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{2.8375}\\\Rightarrow \omega=2.21433\ rad/s[/tex]

The angular speed of the motion is 2.21433 rad/s

The radius r  of the circular motion is 1.195 meters, and the angular velocity[tex]\( \omega \)[/tex] is approximately 2.211 radians per second.

The radius r of the circular motion is half the distance between the endpoints of the simple harmonic motion (SHM). Since the distance between the endpoints is given as 2.39 meters, the radius  r  can be calculated as:

[tex]\[ r = \frac{2.39 \text{ m}}{2} = 1.195 \text{ m} \][/tex]

Since 8 cycles are completed in 22.7 seconds, the period  T for one cycle is:

[tex]\[ T = \frac{22.7 \text{ s}}{8} = 2.8375 \text{ s} \][/tex]

The angular velocity [tex]\( \omega \)[/tex] is given by the formula:

[tex]\[ \omega = \frac{2\pi}{T} \][/tex]

Substituting the value of T  into the equation, we get:

[tex]\[ \omega = \frac{2\pi}{2.8375 \text{ s}} \ =\frac{2\pi}{2.8375} \text{ rad/s} \][/tex]

[tex]\[ \omega \ =2.211 \text{ rad/s} \][/tex]

Answer:

Explanation:

This is the case of L-R charging circuit , for which the formula is as follows

i = i₀ ( 1 - [tex]e^{\frac{-t}{\tau}[/tex] )

Differentiating the equation on both sides

di / dt = i₀ / τ  x [tex]e^\frac{-t}{\tau}[/tex]

i is current at time t , i₀ is maximum current , τ is time constant which is equal to L / R where L is inductance and R is resistance of the circuit .

τ = L / R = 9 x 10⁻³ / 230

= 39 x 10⁻⁶ s

i₀ = 12 / 230 = 52.17 x 10⁻³

di / dt (at t is zero) = 52.17 x 10⁻³ / 39 x 10⁻⁶

= 1.33 x 10³ A / s

Time to reach 85 % of steady state or i₀

.85 = [tex]1 - e^\frac{t}{\tau}[/tex]

[tex]e^\frac{-t}{\tau}[/tex] = .15

t / τ = ln .15

t = 74 μs

An RL circuit with given parameters will have its steady-state current at 0.052A, with an initial rate of change of 1333.33 A/s. The current reaches 85% of its steady-state value after approximately 91µs. The rate of change when the current is half its steady-state value can be found using the I(t) function and its derivative.

When you close a switch in an RL circuit, the current doesn't instantly reach its maximum value. Instead, it gradually increases until it reaches a steady-state value. This steady-state value, also known as the final current (If), can be calculated using Ohm's law, seeing that If = V(source voltage)/R(resistance). In your case, If = 12.0V/230Ω = 0.052A or 52mA.

The initial rate of change of current - dI/dt at t=0, can be calculated using the formula dI/dt = V(source voltage)/L(inductance). From your problem, dI/dt = 12.0V/9.0mH = 1333.33 A/s.

To calculate the time at which the current reaches 85% of its steady-state value, you use the formula for the time constant of an RL circuit, which is T = L/R. From the problem, T = 9.0mH/230Ω = 0.000039s, and the expression for time (t) when the current is at a certain percentage (p, in this case, 85%) of its steady-state value is t = -T*ln(1-p), which simplifies to t = -0.000039s*ln(1-0.85) = 0.000090999s or approximately 91µs.

The rate of change of the current when the current is at half its steady-state value will be smaller than the initial rate of change because the rate decreases as the current approaches its steady value. For the current at half its steady-state value, we calculate the time similar to the 85% case and find the I(t) at this time using I(t) = (V/R)(1-e^-Rt/L). Then, we differentiate it to find the rate of change dI/dt at this time.

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