An ideal parallel-plate capacitor consists of a set of two parallel plates of area Separated by a very small distance 푑. This capacitor is connected to a battery that charges the capacitor such that the energy stored in the capacitor is 푈'. Now the battery is disconnected and the separation between the plates is doubled, how much energy is stored in the capacitor?

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:

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

Answers:

a) [tex]8820 m/s[/tex]

b) [tex]189500 Pa[/tex]

Explanation:

We have the following data:

[tex]t=30 min \frac{60 s}{1 min}=1800 s[/tex] is the time

[tex]h=11 m[/tex] is the height the water reaches vertically

[tex]g=9.8 m/s^{2}[/tex] is the acceleration due gravity

[tex]P_{air}=101.3 kPa=101.3(10)^{3} Pa[/tex] is the pressure of air

[tex]\rho_{water}=1000 kg/m^{3}[/tex] is the density of water

Knowing this, let's begin:

Here we will use the following equation:

[tex]h=h_{o}+V_{o}t-\frac{g}{2}t^{2}[/tex] (1)

Where:

[tex]h_{o}=0 m[/tex] is the initial height of water

[tex]V_{o}[/tex] is the initial speed of water

Isolating [tex]V_{o}[/tex]:

[tex]V_{o}=\frac{1}{t}(h+\frac{g}{2}t^{2})[/tex] (2)

[tex]V_{o}=\frac{1}{1800 s}(11 m+\frac{9.8 m/s^{2}}{2}(1800 s)^{2})[/tex]

[tex]V_{o}=8820.006 m/s \approx 8820 m/s[/tex] (3)

In this part we will use the following equation:

[tex]P=\rho_{water} g d + P_{air}[/tex] (4)

Where:

[tex]P[/tex] is the absolute pressure in the chamber

[tex]d=9 m[/tex] is the depth

[tex]P=(1000 kg/m^{3})(9.8 m/s^{2})(9 m) + 101.3(10)^{3} Pa[/tex]

[tex]P=189500 Pa[/tex] (5)

Answer:

A) τ = 300 N*m

B) θ  = 0.4 rad = 22.9°

Explanation:

Newton's second law:

F = ma has the equivalent for rotation:

τ = I * α   Formula  (1)

where:

τ : It is the moment applied to the body.  (Nxm)

I :  it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)

α : It is angular acceleration. (rad/s²)

Data

I = 0.6 kg*m² : moment of inertia of the bat

Angular acceleration of the bat

We apply the equations of circular motion uniformly accelerated :  

ωf= ω₀ + α*t Formula (2)

Where:  

α : Angular acceleration (rad/s²)  

ω₀ : Initial angular speed ( rad/s)  

ωf : Final angular speed ( rad

t :  time interval (s)

Data

ω₀ = 0

ωf =  20 rad/s

t = 40 ms = 0.04 s

We replace data in the formula (2) :

ωf= ω₀ + α*t

20 = 0 + α* (0.04)

α = 20/ (0.04)

α = 500 rad/s²

Newton's second law to the bat

τ = (0.6 kg*m²) *(500 rad/s²) = 300 (kg*m/s²)* m

τ = 300 N*m

B)  Angle through which the bat moved.

We apply the equations of circular motion uniformly accelerated :  

ωf²= ω₀ ²+ 2α*θ Formula (3)

Where:  

θ : Angle that the body has rotated in a given time interval (rad)

We replace data in the formula (3):  

ωf²= ω₀²+ 2α*θ

(20)²= (0)²+ 2(500 )*θ

400 = 1000*θ

θ  = 400/1000

θ  = 0.4 rad

π rad = 180°

θ  = 0.4 rad *(180°/π rad)

θ  = 22.9°

Answer:

116.1 kgm²/s

1.12718 rad/s

Decreases

Explanation:

m = Mass of girl = 43 kg

M = Mass of roundabout = 120 kg

v = Velocity of roundabout = 2.7 m/s

r = Radius of roundabout = 1 m = R

I = Moment of inertia

Her angular momentum

[tex]L_i=mvr\\\Rightarrow L_i=43\times 2.7\times 1\\\Rightarrow L_i=116.1\ kgm^2/s[/tex]

Magnitude of angular momentum is 116.1 kgm²/s

Here the angular momentum is conserved

[tex]L_f=L_i\\\Rightarrow I\omega=L_i\\\Rightarrow (\frac{1}{2}MR^2+mr^2)\omega=116.1\\\Rightarrow \omega=\frac{116.1}{\frac{1}{2}\times 120\times 1^2+43\times 1^2}\\\Rightarrow \omega=1.12718\ rad/s[/tex]

Angular speed of the roundabout is 1.12718 rad/s

Initial kinetic energy

[tex]K_i=\frac{1}{2}mv^2\\\Rightarrow K_i=\frac{1}{2}43\times 2.7^2\\\Rightarrow K_i=156.735\ J[/tex]

Final kinetic energy

[tex]K_f=\frac{1}{2}I\omega^2\\\Rightarrow K_f=\frac{1}{2}\times (\frac{1}{2}\times 120\times 1^2+43\times 1^2)\times 1.12718^2\\\Rightarrow K_f=65.43253\ J[/tex]

The overall kinetic energy decreases as can be seen. This loss is converted to heat.

To find the angular momentum of the child just before she jumps onto the roundabout, consider her linear momentum and the moment of inertia of the roundabout. The angular speed of the roundabout after the jump can be found using the principle of conservation of angular momentum. The overall kinetic energy of the system remains constant.

To find the angular momentum of the child just before she jumps onto the roundabout, we need to consider her linear momentum and the moment of inertia of the roundabout. The linear momentum of the child is given by the product of her mass and velocity. The angular momentum is then equal to the linear momentum multiplied by the distance from the center of the roundabout.

The angular speed of the roundabout after the jump can be found using the principle of conservation of angular momentum. The initial angular momentum of the system (just the roundabout) is zero, and since angular momentum is conserved, the final angular momentum after the child jumps on is equal to the angular momentum of the child just before she jumps.

The overall kinetic energy of the system remains constant. As the child jumps onto the roundabout, an external force (the ground pushing on the child) does work to change the linear momentum of the child, but no external torque acts on the system. So, the total mechanical energy (kinetic plus potential) is conserved.

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Answer:

W=972.83 J

Explanation:

Given,

angle of inclination of ramp = θ = 24.3°

Force exerted on the block = F= 289 N

Crate is raised to height of = 2.39 m

coefficient of friction = 0.33

where g is the acceleration due to gravity = g = 9.8 m/s²

Work done = ?

let m be the mass of the crate

Gravity force along ramp = m g sinθ

Friction force =  μm g cosθ

now writing all the force

F = m g sinθ  + μm g cosθ

By putting the values

F = m g (sinθ  + μcosθ)              

289 = 9.8 x m  ( sin 24.3°+ 0.33 cos 24.3°)                 ( take g =10 m/s²)

289 = 9.8 m(0.71)

m = 41.53 kg

Work done will be equal to

W= m g h

W= 41.53 x 9.8 x 2.39 J

W=972.83 J

To solve this problem, it is necessary to use the concepts related to the Force given in Newton's second law as well as the use of the kinematic equations of movement description. For this case I specifically use the acceleration as a function of speed and time.

Finally, we will describe the calculation of stress, as the Force produced on unit area.

By definition we know that the Force can be expressed as

F= ma

Where,

m= mass

a = Acceleration

The acceleration described as a function of speed is given by

[tex]a = \frac{\Delta v}{\Delta t}[/tex]

Where,

[tex]\Delta v =[/tex] Change in velocity

[tex]\Delta t =[/tex]Change in time

The expression to find the stress can be defined as

[tex]\sigma=\frac{F}{A}[/tex]

Where,

F = Force

A = Cross-sectional Area

Our values are given as

[tex]v= 80km/h\\t=5.8*10^3s\\m = 3kg \\A = 2.3*10^{-4}m^2[/tex]

Replacing at the values we have that the acceleration is

[tex]a = \frac{\Delta v}{\Delta t}[/tex]

[tex]a = \frac{80km/h(\frac{1h}{3600s})(\frac{1000m}{1km})}{5.8*10^3}[/tex]

[tex]a = 3831.41m/s^2[/tex]

Therefore the force expected is

[tex]F = ma\\F = 3*3831.41m/s^2 \\F = 11494.25N[/tex]

Finally the stress would be

[tex]\sigma = \frac{F}{A}[/tex]

[tex]\sigma = \frac{11494.25N}{2.3*10^{-4}}[/tex]

[tex]\sigma = 49.97*10^6 Pa = 49.97Mpa[/tex]

Therefore the compressional stress that the arm withstands during the crash is 49.97Mpa

Final answer:

The compressional stress is approximately 4.99787 × 10⁷ Pa

Explanation:

The question asks to find the compressional stress that the arm withstands during a crash, where the arm comes to a stop from an initial speed of 80 km/h in 5.8 ms, with an effective mass of 3.0 kg, and the total cross-sectional area of the load-bearing calcified portion of the forearm bones being approximately 2.3 cm².

First, convert the initial speed from km/h to m/s: 80 km/h = 22.22 m/s. To find the acceleration, use the formula a = ∆v / ∆t, where ∆v = -22.22 m/s (as it comes to rest) and ∆t = 5.8 ms or 0.0058 s. This gives an acceleration of approximately -3831.03 m/s².

The force experienced due to this acceleration can be calculated using Newton's second law, F = ma, with m = 3.0 kg and a = -3831.03 m/s², resulting in a force of approximately -11493.09 N. Finally, the compressional stress (σ) is found using the formula σ = F / A, where F = 11493.09 N and A = 2.3 cm² = 0.00023 m². This yields a compressional stress of approximately 4.99787 × 10⁷ Pa.

Answer:

449.37412 N

Explanation:

G = Gravitational constant = 6.67259 × 10⁻¹¹ m³/kgs²

m = Mass of satellite = 438 kg

M = Mass of planet

T = Time period of the satellite = 24 h

r = Radius of planet = [tex]1.94\times 10^8\ m[/tex]

The time period of the satellite is given by

[tex]T=2\pi\sqrt{\frac{r^3}{GM}}\\\Rightarrow M=4\pi^2\frac{r^3}{T^2G}\\\Rightarrow M=4\pi^2\times \frac{(1.94\times 10^8)^3}{(24\times 3600)^2\times 6.67259\times 10^{-11}}\\\Rightarrow M=5.78686\times 10^{26}\ kg[/tex]

The gravitational force is given by

[tex]F=G\frac{Mm}{r^2}\\\Rightarrow F=6.67259\times 10^{-11}\times \frac{5.78686\times 10^{26}\times 438}{(1.94\times 10^8)^2}\\\Rightarrow F=449.37412\ N[/tex]

The force acting on this satellite is 449.37412 N

Gravity, is often known as gravitation. The gravitational force between the planet and the satellite can be written as 0.4332 N.

Gravity, often known as gravitation, is the universal force of attraction that acts between all matter in mechanics. It is the weakest known force in nature, and so has no bearing on the interior properties of ordinary matter.

[tex]F =G\dfrac{m_1m_2}{r^2}[/tex]

As it is given that the value of the gravitational constant is G = 6.67259 × 10−11 N m² /kg², while the radius of the planet is 1.94×10⁵. And the time taken by the satellite to revolve around the planet is 24 hours. therefore, the Mass of the planet can be written as,

The Time period of the satellite is given as:

[tex]T = 2\pi \sqrt{\dfrac{r^3}{GM}}[/tex]

Substitute the values in the formula,

[tex]24 = 2\pi \sqrt{\dfrac{(1.94 \times 10^5)^3}{6.67259 \times 10^{-11}M}}\\\\M = 5.78686 \times 10^{17}\rm\ kg[/tex]

Thus, the mass of the planet can be written as 5.5786×10¹⁷ kg.

Now, the gravitational force can be written as,

[tex]F = G\dfrac{Mm}{r^2}\\\\F = 6.67259\times 10^{-11} \times \dfrac{5.5786 \times 10^{17} \times 438}{1.94 \times 10^5}\\\\ F = 0.4332\rm\ N[/tex]

Hence, the gravitational force between the planet and the satellite can be written as 0.4332 N.

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Answer:

option D.

Explanation:

The correct answer is option D.

When an object is in equilibrium torque calculated at any point will be equal to zero.

An object is said to be in equilibrium net moment acting on the body should be equal to zero.

If the net moment on the object is not equal to zero then the object will rotate it will not be stable.

In a system in equilibrium, the choice of axis about which torques are calculated can be anywhere because for such a system, the sum of torques about any point is zero.

When applying the torque equation, ∑τ = 0, to an object in equilibrium, the chosen axis about which torques are calculated can be located anywhere on or off the object. This principle is a result of the fact that in a system in equilibrium, the sum of torques about any point is zero, not just specific points like the object's pivot, edge, or center of gravity. For example, if you have a seesaw in balance, you could calculate the torques from the center, one of the seats, or even a point suspended above it in the air. As long as the object is in equilibrium and not moving, the torques calculated from any point will sum to zero because they counteract each other in direction and magnitude to maintain the state of balance.

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To solve this exercise it is necessary to use the concepts related to Difference in Phase.

The Difference in phase is given by

[tex]\Phi = \frac{2\pi \delta}{\lambda}[/tex]

Where

[tex]\delta =[/tex] Horizontal distance between two points

[tex]\lambda =[/tex] Wavelength

From our values we have,

[tex]\lambda = 500nm = 5*10^{-6}m[/tex]

[tex]\theta = 1\°[/tex]

The horizontal distance between this two points would be given for

[tex]\delta = dsin\theta[/tex]

Therefore using the equation we have

[tex]\Phi = \frac{2\pi \delta}{\lambda}[/tex]

[tex]\Phi = \frac{2\pi(dsin\theta)}{\lambda}[/tex]

[tex]\Phi = \frac{2\pi(5*!0^{-6}sin(1))}{500*10^{-9}}[/tex]

[tex]\Phi= 1.096 rad \approx = 1.1 rad[/tex]

Therefore the correct answer is C.

The phase difference for waves from the top and bottom of the slit can be calculated by using the formula for calculating phase difference. With the provided values for wavelength, angle and slit width, the calculated phase difference is 0.55 rad.

The phase difference of the waves is directly related to the path difference between them and can be calculated by using the formula:

Φ = 2 π × (d / λ) × sin(θ).

Where φ is the phase difference, π is Pi, d is the slit width, λ is the wavelength and θ is the incident angle.

Let's plug the provided numbers into our formula:

Φ = 2 π × (5.0x10-6 m / 500x10-9 m) × sin(1.0°).

Φ = 2 π × 10 × sin(1.0°)= 0.55 rad.

So, the correct answer is (B) 0.55 rad.

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Final answer:

To find the highest available gain for an op-amp circuit with a required bandwidth, we use the Gain-Bandwidth Product (GBP), which is a constant. For a GBP of 768 kHz and a bandwidth of 32 kHz, the maximum available gain is 24 V/V.

Explanation:

The question concerns determining the highest available gain for an op-amp circuit given a required bandwidth.

The Gain-Bandwidth Product (GBP) is a constant for an op-amp and is found by multiplying the current gain with its corresponding 3-dB frequency.

With an initial gain of 96 V/V and a 3-dB frequency of 8 kHz, the GBP can be calculated as 96 V/V * 8 kHz = 768 kHz.

To meet the requirement of a 32 kHz bandwidth with the same GBP (because GBP is constant), we can calculate the maximum gain as follows:

GBP = Gain * Bandwidth, which gives us

Gain = GBP / Bandwidth.

Plugging in the numbers, we get

Gain = 768 kHz / 32 kHz, resulting in a maximum gain of 24 V/V under the conditions of a 32 kHz bandwidth.

Answer:

Induced emf, [tex]\epsilon=-A\dfrac{B_{max}e^{-t/\tau}}{\tau}[/tex]

Explanation:

The varying magnetic field with time t is given by according to equation as :

[tex]B=B_{max}e^{-t/\tau}[/tex]

Where

[tex]B_{max}\ and\ t[/tex] are constant

Let [tex]\epsilon[/tex] is the emf induced in the loop as a function of time. We know that the rate of change of magnetic flux is equal to the induced emf as:

[tex]\epsilon=-\dfrac{d\phi}{dt}[/tex]

[tex]\epsilon=-\dfrac{d(BA)}{dt}[/tex]

[tex]\epsilon=-A\dfrac{d(B)}{dt}[/tex]

[tex]\epsilon=-A\dfrac{d(B_{max}e^{-t/\tau})}{dt}[/tex]

[tex]\epsilon=A\dfrac{B_{max}e^{-t/\tau}}{\tau}[/tex]

So, the induced emf in the loop as a function of time is [tex]A\dfrac{B_{max}e^{-t/\tau}}{\tau}[/tex]. Hence, this is the required solution.

The emf induced in the loop as a function of time is determined as [tex]emf = \frac{AB_{max}}{\tau} e^{-t/\tau}[/tex].

The emf induced in the rectangular loop is determined by applying Faraday's law of electromagnetic induction.

emf = dФ/dt

where;

Ф = BA

A is the area of the rectangular loop

emf = d(BA)/dt

[tex]emf = \frac{d(BA)}{dt} \\\\emf = A \frac{dB}{dt} \\\\emf = A \times B_{max}(e^{-t/\tau})(1/\tau)\\\\emf = \frac{A B_{max}(e^{-t/\tau})}{\tau}\\\\emf = \frac{AB_{max}}{\tau} e^{-t/\tau}[/tex]

Thus, the emf induced in the loop as a function of time is determined as [tex]emf = \frac{AB_{max}}{\tau} e^{-t/\tau}[/tex].

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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:

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:

0.00288 J

Explanation:

We know that

W= Fds

F = −α∆s^4

α = 45 N/m^4 and ∆s = displacement

W= −α∆s^4ds

integrating both the sides from s= 0 to 0.2

W= 45/5×0.2^5= 0.00288 J

The workdone on the stone-elastic band system is mathematically given as

W= 0.00288 J

Question Parameter(s):

F = −α∆s4 , where α = 45 N m4

∆s is the displacement of the elastic band

Generally, the equation for the Workdone  is mathematically given as

W= Fds

Thereofre

F = −α *ds^4

Where

α = 45  

ds = displacement

In conclusion

W= 45/5*0.2^5

W= 0.00288 J

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Answer:

[tex]\omega_f=-2.6rad/s[/tex]

Since the bar cannot translate, linear momentum is not conserved

Explanation:

By conservation of the angular momentum:

Lo = Lf

[tex]I_B*\omega_o+I_b*(V_o/d)=I_B*\omega_f+I_b*(V_f/d)[/tex]

where

[tex]I_B =1/3*M_B*L^2[/tex]

[tex]I_b=m_b*d^2[/tex]

[tex]M_B=90/g=9kg[/tex]; [tex]m_b=3kg[/tex]; d=1.6m; L=3m; [tex]V_o=-10m/s[/tex]; [tex]V_f=5m/s[/tex]; [tex]\omega_o=0 rad/s[/tex]

Solving for [tex]\omega_f[/tex]:

[tex]\omega_f=m_b*d/I_B*(V_o-V_f)[/tex]

Replacing the values we get:

[tex]\omega_f=-2.6rad/s[/tex]

Since the bar can only rotate (it canno translate), only angular momentum is conserved.

Answer:

True

Explanation:

The angular momentum around the center of the planet and the total mechanical energy will be preserved irrespective of whether the object moves from large R  to small R. But on the other hand the kinetic energy of the planet will not be conserved because it can change from kinetic energy to potential energy.

Therefore the given statement is True.

The correct answer is option A. The statement is true because in celestial mechanics, angular momentum and total energy are conserved under gravity, regardless of whether an object moves from a smaller to a larger radius or vice versa.

Let's explain these concepts:

1. Conservation of Angular Momentum (L):

The angular momentum of an object with respect to the center of the planet is given by the product of its linear momentum (p) and the radius (r) of its orbit:

[tex]\[ L = p \times r = mvr \][/tex]

where m is the mass of the object, v is its velocity, and r is the radius of the orbit. For a central force like gravity, the torque on the object is zero, which means that the angular momentum is conserved.

2. Conservation of Total Mechanical Energy (E):

The total mechanical energy of an object in orbit is the sum of its kinetic energy (K) and potential energy (U):

[tex]\[ E = K + U = \frac{1}{2}mv^2 - \frac{GMm}{r} \][/tex]

where G is the gravitational constant, M is the mass of the planet, and m is the mass of the object.

The complete question is:

The angular momentum about the center of the planet and the total mechanical energy will be conserved regardless of whether the object moves from small R to large R (like a rocket being launched) or from large R to small R (like a comet approaching the earth).

A. True

B. False

The minimum translational speed the ball must have at a height of 1.329 m to complete the loop without falling off the track is approximately 6.61 m/s.

How can you solve the minimum translational speed the ball must have?

E p(top) = K e(bottom)

E p(top) = m * g * (R + H)

where:

m is the mass of the ball (0.6350 kg)

g is the acceleration due to gravity (9.810 m/s²)

R is the radius of the loop (0.8950 m)

H is the height above the bottom of the loop (1.329 m)

Calculate the minimum kinetic energy at the bottom:

Since the ball needs enough speed at the bottom to reach the top again, the minimum kinetic energy at the bottom is equal to the potential energy at the top:

K e(bottom) = E p(top) = m * g * (R + H)

Find the minimum translational speed:

K e = 1/2 * m * vmin²

where vmin is the minimum translational speed we're looking for. Solving for vmin:

v min = √(2 * K e / m) = √(2 * m * g * (R + H) / m)

v min = √(2 * g * (R + H))

Plug in the values and calculate:

v min = √(2 * 9.810 * (0.8950 + 1.329))

v min ≈ 6.61 m/s

Therefore, the minimum translational speed the ball must have at a height of 1.329 m to complete the loop without falling off the track is approximately 6.61 m/s.

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]

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:

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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Answer:

(a) 18667N

(b)280000N

Explanation:

The momentum of both car prior to collision:

M = mv = 1400*20 = 28000 kgm/s

After colliding, the average force exerting on either car to kill this momentum is

(a)[tex] F_1 =\frac{M}{t_1} = \frac{28000}{1.5} = 18667N[/tex]

(b)[tex]F_2 = \frac{M}{t_2} = \frac{28000}{0.1} = 280000N[/tex]

To find the average forces exerted on the cars, use the formula: Force = change in momentum / time interval. For the car hitting the line of water barrels, the average force is -1400 kg * m/s / 1.5 s. For the car hitting the concrete barrier, the average force is -1400 kg * m/s / 0.10 s.

To find the average forces exerted on the cars, we can use the formula:

Force = change in momentum / time interval

(a) For the car that hits a line of water barrels and takes 1.5 s to stop:

The change in momentum is given by: Δp = m * Δv = m * (0 - 20) = -1400 kg * m/s

Then, the average force exerted on the car is: F = Δp / t = -1400 kg * m/s / 1.5 s

(b) For the car that hits a concrete barrier and takes 0.10 s to stop:

The change in momentum is given by: Δp = m * Δv = m * (0 - 20) = -1400 kg * m/s

Then, the average force exerted on the car is: F = Δp / t = -1400 kg * m/s / 0.10 s

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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:

e. All of these statements are false.

Explanation:

As we know that heat transfer take place from high temperature to low temperature.

It is possible to convert all work into heat but it is not possible to convert all heat in to work some heat will be reject to the surrounding.

The first law of thermodynamics is the energy conservation law.

Second law of thermodynamics  states that it is impossible to construct a device which convert all energy into work without rejecting the heat to the surrounding.

By using heat pump ,heat can transfer from cooler body to the hotter body.

Therefore all the answer is False.

The true statement among the given options is that the entropy of a system can be reduced by cooling it.

The correct answer to the student's question regarding true statements about thermodynamics is option (c) It is always possible to reduce the entropy of a system, for instance, by cooling it. This statement aligns with the principles of thermodynamics, which affirm that entropy, a measure of disorder or randomness, can decrease in a system if energy is removed from the system, such as by lowering its temperature. However, this does not violate the second law of thermodynamics because entropy may still increase in the overall process when considering the surroundings.

In contrast, options (a) and (b) are false because they wrongly imply irreversibility in scenarios where reversibility is possible. Specifically, it is possible to reverse an entropy increase by cooling a system and it is also possible to convert some amount of thermal energy back into mechanical energy, although not with 100% efficiency due to inherent thermodynamic losses.

The second law also articulates that heat transfer occurs spontaneously from a higher to a lower temperature body and not in the reverse direction without external work, implying that a spontaneous flow of heat from a colder to a warmer body is impossible, as is complete conversion of heat to work in a cyclical process.

Answer:

a) v = 7,207 m / s

, b) a = 42.3 m / s²

Explanation:

We will solve this exercise using the concept of mechanical energy, We will write it in two points before the car touches the springs and in point of maximum compression

Initial

    Em₀ = K = ½ m v²

Final  

    [tex]Em_{f}[/tex] = 2 Ke = ½ k x²

The two is placed because each barred has two springs and each does not exert the same force

    Emo = [tex]Em_{f}[/tex]

     ½ m v² = 2 ½ k x²

     v = √(2k/m) x

     v = √ (2 3,134 10⁵/4550) 0.614

     v = 7,207 m / s

Let's take this speed to km / h

     v = 5,096 m / s (1km / 1000m) (3600s / 1h)

     v = 25.9 km / h

This speed is common in school zones

Let's use kinematics to calculate the average acceleration

      vf² = v₀² - 2 a x

       0 = v₀² - 2 a x

       a = v₀² / 2 x

       a = 7,207²/2 0.614

       a = 42.3 m / s²

We buy this acceleration with the acceleration of gravity

       a / g = 42.3 / 9.8

       a / g = 4.3

This acceleration is well below the maximum allowed

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

Answers:

a) [tex]8.009(10)^{-7} N[/tex]

b) [tex]1.8(10)^{-8} [/tex]

Explanation:

a) Accoding to the Universal Law of Gravitation we have:

[tex]F_{g}=G\frac{Mm}{d^2}[/tex] (1)

Where:

[tex]F_{g}[/tex] is the gravitational force between the eagle and the throng

[tex]G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex] is the Universal Gravitational constant

[tex]M=4.5 kg[/tex] is the mass of the eagle

[tex]m=(80 kg)(3(10)^{6} people/kg)=240(10)^{6} kg[/tex] is the mass of the throng

[tex]d=300 m[/tex] is the distance between the throng and the eagle

[tex]F_{g}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} \frac{(4.5 kg)(240(10)^{6} kg)}{(300 m)^{2}}[/tex] (2)

[tex]F_{g}=8.009 (10)^{-7} N[/tex] (3) As we can see the gravitational force between the eagle and the throng is quite small.

b) The attraction force between the eagle and Earth is the weight [tex]W[/tex] of the eagle, which is given by:

[tex]W=Mg[/tex] (4)

Where [tex]g=9.8 m/s^{2}[/tex] is the acceleration due gravity on Earth

[tex]W=(4.5 kg)(9.8 m/s^{2})[/tex] (5)

[tex]W=44.1 N[/tex] (6)

Now we can find the ratio between [tex]F_{g}[/tex] and [tex]W[/tex]:

[tex]\frac{F_{g}}{W}=\frac{8.009 (10)^{-7} N}{44.1N}[/tex]

[tex]\frac{F_{g}}{W}=1.8(^{-8})[/tex] As we can see this ratio is also quite small

Final answer:

The minimum time for a simple pendulum of 18.5 meters to swing from maximum displacement to equilibrium position is approximately 2.160 seconds, which is one-fourth the total period of the pendulum.

Explanation:

The minimum time for a pendulum to swing from maximum displacement to the equilibrium position is one-fourth of its period. The period of a simple pendulum, which is the time for one complete to-and-fro swing, can be calculated using the formula T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. Given a length, L, of 18.5 meters and gravity, g, of 9.81 m/s², we find:

T = 2π√(18.5/9.81) ≈ 2π√(1.886) ≈ 2π×1.373≈ 8.639 seconds

Therefore, the least amount of time it takes for the bob to swing from maximum displacement to the equilibrium position would be T/4:

T/4 = 8.639 seconds / 4 ≈ 2.160 seconds.

Final answer:

The least amount of time for a pendulum of length 18.5 m to swing from maximum displacement to equilibrium is approximately 2.17 seconds, which is a quarter of its period.

Explanation:

To find the time taken for a pendulum to swing from maximum displacement to the equilibrium position, we use the formula for the period of a simple pendulum, which is:

T = 2π√(L/g)

Where T is the period of oscillation, L is the length of the pendulum, and g is the acceleration due to gravity. However, the question asks for the time taken to swing from maximum displacement to equilibrium, which is only a quarter of the period. So, we divide the period by 4:

Time = T/4 = (1/2)π√(L/g)

Substitute the given values (L = 18.5 m and g = 9.81 m/s2):

Time = (1/2)π√(18.5/9.81)

After calculating, we find that the time taken is approximately 2.17 seconds.

Answer:

0.35701 T

Explanation:

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

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

[tex]\phi[/tex] = Magnetic flux

t = Time taken = 4.17 ms

N = Number of turns = 510

[tex]\epsilon[/tex] = Induced emf = 10000 V

r = Radius = 0.27 m

A = Area = [tex]\pi r^2[/tex]

Induced emf is given by

[tex]\epsilon=-N\frac{d\phi}{dt}\\\Rightarrow \epsilon=-N\frac{B_fAcos90-B_iAcos0}{dt}\\\Rightarrow \epsilon=N\frac{B_iA}{dt}\\\Rightarrow B_i=\frac{\epsilon dt}{NA}\\\Rightarrow B_i=\frac{10000 \times 4.17\times 10^{-3}}{510\times \pi 0.27^2}\\\Rightarrow B_i=0.35701\ T[/tex]

The magnetic field strength needed is 0.35701 T

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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