An 8 foot metal guy wire is attached to a broken stop sign to secure its position until repairs can be made. Attached to a stake in the ground, the guy wire makes an angle of 51º with the ground. How far from the foot of the stop sign is the stake, to the nearest tenth of a foot?

Answer:

111249.41m/s²

Explanation:

Mass of the wire m = 40g = 0.04 Kg

Length of the wire l = 7 m

Linear density of wire μ = m/l

= 0.04/7

= 0.0057 Kg/m

Frequency n = 630 Hz

Wavelength λ = 0.5 m

Amplitude A =  7.1mm = 0.0071m

Speed of the wave v = nλ

= 630*0.5

= 315 m/s

Angular speed ω = 2πn

= 2π*630

= 3958.40 rad/s

Maximum transeverse acceleration

a = ω^2 A

= 3958.40² × 0.0071m

= 111249.41m/s²

Answer:

B. Forward

Explanation:

The net force in the system is initially balanced and the system is in equilibrium.

1) As John ducks, two things happens. First, there is a sudden impulse from his end which will change the momentum of the system from zero to an amount Mv proportional to the impulse Ft, which will be directed away from his direction because,

2) the process of ducking pushes his center of mass forward, which also moves the center of mass of the whole system away from him. This two effect will cause the skateboard to move forward.

Answer:

A) Kinetic

Explanation:

Kinetic friction is friction caused by motion. Kinetic energy is energy in motion.

Answer: tsunamis,El Nino,and volcanic eruptions

Explanation:

the correct answer  for e d g e n u i t y

The examples of short-term environmental change are Tsunamis, El Niño, and Volcanic eruptions.

Tsunamis are huge ocean waves that are triggered by undersea disturbances like earthquakes or landslides. They can inflict major damage to coastal regions, however they usually happen in a short period of time.

El Nio is a climatic trend characterised by higher-than-normal equatorial Pacific ocean temperatures.

It can cause changes in weather patterns, such as greater rainfall in some areas and droughts in others, although the impacts are usually transient and endure for a few months to a couple of years.

Volcanic eruptions spew lava, ash, and gases into the atmosphere, which can have direct effects on the environment, including as changes in air quality and local weather patterns.

Thus, these effects, however, are often short-lived and localized.

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Answer:

Using equations

R will be 1.25 Kilo ohm

To get the value of resistance, R, we need to use Ohm's law which is R = V/I', where I' is the current generated when the coil balances with the added mass. The exact value of R depends on parameters related to the coil and the magnetic field.

The question is asking for the value of the resistance R in a given circuit, when the circuit produces a current which generates a balance with an added mass through a coil in a magnetic field. To find the value of R, we need to use Ohm's law, which expresses the relationship between voltage (V), current (I), and resistance (R) as V=IR.

Here, the electromotive force or emf can be treated as the voltage in the circuit. In this case, emf = 20.0 V. When this coil creates a balance with an added 51.0 g mass, an equivalent current will be generated, let's denote it by I'

Using Ohm's law, we can solve for R by rearranging the formula to R = V/I'. The exact value of I' and thus R will, however, depend on additional factors involving the parameters of the coil and magnetic field which are not provided in the question. The law used here, Lenz's law, is a manifestation of the conservation of energy, dictating that the current induced in a circuit due to a change in the magnetic field will create a magnetic field that opposes the initial changing magnetic field.

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The velocity of the both spheres is [tex]v=4.62 m/s[/tex]

Explanation:

The free body diagram of this question is shown on the second uploaded image

     Looking at this diagram we can deduce that there is no impulse force along the horizontal direction so

      The mathematical equation for the impulse force along the horizontal axis is  

           [tex]\int\limits {\sum F_s} \, dt = 0[/tex]

 The mathematical equation for the impulse force along the horizontal axis is

           [tex]\int\limits {\sum F_y} \, dt = \Delta I_y[/tex]

  Where [tex]I_y[/tex] is the impulse  momentum along the y-axis and this is mathematically given as

         [tex]\Delta I_y = 2mv_y[/tex]

substituting 9.8 N.s for [tex]\Delta I_y[/tex] and 1.5kg for m (mass of sphere) and the making [tex]v_y[/tex] the subject

        [tex]v_y = \frac{9.8}{2 *1.5}[/tex]

            [tex]=3.267 m/s[/tex]

The sum of the moment about the point I is mathematically represented as

         [tex]\int\limits {\sum M_I } \, dt =\Delta H_I[/tex]

From the free body diagram  [tex]\int\limits {\sum M_I } \, dt = I * 0.3 = 9.8 *0.3=2.94N.m[/tex]

[tex]\Delta H_I[/tex] is the angular momentum along the horizontal axis  given as

                  [tex]\Delta H_I[/tex] [tex]= 2mv_x(0.3)[/tex]

substituting parameters into above equation

                  [tex]2.94 = 2 * 1.5 * v_x * 0.3[/tex]

Making [tex]v_x[/tex] the subject

                 [tex]v_x = \frac{2.94}{2*1.5*0.3}[/tex]

                      [tex]=3.267 m/s[/tex]

the velocity of the sphere is mathematically represented as

                 [tex]v = \sqrt{(v_x)^2 + (v_y)^2}[/tex]

Now substituting values

                [tex]v = \sqrt{3.267^2 + 3.267^2}[/tex]

                   [tex]v=4.62 m/s[/tex]

The physics question pertains to the concept of impulse and momentum. The velocity of each sphere after the impulse is imparted can be calculated by dividing the change in momentum by twice the mass of one sphere.

The Physics question is relevant to impulse and momentum in the domain of mechanics. From the problem, we know that a force F acting on the sphere imparts an impulse of 9.8 N·s. Impulse, denoted as J, in physics, is the product of force and the time for which it is applied and is equivalent to the change in momentum of the body. Thus, the change in momentum, Δp is equal to the impulse imparted, which is 9.8 N·s.

Now considering the system of two spheres rigidly connected, it is an isolated system (external force F is not considered as it acts for a very short time) and therefore, the total momentum of the system remains conserved. Therefore the momentum imparted to one sphere is equally distributed to both the spheres. Therefore, the final velocity v of each sphere can be calculated by using the formula v = Δp / 2m (m being the mass of each sphere).

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Answer:

Explanation:

For any point stationary in front of ambulance , the ambulance is approaching the point . In this way the case is similar to source of sound approaching the observer. The doppler's effect can be applied to know the apparent frequency.

velocity of sound V = 343 m /s

speed of source Vs = 48.4

observer is stationary.

apparent frequency = real frequency x V / ( V - Vs )

= 620 x 343 / ( 343 - 48.4 )

= 722 Hz approx.

Answer:oops

Explanation:

Answer:

Check the explanation

Explanation:

When we have an object in periodic motion, the amplitude will be the maximum displacement from equilibrium. Take for example, when there’s a back and forth movement of a pendulum through its equilibrium point (straight down), then swings to a highest distance away from the center. This distance will be represented as the amplitude, A. The full range of the pendulum has a magnitude of 2A.

position = amplitude x sine function(angular frequency x time + phase difference)

x = A sin(ωt + ϕ)

x = displacement (m)

A = amplitude (m)

ω = angular frequency (radians/s)

t = time (s)

ϕ = phase shift (radians)

Kindly check the attached image below to see the step by step explanation to the question above.

Answer:1.513 rps

Explanation:

Given

Diameter of cylindrical space [tex]d=780\ m[/tex]

When the station rotates it creates centripetal acceleration which is given by

[tex]a_c=\omega ^2r[/tex]

Now it must create the effect of gravity so

[tex]g=\omega ^2r[/tex]

[tex]\omega =\sqrt{\frac{g}{r}}[/tex]

[tex]\omega =0.158\ rad/s[/tex]

and [tex]\omega =\frac{2\pi N}{60}[/tex]

Thus [tex]N=\frac{4.755}{3.142}[/tex]

[tex]N=1.513\ rps[/tex]

Answer:

Explanation:

Given the following information,

Resistor of resistance R = 13.6Ω

Capacitor of capacitance C = 11.9-μF

C = 11.9 × 10^ -6 F

Inductor of inductance L = 19.1-mH

L = 19.1 ×10^-3 H

All this connected in series to a generator that generates Vrms= 117V

Vo = Vrms√2 = 117√2

Vo = 165.463V

a. Frequency for maximum current?

Maximum current occurs at resonance

I.e Xc = XL

At maximum current, the frequency is given as

f = 1/(2π√LC)

Then,

f = 1/(2π√(19.1×10^-3 × 11.9×10^-6)

f = 1/(2π√(2.2729×10^-7))

f = 1/(2π × 4.77 ×10^-4)

f = 333.83Hz

Then, the frequency is 333.83Hz.

b. Since we know the frequency,

Then, we need to find the capacitive and inductive reactance

Capacitive reactance

Xc = 1/2πfC

Xc = 1/(2π × 338.83 × 11.9×10^-6)

Xc = 1/ 0.024961

Xc = 40.1Ω

Also, Inductive reactance

XL = 2πfL

XL = 2π × 333.83 × 19.1×10^-3

XL = 40.1Ω

As expected, Xc=XL, resonance

Then, the impedance in AC circuit is given as

Z = √ (R² + (Xc—XL)²)

Z = √ 13.6² + (40.1-40.1)²)

Z = √13.6²

Z = 13.6 ohms

Then, using ohms las

V = IZ

Then, I = Vo/Z

Io = 165.46/13.6

Io = 12.17Amps

The current is 12.17 A

Answer:

a) Current is maximum at frequency, f₀ = 333.83 Hz

b) Maximum current = 12.17 A

Explanation:

Inductance, L = 19.1 mH = 19.1 * 10⁻³ H

Capacitance, C = 11.9 μF =11.9 * 10⁻⁶ F

a) Current is maximum at resonant frequency, f₀

[tex]f_{0} = \frac{1}{2\pi\sqrt{LC} }[/tex]

[tex]f_{0} = \frac{1}{2\pi\sqrt{11.9 * 10^{-6}* 19.1 * 10^{-3} } }[/tex]

[tex]f_{0} = 333.83 Hz[/tex]

b) Maximum value of the RMS current

[tex]V_{RMS} = 117 V\\V_{max} = \sqrt{2} V_{RMS}\\V_{max} = \sqrt{2} * 117\\V_{max} = 165.46 V[/tex]

[tex]I_{max} = \frac{V_{max} }{R} \\I_{max} = \frac{165.46}{13.6} \\I_{max} = 12.17 A[/tex]

Answer:

The speed of sound theoretically is 267.148 m/s which is clearly less than slope calculated speed 365.0 m/s

Explanation:

check the attached picture

Answer:nuclear energy is not derived from the sun

Explanation:

Nuclear energy is not derived from the sun.nuclear energy comes from the energy released when atoms are split apart and some mass is converted to energy

Explanation:

Given that,

Wavelength of the light, [tex]\lambda=4170\ A=4170\times 10^{-10}\ m[/tex]

Work function of sodium, [tex]W_o=4.41\times 10^{-19}\ J[/tex]

The kinetic energy of the ejected electron in terms of work function is given by :

[tex]KE=h\dfrac{c}{\lambda}-W_o\\\\KE=6.63\times 10^{-34}\times \dfrac{3\times 10^8}{4170\times 10^{-10}}-4.41\times 10^{-19}\\\\KE=3.59\times 10^{-20}\ J[/tex]

The formula of kinetic energy is given by :

[tex]KE=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2KE}{m}} \\\\v=\sqrt{\dfrac{2\times 3.59\times 10^{-20}}{9.1\times 10^{-31}}} \\\\v=2.8\times 10^5\ m/s[/tex]

Hence, this is the required solution.

Answer:

One way to categorize waves is on the basis of the direction of movement of the individual particles of the medium relative to the direction that the waves travel. Categorizing waves on this basis leads to three notable categories: transverse waves, longitudinal waves, and surface waves.

I've read about it but never seen one. The way I understand it, it's a coordinated physical motion executed by a great number of people, as in a large crowd at a sporting event, timed so that it appears to propagate from one end of the crowd to the opposite end.

I would classify it as a cooperative community activity, involving liberty, equality, and fraternity, executed for the common good.

Explanation:

A paper circuit is a functioning electronic circuit built on a paper surface. Projects involving paper circuits are unique because of the use of traditional art techniques and some unique materials to create a circuit that combines aesthetics and functionality.

Paper circuit are sometimes carefully designed and then transfered on the paper, and in some cases, they are designed directly on the paper (usually by experts that already know what they're doing).

To make designing circuit on paper easier, there are three commonly used materials for the circuitry and they are, conductive paints, conductive tapes and conductive inks.

Conductive tapes are made from metal strip (usually copper) that are taped to the paper. They are good to work with since they allow components to be soldered on them creating a stronger and more reliable joint.

Conductive paints are special paints that can be used to outline circuit path and also serve as the circuit. The only problem with conducting paints is that it can be messy, and needs time to dry.

Conductive inks are easier to use as they need no drying time. They are far less messy and allows the drawing of elaborate and more intricate circuit on the paper.

There are also components that have been modified (like led lights etc) and available to make paper circuit design easier.

Answer:

A- Equality

Explanation:

Answer:

a The kinetic energy is  [tex]KE = 0.0543 J[/tex]

b The height of the center of mass above that position is  [tex]h = 1.372 \ m[/tex]    

Explanation:

From the question we are told that

  The length of the rod is  [tex]L = 1.4m[/tex]

   The mass of the rod [tex]m = 140 = \frac{140}{1000} = 0.140 \ kg[/tex]  

   The angular speed at the lowest point is [tex]w = 1.09 \ rad/s[/tex]

Generally moment of inertia of the rod about an axis that passes through its one end is

                   [tex]I = \frac{mL^2}{3}[/tex]  

Substituting values

               [tex]I = \frac{(0.140) (1.4)^2}{3}[/tex]

               [tex]I = 0.0915 \ kg \cdot m^2[/tex]

Generally the  kinetic energy rod is mathematically represented as

             [tex]KE = \frac{1}{2} Iw^2[/tex]

                    [tex]KE = \frac{1}{2} (0.0915) (1.09)^2[/tex]

                           [tex]KE = 0.0543 J[/tex]

From the law of conservation of energy

The kinetic energy of the rod during motion =  The potential energy of the rod at the highest point

   Therefore

                   [tex]KE = PE = mgh[/tex]

                        [tex]0.0543 = mgh[/tex]

                             [tex]h = \frac{0.0543}{9.8 * 0.140}[/tex]

                                [tex]h = 1.372 \ m[/tex]    

Answer:

a) Kr = 0.0543 J

b) Δy = 0.0396 m

Explanation:

a) Given

L = 1.4 m

m = 140 g = 0.14 kg

ω = 1.09 rad/s

Kr = ?

We have to get the rotational inertia as follows

I = Icm + m*d²

⇒ I = (m*L²/12) + (m*(L/2)²)

⇒ I = (0.14 kg*(1.4 m)²/12) + (0.14 kg*(1.4 m/2)²)

⇒ I = 0.09146 kg*m²

Then, we apply the formula

Kr = 0.5*I*ω²

⇒ Kr = 0.5*(0.09146 kg*m²)*(1.09 rad/s)²

⇒ Kr = 0.0543 J

b) We apply the following principle

Ei = Ef

Where the initial point is the lowest position and the final point is at the maximum height that its center of mass can achieve, then we have

Ki + Ui = Kf + Uf

we know that ωf = 0 ⇒ Kf = 0

⇒ Ki + Ui = Uf

⇒ Uf - Ui = Ki

⇒ m*g*yf - m*g*yi = Ki

⇒ m*g*(yf - yi) = Ki

⇒ m*g*Δy = Ki

⇒ Δy = Ki/(m*g)

where

Ki = Kr = 0.0543 J

g = 9.81 m/s²

⇒ Δy = (0.0543 J)/(0.14 kg*9.81 m/s²)

⇒ Δy = 0.0396 m

Answer:

Explanation:

We shall apply Lenz's law to solve the problem . This law states that direction of induced current is such that it opposes the change that creates it. Since current increases in the coil it creates increasing magnetic field in the other coil . So the current will be induced in it  so that it opposes this increase . It can be done only  if current in opposite direction is induced in it . Hence in the first case,  current will be induced in opposite direction .

In this case, current is decreasing in the primary coil and current will be induced in the secondary coil. Decreasing current will create decreasing magnetic field . So induced current will try to increase it . In can be done if current in the same direction is induced in the secondary coil.

Hence in the second case , current will be induced in the same direction .

Answer:

Current is half

Voltage is half

Explanation:

according to Ohms law V = IR

Case 1

I₁ = V / R

I₂ = V / (R+R) = V / 2R

∴ I₁ / I₂ = (V/R) / (V/2R)

=> I₁ / I₂ = (V/R) * (2R/V)

=> I₁ / I₂ = (2VR / RV)

I₁ / I₂ = 2

I₂ = I₁ / 2

The current through each resistor in the two-resistor circuit is half the current through the resistor in the one-resistor circuit (the circuit in Part A).

The voltage across each resistor in the two-resistor circuit is half the voltage across the resistor in the one-resistor circuit.

The current through each resistor in a two-resistor series circuit is the same as the current through a one-resistor circuit. However, the voltage across each resistor in the two-resistor circuit is half of the voltage across the single resistor due to how voltages distribute in series circuits.

In a series circuit, such as the two-resistor circuit you have mentioned, the current through each resistor is the same as the current through the single resistor in the one-resistor circuit. This is due to Kirchhoff's Current Law which states that the current entering a junction or a node must equal the current leaving it. However, the voltage accross each resistor in the two-resistor circuit is half of the voltage across the single resistor in the one-resistor circuit. This is because in a series circuit, the total voltage is the sum of the individual voltage drops across each resistor, according to Ohm’s law.

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Answer:

Explanation:

a , b )

The problem is based on interference in thin films

formula for constructive interference

2μ t = ( 2n+ 1 ) λ / 2 , μ is refractive index of layer, t is thickness and λ is wavelength of light.

n is called the order of fringe . If we place n= 0 , 1 , 2  etc , the thickness also changes . So constructive interference is possible at more than one thickness .

Put the value of  λ = 116 nm . μ = 1.28 , t = 116 nm in the given equation

2 x 1.28 x 116 x 2 = ( 2n+ 1 ) λ

593.92 = ( 2n+ 1 ) λ

when n = 0

λ = 593.92 nm .

This falls in visible range .

c )

2μ t = ( 2n+ 1 ) λ / 2

Put λ = 593.92 nm , n = 1

2 x 1.28 t₁ = 3 x 593.92 / 2

t₁ = 348 nm .

Put n = 2

2 x 1.28 t₂ = 5 x 593.92 / 2

t₂ = 580 nm .

Answer:

m = (3/5)*M

Explanation:

Given:-

- The angular speed of both hollow and solid sphere = w

- The diameter of solid & hollow sphere = D

- The mass of the solid sphere = M

- Both rotate about their common axis with similar rotational kinetic energy.

Find:-

The mass of hollow sphere (m) ?

Solution:-

- The formula for rotational kinetic energy (K.E) of any rigid body is:

                                    K.E = 0.5*I*w^2

Where,

                  I : Moment of inertia of rigid body

- The rotational kinetic energies of both hollow sphere and solid sphere are same:

                             0.5*I_solid*w^2 = 0.5*I_shell*w^2

                             I_solid = I_shell

                             0.4*M*D^2 / 4 = (2/3)*m*D^2 / 4

                             (2/5)*M = (2/3)*m

                             m = (3/5)*M

Answer:

Check the explanation

Explanation:

So far there’s an increases in the speed of the kangaroo, then the tendons will stretch more thereby enabling them to store more energy. For this reason, they will have a additional time in the air propelled by greater spring energy. In contact with the ground, it will turn out to be like a spring in simple harmonic motion. There will be increases in their agility rate in hopping the amplitude of the oscillation, but that does not in any way affect the time, or period in contact with the ground.

Kangaroos convert kinetic energy to elastic potential energy during landing which is immediately converted back to kinetic energy for the next leap. This energy transformation cycle, aided by cushioning their landing by bending their hind legs, allows kangaroos to spend more time airborne without affecting the ground contact time, irrespective of their speed.

The phenomenon in question about kangaroo jumping is rooted in physics and biomechanics. As kangaroos gain speed, more kinetic energy is stored and thus converted into potential energy when they're in the air. This essentially means the kangaroo uses most of their energy while airborne rather than contacting the ground. Hence, ground contact time stays the same no matter what the kangaroo's speed is.

Firstly, the kangaroo's tendons work like a spring; as it lands, much of its kinetic energy is converted to elastic potential energy in the tendons. This act transforms the downward motion (gravitational potential energy) into an upward motion (kinetic energy) through elastic potential energy, which is stored as the tendons stretch. This is similar to the motion observed in a bungee jumper or an elastic band when stretched.

Additionally, the kangaroo's landing is cushioned by bending its hind legs, reducing the impact force over the contact time. With each hop, the kangaroo's kinetic energy increases when it leaves the ground while potential energy increases during 'flight' due to gravity. This establishes a cycle between kinetic and potential energy, with energy oscillating back and forth as the kangaroo continues to hop. Such energy efficiency is what enables the kangaroo to maintain the same ground contact time irrespective of its speed.

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Answer:

[tex]8.0\mu C[/tex]

Explanation:

We are given that

[tex]f=1.6 Hz[/tex]

[tex]q=3.0\mu C=3.0\times 10^{-6} C[/tex]

[tex]1\mu C=10^{-6} C[/tex]

Current,I=[tex]75\mu A=75\times 10^{-6} A[/tex]

[tex]1\mu A=10^{-6} A[/tex]

We have to find the maximum charge of the capacitor.

Charge on the capacitor,[tex]q=q_0cos\omega t[/tex]

[tex]\omega=2\pi f=2\pi\times 1.6=3.2\pi rad/s[/tex]

[tex]3\times 10^{-6}=q_0cos3.2\pi t[/tex]....(1)

[tex]I=\frac{dq}{dt}=-q_0\omega sin\omega t[/tex]

[tex]75\times 10^{-6}=-q_0(3.2\pi)sin3.2\pi t[/tex]....(2)

Equation (2) divided by equation (1)

[tex]-3.2\pi tan3.2\pi t=\frac{75\times 10^{-6}}{3\times 10^{-6}}=25[/tex]

[tex]tan3.2\pi t=-\frac{25}{3.2\pi}=-2.488[/tex]

[tex]3.2\pi t=tan^{-1}(-2.488)=-1.188rad[/tex]

[tex]q_0=\frac{q}{cos\omega t}=\frac{3\times 10^{-6}}{cos(-1.188)}=8.0\times 10^{-6}=8\mu C[/tex]

Hence, the maximum charge of the capacitor=[tex]8.0\mu C[/tex]

Final answer:

The tension in the cable connecting the drum to the bucket is equal to the mass of the bucket times the difference between the acceleration due to gravity and the bucket's downward acceleration.

Explanation:

To determine the tension in the cable between the drum and the bucket, we can apply Newton's second law to the hanging mass. The net force acting on the bucket is the difference between the weight of the bucket and the tension in the rope:

[tex]F_{net}[/tex] = mg - T

Since we know the bucket has a downward acceleration a, we can write Newton's second law as:

ma = mg - T

Where m is the mass of the bucket, g is the acceleration due to gravity, and T is the tension in the rope. We can rearrange the equation to solve for T:

T = mg - ma

T = m(g - a)

So the tension in the cable is equal to the mass of the bucket times the difference between the acceleration due to gravity and the downward acceleration of the bucket.

The question is not complete, so i have attached an image with the complete question.

Answer:

A) speed of block a before it hits block B;v = √2gR

B) speed of combined blocks after collision;V_ab = ½√2gR

C) kinetic energy lost = ½MgR

D) temperature change; ∆t = ½gR/c

E) Additional thermal energy;W_f = 2μMgL

Explanation:

A) From conservation of energy, potential energy = kinetic energy.

Thus; P.E = K.E

So, mgr = ½mv²

m will cancel out to give;

gr = v²

So, v = √2gR

B) from conservation od momentum, momentum before collision = momentum after collision.

Thus;

M_a•V_a = M_ab•V_ab

Where;

M_a is the mass of block A

V_a is speed of block A

M_ab is the mass after collision

V_ab is speed after collision

So, making V_ab the subject, we have;

V_ab = M_a•V_a/M_ab

Now from answer in a above,

V_a = V = √2gR

Also, M_a = M and M_ab = 2M because it's the sum of 2 masses after collision.

Thus;

V_ab = (M√2gR)/(2M)

M will cancel out to give;

V_ab = ½√2gR

C) From the work-kinetic energy theorem, the net work done on the object is equal to the change in the kinetic energy of the object.

Thus;

W_net = K_f - K_i = ½m_ab•v_ab²- ½m_a•v_a²) = ∆K.

We have seen that:

M_a = M and M_ab = 2M

V_a = √2gR and V_ab = ½√2gR

Thus;

∆K = ½(2M•(½√2gR)²) - ½(M•(√2gR)²)

∆K = ½MgR - MgE

∆K = -½MgR

Negative sign means loss of energy.

Thus kinetic energy lost = ½MgR

D) The formula for heat energy is given by;

Q = m•c•∆t

Thus, change in temperature is;

∆t = Q/Mc

Q is the heat energy, thus Q = ½MgR

Thus;∆t = ½MgR/(Mc)

M will cancel out to give;

∆t = ½gR/c

E) Additional thermal energy is gotten from;

Work done by friction;W_f = F_f x d

Where;

F_f is frictional force given by μF_n. F_n is normal force

d is distance moved

Thus;

W_f = μF_n*d

Mass after collision was 2M,thus, F_n = 2Mg

We are told to express distance in terms of L

Thus;

W_f = μ2Mg*L

W_f = 2μMgL

The problem involves principles of energy conservation and kinetic friction. As the block slides down, it converts potential energy to kinetic, collides with another block, and moves together along a rough surface. The work done against friction equals energy at the collision point yielding the expression for velocity after collision V' = sqrt(2µgl).

The important concept here is conservation of energy. As block A slides down, it is converting potential energy into kinetic energy. At point X, the potential energy is maximum because it's at the highest point. The potential energy is given by M*g*R. By the time it reaches point Y, the potential energy has turned into kinetic energy which results in the block's speed (½MV²) at point Y.

After the collision, the two blocks move together with a new mass of 2M and new velocity V': MV = 2MV'. In the horizontal section YZ, due to kinetic friction, the blocks lose energy as work done against friction which causes them to stop.

The work done against friction is: F *d = µ*2M*g*l. Equating the energy at point Y after collision to work done against friction, we get: 2*½MV'² = µ*2M*g*l. Resolving gives the expression V' = sqrt(2µgl).

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The final temperature of the refrigerant in the piston-cylinder device can be determined via lookup and interpolation from standard tables, based on the final pressure of 0.4 MPa. The work done by the refrigerant as it expands and drops in pressure can be calculated via the first law of thermodynamics, specifically accounting for changes in internal energy as work done by the system.

This is an example of a reversible adiabatic expansion process involving a refrigerant (refrigerant-134a) in a piston-cylinder device. The process follows the behavior of ideal gases, and so we can make use of the first law of thermodynamics and other gas laws to determine the final temperature and the work done by the refrigerant.

To find the final temperature, we incorporate the specifics of the refrigerant-134a, which is defined by specific tables found in thermodynamics textbooks, providing details of temperature at various pressures. Interpolation between table values may be necessary to find the exact temperature at 0.4 MPa. Generally, the temperature will fall as the refrigerant expands and the pressure lowers, as defined by the ideal gas law (pV=nRT).

For the work done by the refrigerant, we'd need to apply the first law of thermodynamics, which in this case can be simplified as ΔU = W because there's no heat transfer in an adiabatic process (dQ=0). Changes in the internal energy of the gas would translate directly into work done on the piston. We calculate work using the formula -P(ΔV), where the negative sign indicates work being done by the system.

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(a) The final temperature in the cylinder is approximately 18.0°C.

(b) The work done by the refrigerant during the expansion is approximately 7.72 kJ.

To solve this problem, we need to determine two things: the final temperature of the refrigerant-134a vapor and the work done by the refrigerant during the expansion.

Given that the process is reversible and the system is insulated, this suggests that the expansion process is isothermal for an ideal gas, but since refrigerant-134a is not an ideal gas, we will use the refrigerant tables to find the properties.

Given Data:

Step-by-Step Solution:

(a) Determine the Final Temperature in the Cylinder:

1. Find the Initial and Final States Using Refrigerant-134a Tables:

Look up the refrigerant-134a property tables for [tex]\( P = 0.8 \)[/tex] MPa.

At [tex]\( P = 0.8 \)[/tex] MPa, the refrigerant-134a is in the saturated vapor region. The saturated temperature and properties at this pressure are:

Similarly, look up the refrigerant-134a property tables for [tex]\( P = 0.4 \)[/tex] MPa.

At [tex]\( P = 0.4 \)[/tex] MPa, the refrigerant-134a is also in the saturated vapor region. The properties at this pressure are:

The final temperature of the refrigerant-134a after the expansion, which is at 0.4 MPa, is approximately [tex]\( 18.0^\circ \text{C} \)[/tex].

(b) Determine the Work Done by the Refrigerant:

1. Calculate the Initial and Final Mass:

Using the specific volume [tex]\( v_g \)[/tex] to find the mass:

[tex]\[ m = \frac{V_1}{v_g (P_1)} \][/tex]

For [tex]\( P_1 = 0.8 \)[/tex] MPa:

[tex]\[ m = \frac{0.05 \text{ m}^3}{0.0769 \text{ m}^3/\text{kg}} \approx 0.650 \text{ kg} \][/tex]

For [tex]\( P_2 = 0.4 \)[/tex] MPa, verify if the mass is the same (which it is for an ideal gas in a closed system):

[tex]\[ V_2 = m \times v_g (P_2) = 0.650 \text{ kg} \times 0.0917 \text{ m}^3/\text{kg} \approx 0.0597 \text{ m}^3 \][/tex]

The new volume [tex]\( V_2 \)[/tex] can be verified as:

[tex]\[ V_2 = \text{initial volume} \times \frac{v_g (P_2)}{v_g (P_1)} \][/tex]

Which confirms our calculation.

2. Calculate the Work Done During Expansion:

For a reversible expansion in an insulated system (isoenthalpic process), the work done \( W \) can be approximated using:

[tex]\[ W = \int_{V_1}^{V_2} P \, dV \][/tex]

The work done for an ideal gas (but approximate here for refrigerants) can be simplified using:

[tex]\[ W = \text{P}_1 \times V_1 \ln \left( \frac{V_2}{V_1} \right) \][/tex]

For refrigerants, use specific volume:

[tex]\[ W = \text{P}_1 \times V_1 \left[ \frac{v_g (P_2)}{v_g (P_1)} - 1 \right] \][/tex]

Substituting:

[tex]\[ W = 0.8 \text{ MPa} \times 0.05 \text{ m}^3 \left( \frac{0.0917}{0.0769} - 1 \right) \][/tex]

[tex]\[ W = 0.8 \times 10^3 \text{ Pa} \times 0.05 \text{ m}^3 \left( 1.193 - 1 \right) \][/tex]

[tex]\[ W = 40 \text{ kJ} \times 0.193 \approx 7.72 \text{ kJ} \][/tex]

Answer:

Explanation:

Given that,

Three children of masses and their position on the merry go round

M1 = 22kg

M2 = 28kg

M3 = 33kg

They are all initially riding at the edge of the merry go round

Then, R1 = R2 = R3 = R = 1.7m

Mass of Merry go round is

M =105kg

Radius of Merry go round.

R = 1.7m

Angular velocity of Merry go round

ωi = 22 rpm

If M2 = 28 is moves to center of the merry go round then R2 = 0, what is the new angular velocity ωf

Using conservation of angular momentum

Initial angular momentum when all the children are at the edge of the merry go round is equal to the final angular momentum when the second child moves to the center of the merry go round

Then,

L(initial) = L(final)

Ii•ωi = If•ωf

So we need to find the initial and final moment of inertia

NOTE: merry go round is treated as a solid disk then I= ½MR²

I(initial)=½MR²+M1•R²+M2•R²+M3•R²

I(initial) = ½MR² + R²(M1 + M2 + M3)

I(initial) = ½ × 105 × 1.7² + 1.7²(22 + 28 + 33)

I(initial) = 151.725 + 1.7²(83)

I(initial) = 391.595 kgm²

Final moment of inertial when R2 =0

I(final)=½MR²+M1•R²+M2•R2²+M3•R²

Since R2 = 0

I(final) = ½MR²+ M1•R² + M3•R²

I(final) = ½MR² + (M1 + M3)• R²

I(final)=½ × 105 × 1.7² + ( 22 +33)•1.7²

I(final) = 151.725 + 158.95

I(final) = 310.675 kgm²

Now, applying the conservation of angular momentum

L(initial) = L(final)

Ii•ωi = If•ωf

391.595 × 22 = 310.675 × ωf

Then,

ωf = 391.595 × 22 / 310.675

ωf = 27.73 rpm

So, the final angular momentum is 27.73 revolution per minute

Answer:

a

The velocity is  [tex]v =17.98 \ m/s[/tex]

b

The diameter is  [tex]d = 0.00184m[/tex]

Explanation:

The diagram of the set up is shown on the first uploaded image  

From the question we are told that

    The height of the water tank is [tex]h = 20.0 \ m[/tex]

    The position of the hole [tex]p_h = 16.5m[/tex] below  water level

     The  rate  of water flow [tex]\r V = 2.90 *10^{-3} m^3 /min = \frac{2.90 *10^{-3}}{60} = 0.048*10^{-3} m^3/s[/tex]

  According to Bernoulli's theorem position of the hole

              [tex]\frac{P_o + h \rho g}{\rho} + \frac{1}{2} u^2 = \frac{P_o}{\rho } + \frac{1}{2} v^2[/tex]

Where  u is  the initial speed the water through the hole = 0 m/s

              [tex]P_o[/tex] is the atmospheric pressure

            [tex]\frac{P_o }{\rho} + \frac{ h \rho g}{\rho} + 0 = \frac{P_o}{\rho } + \frac{1}{2} v^2[/tex]

                   [tex]v = \sqrt{2gh}[/tex]

Substituting value

           [tex]v = \sqrt{2 * 9.8 * 16.5 }[/tex]

              [tex]v =17.98 \ m/s[/tex]

The Volumetric flow rate is mathematically represented as

          [tex]\r V = A * v[/tex]

     Making A the subject

              [tex]A = \frac{\r V}{v}[/tex]

 substituting value  

             [tex]A = \frac{0.048 *10^{-3}}{17.98}[/tex]

                 [tex]= 2.66*10^{-6}m^2[/tex]

Area is mathematically represented as

        [tex]A = \frac{\pi d^2}{4}[/tex]

  making d the subject

         [tex]d = \sqrt{\frac{4*A}{\pi} }[/tex]

  Substituting values

        [tex]d = \sqrt{\frac{4 * 2.67 *10^{-6}}{3.142} }[/tex]

          [tex]d = 0.00184m[/tex]

Final answer:

Kinetic energy is the energy caused by the vibration of particles in a medium. It is calculated using the formula KE = 0.5mv², where KE is the kinetic energy, m is the mass of the particle, and v is its velocity.

Explanation:

Kinetic energy is the energy an object has because of its motion. It is caused by the vibration of particles in a medium such as steel, water, or air. Kinetic energy is calculated as one-half the product of the mass of the particle and the square of its speed.

For example, when a rock is thrown into a pond or when a swimmer splashes the water's surface, the kinetic energy of the vibrating water particles is generated. Similarly, mechanical sound waves also have kinetic energy due to the movement of air particles and the potential energy caused by the elasticity of the material through which the sound propagates.

In summary, kinetic energy arises from the motion of particles in a medium and can be calculated using the formula KE = 0.5mv², where KE is the kinetic energy, m is the mass of the particle, and v is its velocity.