What must the charge (sign and magnitude) of a particle of mass 1.41 gg be for it to remain stationary when placed in a downward-directed electric field of magnitude 670 N/CN/C ?

Explanation:

When copper rod is heated , its length increases

The increase in length can be found by the relation

L = L₀ ( 1 + α ΔT )

here L is the increased length and L₀ is the original length

α  is the coefficient of linear expansion and ΔT is the increase in temperature .

The increase in length = L - L₀ = L₀ x α ΔT

Substituting all these value

Increase in length = 27.5 x 1.7 x 10⁻⁵ x 35.9

= 1.87 x 10⁻² m

Answer:0.01678325m

Explanation:

Original length(L1)=27.5m

Temperature rise(@)=35.9°C

Linear expansivity(α)=0.000017

Length increase(L)=?

L=α x L1 x @

L=0.000017 x 27.5 x 35.9

L=0.01678325m

Answer:

None of the above.

The correct answer would be momentum

Answer:

Explanation:

The two objects free-fall at the same rate of acceleration, thus giving them the same speed when they hit the ground. The heavier object however has more momentum since momentum takes into account both the speed and the mass of the object (p=m*v).

Answer:

0.001067 Wb

Explanation:

Parameters given:

Magnetic field, B = 0.77 T

Angle, θ = 30º

Width = 0.4cm = 0.04m

Length = 0.4cm = 0.04m

Magnetic flux, Φ(B) is given as:

Φ(B) = B * A * cosθ

Where A is Area

Area = length * width = 0.04 * 0.04 = 0.0016 m²

Φ(B) = 0.77 * 0.0016 * cos30

Φ(B) = 0.00167 Wb

Answer:

1.07×10⁻⁵ Wb

Explanation:

Using

Φ = BAcosθ.................. Equation 1

Where Φ = magnetic Flux, B = magnetic Field, A = Area of the rectangular loop, Angle between the loop and the Field.

But

A = L×W........................ Equation 2

Where L = Length, W = Width.

Substitute equation 2 into equation 1

Φ = BLWcosΦ................ Equation 3

Given: B = 0.77 T, L = 0.4 cm = 0.004 m, W = 0.4 cm = 0.004 m, Ф = 30°

Substitute into equation 3

Ф = 0.77(0.004)(0.004)cos30

Ф  = 1.07×10⁻⁵ Wb.

Hence the magnetic Field through the loop = 1.07×10⁻⁵ Wb.

Answer:

0

Explanation:

       Q +(-Q) =0.

Final answer:

When identical conducting spheres A and B with charges of –5 nC and –3 nC respectively are brought into contact, the total initial charge of –8 nC is shared equally. Upon separation, each sphere holds a final charge of –4 nC.

Explanation:

When two conducting spheres, A and B, with different charges are brought into contact, the total charge is shared equally between them. Given sphere A has a charge of –5 nC and sphere B has a charge of –3 nC, the total initial charge is –8 nC. This charge will be evenly distributed between both spheres when they touch.

Therefore, after spheres A and B are brought into contact and then separated, the final charge on each sphere will be half of the total charge. This final charge will be calculated as follows:

Thus, after separation, both sphere A and B will have a charge of –4 nC each.

Answer:0.01678325m

Explanation:

Original length(L1)=27.5m

Temperature rise(@)=35.9°C

Linear expansivity of copper(α)=0.000017

Length increment(L)=?

L=α x L1 x @

L=0.000017 x 27.5 x 35.9

L=0.01678325m

Answer:

0.0168

Explanation:

Remember about significant digits (there must be 3)

Final answer:

The temperature of the cold reservoir for an engine with a second-law efficiency of 85% that absorbs 480J of heat from a hot reservoir at 300C and dumps 300J into the cold reservoir is 358.125 K.

Explanation:

The question asks for the temperature of the cold reservoir for an engine with a second-law efficiency of 85% that absorbs 480J of heat from a hot reservoir at 300C and dumps 300J into the cold reservoir.

The second-law efficiency η of a heat engine is defined as the ratio of the work output W to the heat input Qh at the high temperature, while for a Carnot engine, the efficiency can also be related to the temperatures of the hot (Th) and cold (Tc) reservoirs as:

η = 1 - (Tc/Th)

Given the heat absorbed (Qh = 480 J) and the heat rejected (Qc = 300 J), we can calculate the work done (W = Qh - Qc) which is 180 J here. We know that Th is the temperature of the hot reservoir in kelvin, which we obtain by converting 300C to kelvin (Th = 573 K). Note that 0 degrees Celsius is equivalent to 273 K.

Using the given second-law efficiency:

η = W / Qh = 180 J / 480 J = 0.375

For a Carnot engine:

η = 1 - (Tc/Th)

0.375 = 1 - (Tc/573 K)

Tc = 573 K * (1 - 0.375)

Tc = 358.125 K

The temperature of the cold reservoir for this engine is therefore 358.125 K.

Answer:

[tex]F=2187\ N[/tex]

Explanation:

Given:

Here as given in the question the bullet penetrates the target by the given displacement of the bullet into it. During this process it faces deceleration and hence it comes to rest.

Now using the equation of motion:

[tex]v^2=u^2+2a.s[/tex]

where:

[tex]a=[/tex] acceleration of the bullet

[tex]0^2=270^2+2a\times 0.25[/tex]

[tex]a=145800\ m.s^{-2}[/tex]

Now the force of resistance offered by the target in stopping it:

[tex]F=m.a[/tex]

[tex]F=0.015\times145800[/tex]

[tex]F=2187\ N[/tex]

Answer:

102.8 μH

Explanation:

The (self) inductance of a coil based on its own geometry is given as

L = (μ₀N²A)/l

where

μ₀ = magnetic constant = (4π × 10⁻⁷) H/m

N = number of turns = 189

A = Cross sectional Area = (πD²/4) = (π×0.00799²/4) = 0.00005014 m²

l = length of the solenoid = 2.19 cm = 0.0219 m

L = (4π × 10⁻⁷ × 189² × 0.00005014)/0.0219

L = 0.0001028131 H = (1.028 × 10⁻⁴) H = (102.8 × 10⁻⁶) H = 102.8 μH

Hope this Helps!!!

Answer:

Heat flux = (598.3î + 204.3j) W/m²

a) Magnitude of the heat flux = 632.22 W/m²

b) Direction of the heat flux = 18.85°

Explanation:

- The correct question is the first image attached to this solution.

- The solution to this question is contained on the second and third images attached to this solution respectively.

Hope this Helps!!!

Answer:

126.03 ft/sec

Explanation:

From the question above,

V₁ = VsinФ............. Equation 1

Where V₁ = vertical component of the velocity, V = Velocity acting on the x-y plane, Ф = angle to the horizontal.

Given: V = 1446 ft/sec, Ф = 5°

Substitute into equation 1

V₁ = 1446sin(5)

V₁  = 1446(0.0872)

V₁  = 126.03 ft/sec.

Hence the vertical component of the velocity = 126.03 ft/sec

Answer:

27588

Explanation:

Given,

Speed of the centrifuge = 15000 rpm

time, t = 220 s

Revolution = ?

1 min = 60 s

Speed of centrifuge = [tex]\dfrac{15000}{60}[/tex]

                                 = 250 rps

Initial angular speed = 0 rps

[tex]\alpha = \dfrac{250-0}{220}[/tex]

[tex]\alpha = 1.14\ rev/s^2[/tex]

Now,

revolution

[tex]\theta = \omega_0 t - \dfrac{1}{2}\alpha t^2[/tex]

[tex]\theta = 0 + \dfrac{1}{2}\times 1.14\times 220^2[/tex]

[tex]\theta = 27588\ rev[/tex]

Hence, the number of turns is equal to 27588.

Answer:

1. the electromagnetic wave.

Explanation:

Mathematically,

wavelength = velocity ÷ frequency

A mechanical wave is a wave that is not capable of transmitting its energy through a vacuum. Mechanical waves require a medium in order to transport their energy from one location to another. A sound wave is an example of a mechanical wave. Sound waves are incapable of traveling through a vacuum.

Electromagnetic waves of different frequency are called by different names since they have different sources and effects on matter, increasing frequency decreases wavelength.

Sound waves (which obviously travel at the speed of sound) are much slower than electromagnetic waves (which travel at the speed of light.)  

Electromagnetic waves are much faster than sound waves and If the Velocity of the wave increases and the frequency is constant, the wavelength also increases.

Answer: because 1.8 is an outlier.

Explanation: it can been seen that from the data given to us, 1.7, 1.8, 1.9, 1.9 are all close to each other with a difference of 0.2 or 0.1.

By considering 0.8, this value (0.8) is at a very far distance away from other observational data with a difference of at least 0.9.

This henceforth makes 0.8 an outlier ( a data which is at a far distance away from other observational data)

The 0.8-second measurement is likely inconsistent with the others due to its significant deviation from the clustered times around 1.8 seconds, pointing to it as an outlier potentially caused by an experimental error.

One quantitative reason to suspect that the time value of 0.8 seconds is inconsistent with the other measurements of 1.8, 1.7, 1.9, and 1.9 seconds is the concept of measurement uncertainties and standard deviation. When comparing the given times, we see that the majority of the values are clustered around 1.8 seconds, suggesting a certain range of natural variability in repeated measurements. However, 0.8 seconds is significantly lower than the others, indicating that it is an outlier and might have been affected by experimental error or a faulty measurement. The use of mean and standard deviation can give us a formal way to assess consistency among data. If we calculated the mean and standard deviation, we would likely find 0.8 seconds to fall outside an acceptable range based on the precision of our measurements, reinforcing the idea that this result is inconsistent with the others.

Answer:

Distance of the object from eye is approx 4.52 m

Explanation:

As we know that the object subtend a small angle on both the eyes which is given as

[tex]\theta = 0.995 degree[/tex]

now we know that the distance between two eyes is given as

d = 7.50 cm

so we have

[tex]angle = \frac{arc}{Radius}[/tex]

so here the radius is same as the distance from eye while arc is the distance between two eyes

so we have

[tex]0.95 \frac{\pi}{180} = \frac{7.50}{R}[/tex]

[tex]R = 452 cm[/tex]

Answer:

T(max) = 1.17 × 10⁻⁴Nm

= 117μNm

Explanation:

T = BIA sinθ

A = area enclosed

θ = angle between normal plane

for max. torque θ = 90, (sin90° =1)

T = BIA sin90°

T= BI (πd/4)

T = [tex]T_m_a_x = \frac{1}{4} (2.98 * 10^-^3)(5)(\pi )\\T_m_a_x = 1.17 * 10^-^4Nm\\T_m_a_x = 117UNm[/tex]

Answer:

The metal wire will stretch by [tex]2 \delta L[/tex]

Explanation:

[tex]T = \frac{kA \delta L}{L}[/tex]......................................(1)

Where T = Tension applied

ΔL = Extension

L = length

k = constant

T₁ = T₂ = T

A₁ = A₂ =A

L₁ = L

L₂ = 2L

(ΔL)₁ = ΔL

(ΔL)₂ = ?

From equation (1)

[tex]TL/kA = \delta L[/tex].....................(2)

[tex]TL/kA = (\delta L) ........................(3)\\ 2TL/kA = (\delta L)_{2} ........................(4)[/tex]

Divide (4) by (3)

[tex]\frac{(\delta L)_{2} }{\delta L} =\frac{\frac{2TL}{kA} }{\frac{TL}{kA} } \\\frac{(\delta L)_{2} }{\delta L} = 2\\ (\delta L)_{2} = 2\delta L[/tex]

If the starting length is twice, the extension is doubled as well.

Given that;

Length of metal wire = L

Stretch = ΔL

So,

It will stretched to a length of 2L.

Although when tension is continuous, the length of the extension is proportional to the size of the initials.

As a result, if the starting length is doubled, the extension will be twice as well.

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Answer:

THE KINETIC ENERGY OF THE SMALLER CHILD IS LESS THAN THAT OF THE BIGGER CHILD

Explanation: Kinetic energy is the energy that is exerted on a body that is in motion, kinetic energy is affected by both the mass of the object and the velocity of the object.

Mathematically,Kinetic energy is represented as follows;

K.E=1/2M[tex]V^{2}[/tex]

Where M represents the mass of the object in kilograms and V represents velocity of the moving object measured in meters per seconds.

The higher the weight of the object the higher the kinetic energy of the object which means the bigger child will have a higher kinetic energy than the smaller child.

Both children start with the same potential energy, which is converted into kinetic energy as they move; hence, both have the same kinetic energy upon landing in the water. However, the distribution of this energy according to the mass and velocity of each child results in a higher velocity for the smaller child and a lower velocity for the larger child.

The subject of this question is in the field of Physics, specifically on the concept of conservation of energy. Both children start from the same height, so they both have the same potential energy. When they land in the pool, this potential energy has been converted into kinetic energy.

The kinetic energy of an object in motion is given by the formula [tex]KE = 1/2 mv^2[/tex], where 'm' is the mass of the object, and 'v' is its velocity. Since we're told to ignore factors like friction and air resistance, we can say that when both children land in the pool they have the same kinetic energy, and the only difference is how this energy is distributed between the mass and the velocity.

The smaller child, having less mass, will have to have a greater velocity to account for the same amount of kinetic energy. The bigger child, with more mass, will have a smaller velocity. So in terms of kinetic energy, it is the same for both children when they reach the water.

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Complete question:

A spring of negligible mass has force constant k = 1600 N/m. (a) How far must the spring be compressed for 3.20 J of potential energy to be stored in it? (b) You place the spring vertically with one end on the floor. You then drop a 1.40-kg book onto it from a height of 0.800 m above the top of the spring. Find the maximum distance the spring will be compressed.

Answer:

(a) 0.063 m

(b) 0.126 m

Explanation:

Given;

force constant, K =  1600 N/m

Part (a)

Elastic potential energy is given as;

U = ¹/₂Kx²

where;

x is the extension in the spring

[tex]x = \sqrt{\frac{2U}{K} } = \sqrt{\frac{2*3.2}{1600} } = 0.063 \ m[/tex]

Part (b)

given;

mass of the book, m = 1.4 kg

height above the spring from which the book was dropped, h = 0.8 m

From the principle of conservation of energy;

Gravitational potential energy = Elastic potential energy

mgH = ¹/₂Kx²

H is the total vertical distance from floor to 0.8 m =  maximum distance the spring will be compressed + h

let the maximum distance = A

mg(A+h) = ¹/₂KA²

1.4 x 9.8(A + 0.8) = ¹/₂ x 1600A²

13.72 (A + 0.8) = 800A²

13.72A + 10.976 = 800A²

800A² - 13.72A -  10.976 = 0

This is a quadratic equation, and we solve using formula method, where a = 800, b = - 13.72 and c = - 10.976

A = 0.126 m

Final answer:

The question involves determining the maximum compression of a spring by using energy conservation to equate the gravitational potential energy of a falling book to the elastic potential energy of the spring. The calculation needs the spring constant, which is not given in the question.

Explanation:

The student's question involves finding the maximum compression of the spring after dropping a 1.40 kg book on it from a certain height. This is a classic energy conservation problem in physics, where the gravitational potential energy of the book is converted into the elastic potential energy of the spring upon impact.

Firstly, the initial gravitational potential energy (Ug) of the book can be calculated using Ug = mgh, where m is the mass of the book, g is the acceleration due to gravity (approximately 9.81 m/s2), and h is the height from which the book is dropped.

The elastic potential energy stored in the spring (Ue) at maximum compression can be expressed as Ue = (1/2)kx2, where k is the spring constant and x is the compression of the spring.

Assuming no energy is lost due to friction or air resistance, energy conservation dictates that the initial gravitational potential energy will equal the elastic potential energy at the point of maximum compression. Therefore, mgh = (1/2)kx2. To find the maximum compression, x, you would rearrange this equation to solve for x and plug in the values for m, g, h, and k.

The actual computation requires the value of the spring constant k, which was not provided in the question. If it were provided, one would simply calculate the maximum compression using the specified formula.

Answer: a) [tex]L = 10206 kg \cdot \frac{m^{2}}{s}[/tex], b) [tex]\omega_{f} = 20.25 rad/s[/tex], c) [tex]L = 10206 kg \cdot \frac{m^{2}}{s}[/tex]

Explanation:

The angular momentum of a system of particles rotating around an axis is:

[tex]L = \sum_{i=1}^{7} r_{i}^{2} \cdot m_{i} \cdot \omega[/tex]

a) The previous expression can be simplified to find the initial angular momentum:

[tex]L = 7 \cdot r^{2} \cdot m \cdot \omega\\L = 7 \cdot (9 m)^2 \cdot (2kg) \cdot (9 \frac{rad}{s} )\\L = 10206 kg \cdot \frac{m^{2}}{s}[/tex]

b) The new angular speed is calculated from the Principle of Angular Momentum Conservation, since there are no external forces influencing over the system:

[tex]L = 7 \cdot r_{f}^2 \cdot (m) \cdot \omega_{f}[/tex]

[tex]\omega_{f}=\frac{L}{7 \cdot r_{f}^2\cdot m}[/tex]

[tex]\omega_{f} = 20.25 rad/s[/tex]

c) The angular momentum remains constant, since there is no information that indicates the presence of external forces influencing the system.

(A)  The initial angular momentum of the object is 10206 kgm²/s

(B)  The new angular speed is 20.25 rad/s

(C)  The angular momentum does not change

All 7 objects have a mass of m = 2kg. The length of the spokes is R = 9m.

(A) The angular momentum of a system is given by:

L = Iω

where I is the moment of inertia of the system

ω is the angular speed = 9 rad/s (given)

L = 7×mR²×ω

L = 7×(2×9²)×9 kgm²/s

L = 10206 kgm²/s

(B) According to the law of conservation of angular momentum, the angular momentum of the system must remain conserved.

So, the new angular momentum is:

L' =  7×(2×6²)×ω'

where ω' is the new angular momentum

10206 = 7×(2×6²)×ω'

ω' = 20.25 rad/s

(C) The new angular momentum of the object is same as the original angular momentum of the object since there is no dissipative force so the angular momentum must be conserved.

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Answer:

Explanation:is in the picture below

Torque is the force's twisting action about the axis of rotation magnitude of the net torque acting on the cylinders about the rotation axis will be 331.402 Nm.

Torque is the force's twisting action about the axis of rotation. Torque is the term used to describe the instant of force. It is the rotational equivalent of force. Torque is a force that acts in a turn or twist.

The amount of torque is equal to force multiplied by the perpendicular distance between the point of application of force and the axis of rotation.

In the first situation, force is delivered in the right direction, hence torque is applied upwards.

The value of torque for case 1

[tex]\rm \tau_1=4.49\times 2.50\\\\\rm \tau_1=11.25 Nm[/tex]

The value of torque for case 2

[tex]\rm \tau_2=91.3\times 1.14\\\\\rm \tau_2=104.08[/tex]

The resultant value of torque will be;

[tex]\rm \tau =\sqrt{(11.25)^2+(41.90)^2} \\\\\ \rm \tau =331.402 Nm[/tex]

Hence the magnitude of the net torque acting on the cylinders about the rotation axis will be 331.402 Nm.

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Final answer:

The magnitude of the net torque acting on the system of two cylinders is 21.633 N⋅m.

Explanation:

The question asks us to calculate the net torque on a system of two cylinders with ropes exerting forces at different radii. Torque (τ) is calculated by the product of the radius (r), the force (F), and the sine of the angle (θ) between them, τ = rFsinθ. In this case, as both forces are perpendicular to the radii, sinθ = 1, simplifying the torque to τ = rF.

For the cylinder with radius 2.50 m, the torque exerted by the 4.49 N force is τ = 2.50 m * 4.49 N = 11.225 N⋅m. For the cylinder with radius 1.14 m, the torque from the 9.13 N force is τ = 1.14 m * 9.13 N = 10.4082 N⋅m. Since both torques are acting to rotate the system in the same direction (counterclockwise), we can sum them to find the net torque.

The magnitude of the net torque is 11.225 N⋅m + 10.4082 N⋅m = 21.6332 N⋅m, or to three decimal places, 21.633 N⋅m.

Explanation:

When any body rotates about its axis , the angular momentum of the body remains constant .

Thus the product of moment  of inertia and its angular velocity remains constant .

In first case

The moment of inertia of cylinder = 41.8 kg m²

and Angular velocity = 2.27 rad/s

Thus angular momentum L₁ = 41.8 x 2.27 N-m s

In the second case

The moment of inertia = 41.8 + 38.0 = 79.8 kg m²

Suppose the angular velocity = ω

Thus angular momentum L₂ = 79.8 x ω

But according to principle of conservation of momentum

L₁ = L₂

41.8 x 2.27 = 79.8 x ω

Thus ω = [tex]\frac{41.8x2.27}{79.8}[/tex]

ω = 1.19 rad/s

Answer:

Final Angular Velocity=[tex]\omega[/tex]=1.189 rad/s

Explanation:

Given Data:

Moment of Inertia of 1st cyclinder=[tex]I_1=41.8 kg/m^{2}[/tex]

Angular Velocity of 1st cyclinder=[tex]\omega_1[/tex]=2.27 rad/s

Moment of Inertia of After contact=[tex]I_2=(41.8+38) kg/m^{2}=79.8 kg/m^{2}[/tex]

Required:

Final Angular velocity =[tex]\omega[/tex]=?

Formula:

Angula Momentum=L=[tex]I\omega[/tex]

Solution:

According to the conservation of angular momentum:

[tex]L_1=L_2[/tex]

[tex]I_1 \omega_1=I_2\omega\\41.8*2.27=(41.8+38)*\omega\\\omega=\frac{41.8*2.27}{79.8}\\\omega= 1.189\ rad/s[/tex]

Final Angular Velocity=[tex]\omega[/tex]=1.189 rad/s

Answer:

The wavelength is 503 nm  

Explanation:

Considering constructive interference , this means that route(path) difference is equal to the product of order of fringe and wavelength of the light

 i.e   dsinθ  = m[tex]\lambda[/tex]

Where [tex]\lambda[/tex]  is the wavelength of light  and m is the order of the fringe

    Looking at θ to be very small , sin θ can be approximated to  θ

              and   [tex]\theta \approx \frac{x}{l}[/tex]

Substituting this into the above equation

           [tex]d[\frac{x}{l} ] =m\lambda[/tex]

      making x the subject

                    [tex]x =\frac{m\lambda l}{d}[/tex]

      This above equation will give the value of the distance of the [tex]m^{th}[/tex] order fringe of the wavelength [tex]\lambda[/tex] from the central fringe

       Replacing with the value given in the question we have

  [tex]\lambda[/tex] = 710 nm  m = 2  d =0.66 mm , l = 2.0 m

                  [tex]x = \frac{(2)(710nm)(2.0m)[\frac{10^9}{1m} ]}{(0.66mm)(\frac{10^6}{1mm} )}[/tex]

                   [tex]=(4.303*10^6nm)[\frac{\frac{1}{10^6}mm }{1nm} ][/tex]

                    [tex]=4.303mm[/tex]

 The separation of the second fringe from central maximum is 4,303 mm    

To obtain the separation of the second order fringe of the unknown light from central maximum

       [tex]x' = 4.303mm - 1.25 mm = 3.053mm[/tex]

Now to obtain the wavelength of this second source

                   from [tex]x = \frac{m\lambda l}{d}[/tex]

                       [tex]\lambda' = \frac{x'd}{ml}[/tex]

Now substituting 3,053 mm for [tex]x'[/tex] 2.0 mm for l , 0.66 mm  for d and 2 for m in the above formula

           [tex]\lambda' =\frac{(3.053mm)(0.66mm)}{(2)(2.0)(\frac{10^3mm}{1m} )}[/tex]

                  [tex]= (503.7*10^{-6}mm)(\frac{10^6nm}{1mm} )[/tex]

                   [tex]=503.7nm[/tex]

In scenario (a), the temperature of the gas decreases since it is an adiabatic process. In scenario (b), the final temperature depends on the initial and final volumes and the presence of heat exchange.

In scenario (a), the gas is slowly drawn into cylinder B by pulling out the piston B while cylinder A remains stationary. Since the cylinders are thermally insulated, there is no heat exchange with the surroundings, and the process is adiabatic. As a result, the temperature of the gas decreases.

In scenario (b), the gas is driven as far as it will go into cylinder B by pushing the piston A at a rate that maintains constant pressure in cylinder A. In this case, the process is isobaric, and the gas expands while exerting work. Since there is thermal contact between the cylinders, heat can be exchanged between the gas and the surroundings, leading to a change in temperature.

To calculate the final temperatures in both scenarios, it is necessary to know the initial pressure and volume of the gas in cylinder A, as well as the final volume of the gas in cylinder B, in each case.

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For the adiabatic free expansion in scenario (a), the final temperature remains the same as the initial temperature. In scenario (b), the gas undergoes adiabatic compression followed by isothermal expansion, resulting in the final temperature being the same as the initial temperature, T.

Final Temperature Calculation in Two Scenarios

Let's explore the gas dynamics in two scenarios involving thermally insulated cylinders A and B containing a perfect monatomic gas at initial temperature T.

Scenario (a): Valve is Fully Opened and Gas is Drawn into B

Initial conditions:

Since cylinder A is insulated and its piston remains stationary, there is no work done on or by the gas in cylinder A. The gas expands into cylinder B, which is an adiabatic free expansion:

The final temperature (T') will be the same as the initial temperature (T). Because the process is adiabatic and involves no work, the internal energy (and thus temperature) of the gas remains unchanged.

Scenario (b): Adiabatic Compression of A and Isothermal Expansion into B

Initial conditions:

We perform an adiabatic process on cylinder A and isothermal expansion into B. The adiabatic compression affects the temperature of the gas in A before the gas is allowed to expand isothermally:

1. Adiabatic compression in A:
For adiabatic processes, TVγ-1 = constant, where γ = 5/3 for a monatomic ideal gas. The final volume of gas in A is V/2 because the gas expands equally into B.

2. Isothermal expansion into B:
After the compression, we allow the gas to expand isothermally into B at constant temperature T. Hence, the final temperature in both cylinders will be T because the gas reaches thermal equilibrium with the environment.

Answer:

mass flow rate = 11.464 kg/sec

Explanation:

given data

P = 200 kPa = 2 × [tex]10^{5}[/tex]  Pa

temperature = 150°C  = 423 k

volumetric flow rate = 7000 liters/s

solution

first we get here molar flow rate that is express as

molar flow rate  = [tex]\frac{Pv}{RT}[/tex]   .................1

put here value

molar flow rate  = [tex]\frac{2 \times 10^5 \times 7}{8.314\times 423}[/tex]  

molar flow rate  = 398.06 mol/sec

and

now we get mass flow rate

mass flow rate = molar flow rate × average  mol weight ..............2

mass flow rate = 398.06 × 28.8

mass flow rate = 11464.3 g/sec

mass flow rate = 11.464 kg/sec

Final answer:

The mass flow rate of air exiting the turbine is calculated using the ideal gas equation and the given conditions of pressure, temperature, and volumetric flow rate. By substituting the values into the equation, the mass flow rate is found to be 11.6 kg/s.

Explanation:

The problem involves finding the mass flow rate of air using the ideal gas model, given the exiting conditions from a turbine. The known variables are the pressure (200 kPa), temperature (150°C which is 423.15 K after conversion from Celsius to Kelvin), and volumetric flow rate (7000 liters/s or 7 m³/s). To find the mass flow rate, we can use the ideal gas equation PV = mRT, where P is the absolute pressure, V is the volume flow rate per unit time, m is the mass flow rate, R is the specific gas constant for air, and T is the absolute temperature.

The specific gas constant for air (R) is 287 J/kg·K, and it can be solved for m as:

m = PV / (RT)

Substituting the given values:

m = (200,000 Pa)(7 m³/s) / (287 J/kg·K)(423.15 K)

m = 11.599 kg/s (rounded to three significant figures)

Therefore, the mass flow rate of the air exiting the turbine is 11.6 kg/s.

Answer:

(A). The the time interval between signals according to an observer on A is 1.09 h.

(B). The time interval between signals according to an observer on B is 1.09 h.

(C). The speed is 0.866c.

Explanation:

Given that,

Time interval = 1.00 h

Speed = 0.400 c

(A). We need to calculate the the time interval between signals according to an observer on A

Using formula of time

[tex]\Delta t=\dfrac{\Delta t_{0}}{\sqrt{1-(\dfrac{v}{c})^2}}[/tex]

Put the value into the formula

[tex]\Delta t=\dfrac{1.00}{\sqrt{1-(\dfrac{0.400c}{c})^2}}[/tex]

[tex]\Delta t=\dfrac{1.00}{\sqrt{1-(0.400)^2}}[/tex]

[tex]\Delta t=1.09\ h[/tex]

(B). We need to calculate the time interval between signals according to an observer on B

Using formula of time

[tex]\Delta t=\dfrac{\Delta t_{0}}{\sqrt{1-(\dfrac{v}{c})^2}}[/tex]

Put the value into the formula

[tex]\Delta t=\dfrac{1.00}{\sqrt{1-(\dfrac{0.400c}{c})^2}}[/tex]

[tex]\Delta t=\dfrac{1.00}{\sqrt{1-(0.400)^2}}[/tex]

[tex]\Delta t=1.09\ h[/tex]

(C). Here, time interval of 2.00 h between signals.

We need to calculate the speed

Using formula of speed

[tex]\Delta t=\dfrac{\Delta t_{0}}{\sqrt{1-(\dfrac{v}{c})^2}}[/tex]

Put the value into the formula

[tex]2.00=\dfrac{1.00}{\sqrt{1-(\dfrac{v}{c})^2}}[/tex]

[tex]\sqrt{1-(\dfrac{v}{c})^2}=\dfrac{1.00}{2.00}[/tex]

[tex]1-(\dfrac{v}{c})^2=(\dfrac{1.00}{2.00})^2[/tex]

[tex](\dfrac{v}{c})^2=\dfrac{3}{4}[/tex]

[tex]v=\dfrac{\sqrt{3}}{2}c[/tex]

[tex]v=0.866c[/tex]

Hence, (A). The the time interval between signals according to an observer on A is 1.09 h.

(B). The time interval between signals according to an observer on B is 1.09 h.

(C). The speed is 0.866c.

Answer:

The correct answer is a

Explanation:

At projectile launch speeds are

X axis     vₓ = v₀ = cte

Y axis     [tex]v_{y}[/tex] = v_{oy} –gt

The moment is defined as

         p = mv

For the x axis

         pₓ = mvₓ = m v₀ₓ

As the speed is constant the moment is constant

For the y axis

        p_{y} = m v_{y} = m (v_{oy} –gt) = m v_{oy} - m (gt)

Speed ​​changes over time, so the moment also changes over time

Let's examine the answer

i   True

ii False.  The moment changes with time

The correct answer is a

In projectile motion without air resistance, the horizontal component of momentum remains constant while the vertical component changes due to gravity. A projectile with initial velocity 2i + j will have a final velocity of 2i - 2j before striking the ground. A projectile with velocity 2i + 3j m/s is ascending towards its maximum height.

The student's question regards the momentum components of a projectile fired at a 45-degree angle assuming no air resistance. The correct answer to the student's multiple-choice question is that statement (i) is True and (ii) is False. This is because, in projectile motion, the horizontal component of momentum, which is dependent on the horizontal component of velocity (Vx), remains constant due to the absence of horizontal forces acting on the projectile (assuming no air resistance). On the other hand, the vertical component of momentum changes because gravity acts in the vertical direction, affecting the vertical component of velocity (Vy) over time.

Considering the examples provided, the velocity of a projectile with an initial velocity of 2i + j (in terms of unit vectors i and j) before striking the ground is 2i - 2j, considering that gravity (g = 10 m/s²) acts downward, affecting only the vertical component of velocity. For the second example, a projectile with a velocity of 2i + 3j m/s could be either ascending or descending, but since the vertical component of velocity is still positive, it suggests that the projectile is ascending to the maximum height. So, the answer would be option (d).

Answer:

A) vᵣ = 1.2 m/s

B) vₜ = 0.4π m/s = 1.257 m/s

C) aᵣ = - 1.2π m/s² = - 3.771 m/s²

D) aₜ = 7.2 m/s²

Explanation:

Given,

θ' = 3 rad/s

r = 0.4 θ

Note that

θ' = (dθ/dt) = 3 rad/s

θ" = (d²θ/dt²) = (d/dt) (3) = 0 rad/s²

r' = (dr/dt) = (d/dt) (0.4θ) = 0.4 (dθ/dt) = 0.4 × θ' = 0.4 × 3 = 1.2 m/s

r" = (d²r/dt²) = 0.4 (d²θ/dt²) = 0 m/s²

A) Radial component of the velocity of P at the instant θ=π/3rad.

From the kinematics of a particle in a plane,

vᵣ = r' = 1.2 m/s

B) Transverse component of the velocity and acceleration of P at the instant θ=π/3rad.

vₜ = rθ' = (0.4 θ) (3) = 1.2 θ = 1.2 (π/3) = 0.4π m/s = 1.257 m/s

C) Radial component of the acceleration of P at the instant θ=π/3rad.

aᵣ = r" - r(θ'²) = 0 - (0.4θ)(3²) = - 3.6θ = - 3.6 (π/3) = - 1.2π m/s² = - 3.771 m/s²

D) Transverse component of the acceleration of P at the instant θ=π/3rad

aₜ = rθ" + 2r'θ' = (0.4θ)(0) + (2)(1.2)(3) = 0 + 7.2 = 7.2 m/s²

Hope this Helps!!!

The response requires understanding rotational kinematics and dynamics to determine the velocity and acceleration of a particle moving along a spiral guide, involving calculus and principles of rotational motion.

The question pertains to kinematics in the context of rotational and spiral motion. In particular, it involves the motion of a particle along a spiral guide with a given radial dependency on the angle turned through (in radians). To find the velocity and acceleration of the particle, one must apply the equations of motion that relate linear and angular quantities and consider any given geometric relationships and constraints, such as the provided relationship between radial distance and angular position. To solve the tasks posed, one would typically employ differential calculus and the principles of rotational dynamics including tangential and radial components of velocity and acceleration.

For example, the velocity of the particle can be found by taking the time derivative of the radial position r, which in turn depends on the angular velocity ω. The acceleration would then be found by differentiating the velocity with respect to time or by explicitly using the tangential and radial acceleration components for rotational motion, which are related to ω and its time derivative (angular acceleration).

Answer:

a) C = 1.065 * 10^-10 F

b) 7.775 * 10^-9

c) 7.444 * 10^-9 C

Explanation:

A spherical capacitor, with inner radius of a = 1.2 cm and outer radius  

of b = 1.7 cm is filled with a dielectric material with dielectric constant of  

K = 27 and connected to a potential difference of V = 64.5 V.  

(a) The capacitance of a filled air spherical capacitor is given by equation :

                C = 4*π*∈o*(a*b/b-a)

if the capacitor is filled with a material with dielectric constant K, we need  

to modify the capacitance as ∈o ---->k∈o , thus:  

                C = 4*π*∈o*(a*b/b-a)

substitute with the given values to get:  

    C = 4*π*(27)*(8.84*10^-12)[(1.2*10^-2)*(1.7*10^-2)/(1.7*10^-2)-(1.2*10^-2)*]

    C = 1.065 * 10^-10 F

(b) The charge on the capacitor is given by q = CV, substitute to get:

   q = (1.065 * 10^-10)*64.5 V

      = 7.775 * 10^-9

(c) The induced charge on the dielectric material is given by equation as:  

   q' = q(1-1/k)

  substitute with the given values to get:

    q' = (7.775 * 10^-9)*(1-1/27)

        = 7.444 * 10^-9 C

note:

calculation maybe wrong but method is correct. thanks

The capacitance is 3.36 nF, the free charge is 216.72 nC and the induced charge is zero.

Given information:

Radius, a = 0.017 m

b = 0.012 m

Potential difference, V = 64.5 V

(a)

The capacitance is given by:

C = (4πε₀ / (1/b - 1/a))

C = (8.85*10⁻²×3.14×4)/(1/0.012-1/0.017)

C = (4π(8.85 x 10^-12) / 293.3)

C = 3.36 nF

Hence, the capacitance is 3.36 nF.

(b)

The free charge can be calculated from the relation of charge, capacitance, and voltage:
q = CV

q = 3.36×64.5

q= 216.72 nC

Hence, the free charge is 216.72 nC.

(c)

The induced charge is given by:

q' = q - C × V

q' =  0 nC

Hence, the charge is 0 nC.

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Answer:

2321 N

Explanation:

Let g = 9.807 m/s2

Assume this is a uniform beam and the center of mass it at the geometric center, which is half way between the 2 ends, or h = 13/2 = 6.5 m from the left end (support A).

For the system to remain at rest, then the total moments around a point (let pick support A) must be 0. Moments created by each force is the product of the force magnitude and the moment arm, aka distance from the force to support A

[tex]\sum M_A = 0[/tex]

[tex]M_w + M_b + M_B = 0[/tex]

[tex]F_wx + F_bh + F_BL = 0[/tex]

[tex]m_wgx + m_bgh = -F_BL[/tex]

If we take upward direction be the positive direction, that means all the gravity acting downward are negative

[tex]74*(-9.807)*7.5 + 388(-9.807)*6.5 = -F_B13[/tex]

[tex]-13F_B = -30176.139[/tex]

[tex]F_B = -30176.139/-13 = 2321 N[/tex]

So the force at support B has a magnitude of 2321 N acting upward

The force that support B must apply to keep the beam balanced when the worker sits at a distance of 7.5m from support A is approximately 1597.92 N.

This problem involves understanding of statics, specifically the principle of moments or torques. Assuming the system is in equilibrium (i.e., the beam does not rotate or translate), the sum of the forces and the sum of the torques must both be zero.

We start by calculating the total weight of the beam, which is simply its mass multiplied by the acceleration due to gravity, g. This gives mb * g = 388 kg * 9.81 m/s^2 = 3806.68 N, this force acts in the middle of the beam, that is at a distance of L/2 = 13m/2 = 6.5m from each support.

Similarly, the weight of the worker is given by mw * g = 74 kg * 9.81 m/s^2 = 725.94 N, this force acts at a distance of x = 7.5m from the left support (A) and L - x = 13m - 7.5m = 5.5m from the right support (B).

Now we can calculate the net force (Fn) and the net torque (Tn). We choose the left support (A) as the pivot point. The total force acting on the beam is the sum of the weights of the beam and the worker, and this must be supported by the two supports. Since sum of the forces must be zero: Fn = F_A + F_B = mass_beam * g + mass_worker * g.

The net torque is calculated from the forces acting at a distance from the pivot point. So we get, Tn = beam_weight * distance_beam - worker_weight * distance_worker + F_B * L = 0. From this, we can solve for F_B (force at support B), we find: F_B = (beam_weight * distance_beam - worker_weight * distance_worker) / L.

Plugging into these formulas, we get the force that support B must apply to keep the system in balance: F_B = (3806.68 N * 6.5m - 725.94 N * 7.5m) / 13m = 1597.92 N.

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