A 2500-lb vehicle has a drag coefficient of 0.35 and a frontal area of 20 ft2. What is the minimum tractive effort required for this vehicle to maintain a 70 mi/h speed on a 5% upgrade through an air density of 0.002045 slugs/ft3?

Answer:

[tex]final-temperature = T_{f} = 252.51K[/tex]

Explanation:

we can solve this problem by using the first law of thermodynamics.

    [tex]\Delta U= Q-W[/tex]

Q= heat added

U= internal energy

W= work done by system

                        [tex]E_{final}= E_{initial}[/tex]

[tex]C_{v} (N_{2}) C_{v}(He) T_{f} =C_{v}( N_{2}) T_{N_{2} } C_{v} (He) T_{He}[/tex]    (1)

[tex]C_{v}(N_{2})=1.04\frac{KJ}{Kg K}[/tex]

[tex]C_{v}(He)=5.193\frac{KJ}{Kg K}[/tex]

now

From equation 1

[tex]T_{f}=\frac{C_{v}( N_{2}) T_{N_{2} } C_{v} (He) T_{He}}{C_{v} (N_{2}) C_{v}(He)}[/tex]

[tex]T_{f} = \frac{315\times1.04+5.193\times240}{1.04+5.193}[/tex]

[tex]T_{f} = 252.51K[/tex]

Answer:

Explanation:

Given

thermal conductivity of wood [tex] K_w=0.08 W/m^2-K[/tex]

thermal conductivity of insulation [tex]K_i=0.01 W/m^2-K[/tex]

thickness of wood [tex]t_2=3 cm[/tex]

thickness of insulation [tex]t_1=2.4 cm[/tex]

[tex]T_i=20^{\circ}C[/tex]

[tex]T_o=-13^{\circ}C[/tex]

we know heat Flow is given by

[tex]Q=kA\frac{dT}{dx}[/tex]

[tex]dT=[/tex] change in temperature

[tex]dx=[/tex] thickness

K=thermal conductivity

A=Area of cross-section

A is same

Suppose T is the temperature of Junction

as heat Flow is same thus

[tex]\frac{k_w(20-T)}{3}=\frac{k_i(T-(-13))}{2.4}  [/tex]      

[tex]\frac{0.08(20-T)}{3}=\frac{0.01(T+13)}{2.4}[/tex]

[tex]T=19.36 ^{\circ}C[/tex]

(b)Rate of heat flow

[tex]Q=\frac{k_w(T+13)}{3\times 10^{-2}}[/tex]

[tex]Q=\frac{0.08\times 32.36}{0.03}[/tex]

[tex]Q=86.303 W/m^2[/tex]                                            

The temperature at which the wood meets the Styrofoam is determined using the proportional temperature drops across each material, considering their respective thermal resistances. The rate of heat flow per square meter through the wall is calculated using Fourier's law, considering the combined thermal resistance of both layers.

Temperature at the Wood-Styrofoam Plane and Heat Flow Rate

To find the temperature at the plane where the wood meets the Styrofoam, we need to use the concept of thermal resistance and the fact that the heat flow through both materials is the same. With the given thermal conductivity (k values) and thicknesses of the wood and Styrofoam, we can calculate their respective thermal resistances (R = thickness/k). Then, by setting up a proportion, we can find the temperature drop across the wood (ΔTw) and the Styrofoam (ΔTs) using the formula ΔT = (k*A* ΔT)/(thickness), where A is the area (which cancels out as it's the same for both layers). Knowing the temperature drop across each and the outer and inner surface temperatures, we can find the temperature at the plane where the layers meet.

To calculate the rate of heat flow per square meter through this wall, we'll use Fourier's law of thermal conduction, which is given by Q = k*A* ΔT/d, where Q is the heat flow rate, A is the area, ΔT is the temperature difference, and d is the thickness. Since the wall consists of two layers with different thermal conductivities and thicknesses, we need to calculate the equivalent thermal resistance for the combined wall and then use the overall temperature difference to find the heat flow rate.

Answer:

Transverse wave formation.

Explanation:

This type of wave formation is called transverse wave formation.

Transverse wave is a wave arising when the medium's pulse is perpendicular  to the wave's path of propagation. The crowd was standing vertically, but they were arranged horizontally in the stadium, according to the question. Hence the wave form was transverse.

Answer:

280N

Explanation:

Pascal's law states that the pressure in a fluid is transmitted across every point in the fluid system.

Hence, the pressure in both tubes must remain same;

So, pressure = F1/A1 = F2/A2

where F1 = initial force on small pipe, A1 = area of small pipe

F2 = force on larger piston, A2 = area of larger piston

Given:

F1 = 11.2N

D1 = diameter of smaller pipe = 6 cm

D2 = diameter of larger piston = 15 cm

A1 = π*(r1)² = π*(6/2)² =π*9 (As radius, r = diameter/2)

A2 = π*(r2)² = π*(30/2)² =π*225

Hence 11.2/π*9 = F2/π*225

solving, we have

F2 = 280N

Answer:

(a) The length of the pendulum on Earth is 36.8cm

(b) The length of the pendulum on Mars is 13.5cm

(c) Mass suspended from the spring on Earth is 0.37kg

(d) Mass suspended from the spring on Mars is 0.36kg

Explanation:

Period = 1.2s, free fall acceleration on Earth = 9.8m/s^2, free fall acceleration on Mars = 3.7m/s^2

( a) Length of pendulum on Earth = [( period ÷ 2π)^2] × acceleration = (1.2 ÷ 2×3.142)^2 × 9.8 = 0.0365×9.8 = 0.358m = 35.8cm

(b) Length of the pendulum on Mars = (1.2÷2×3.142)^2 × 3.7 = 0.0365×3.7 = 0.135cm = 13.5m

(c) Mass suspended from the spring on Earth = (force constant×length in meter) ÷ acceleration = (10×0.358) ÷ 9.8 = 0.37kg

(d) Mass suspended from the spring on Mars = (10×0.135)÷3.7 = 0.36kg

The length of a pendulum with a period of 1.2 s on Earth is approximately 36.95 cm, while on Mars it is around 16.99 cm. The mass suspended from a spring that would result in a period of 1.2 s on Earth is approximately 0.722 kg, and on Mars it is approximately 0.329 kg.

(a) What length of pendulum has a period of 1.2 s on Earth? cm

Using the equation for the period of a pendulum, T = 2π √(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity, we can solve for L. Rearranging the equation, we have L = (T/2π)² * g.

Given that the free-fall acceleration on Earth is approximately 9.8 m/s², substituting the values into the equation, we have:

L = (1.2/2π)² * 9.8 = 0.3695 m = 36.95 cm

(b) What length of pendulum would have a 1.2-s period on Mars? cm

Using the same equation, L = (T/2π)² * g, we can substitute the values for the period and acceleration due to gravity on Mars:

L = (1.2/2π)² * 3.7 = 0.1699 m = 16.99 cm.

(c) Find the mass suspended from this spring that would result in a period of 1.2 s on Earth. kg

For a spring-mass system, the period is given by T = 2π √(m/k), where T is the period, m is the mass, and k is the spring constant. Rearranging the equation, we have m = (T/2π)² * k.

Given that the spring constant is 10 N/m, substituting the values into the equation, we have:

m = (1.2/2π)² * 10 = 0.722 kg.

(d) Find the mass suspended from this spring that would result in a period of 1.2 s on Mars. kg

Using the same equation, m = (T/2π)² * k, we can substitute the values for the period and spring constant:

m = (1.2/2π)² * 10 = 0.329 kg.

Answer: this one is tough. There is a estimated 1.2 billion drivers to help best I could. I did the math the best I could and didn’t get an answer close to the choices. Just guessing I’d guess C. 20,000 because of logic.

Explanation: I can’t help much but would love to hear how to work this out.

To develop this problem it is necessary to apply the concepts related to the conservation of Energy. In this case the definition concerning kinetic energy from the simple harmonic movement.

From the conservation of energy we know that the kinetic energy would be conserved through the work done by the frictional force and the simple harmonic potential energy, in other words:

[tex]KE = PE_s +W_f[/tex]

[tex]KE = \frac{1}{2}kA^2 + W_f[/tex]

Where,

[tex]KE =[/tex] Kinetic Energy

[tex]PE_s =[/tex]Potential Harmonic Simple Energy

[tex]W_f =[/tex] Work made by friction.

Our values are given as,

[tex]m = 10Kg  \rightarrow[/tex] mass

[tex]k = 1400N/m \rightarrow[/tex] Spring constant

[tex]A = 0.1m \rightarrow[/tex]Amplitude

[tex]f_f = 30N \rightarrow[/tex] Frictional Force

Replacing we have,

[tex]KE = \frac{1}{2}kA^2 + W_f[/tex]

[tex]KE =\frac{1}{2} 1400 * 0.1^2 + ( - 30 * 0.1)[/tex]

[tex]KE = 4 J[/tex]

Therefore the Kinetic Energy of the block as it passes through its equlibrium position is 4J.

The kinetic energy of the 10-kg block as it passes through its equilibrium position is 4 Joules. This is calculated by converting the potential energy stored in the spring to kinetic energy and then subtracted the energy lost due to friction.

The first step in solving this problem is to understand the two forces acting on the block in this question: the spring force and the frictional force. The spring potential energy when the block is pulled 10 cm to the right is given by the formula U = 1/2kx^2, where k is the force constant and x is the displacement. Substituting the given values, we have U = 1/2(1400 N/m)(0.1 m)^2 = 7 Joules. This is the initial potential energy stored in the spring. As the block passes through its equilibrium position, this potential energy is fully converted to kinetic energy.

We also need to take into account the work done against frictional force which is equal to the frictional force times the displacement, i.e., W_friction = Friction * displacement = 30N * 0.1m = 3 Joules. This is the energy lost due to friction.

Finally, we subtract the work done by the frictional force from the potential energy to achieve the kinetic energy. Therefore, the kinetic energy of the block as it passes through its equilibrium position is K = U - W_friction = 7J - 3J = 4 Joules.

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Answer:

On the standing waves on a string, the first antinode is one-fourth of a wavelength away from the end. This means

[tex]\frac{\lambda}{4} = 0.275~m\\\lambda = 1.1~m[/tex]

This means that the relation between the wavelength and the length of the string is

[tex]3\lambda/2 = L[/tex]

By definition, this standing wave is at the third harmonic, n = 3.

Furthermore, the standing wave equation is as follows:

[tex]y(x,t) = (A\sin(kx))\sin(\omega t) = A\sin(\frac{\omega}{v}x)\sin(\omega t) = A\sin(\frac{2\pi f}{v}x)\sin(2\pi ft) = A\sin(\frac{2\pi}{\lambda}x)\sin(\frac{2\pi v}{\lambda}t) = (2.45\times 10^{-3})\sin(5.7x)\sin(59.94t)[/tex]

The bead is placed on x = 0.138 m. The maximum velocity is where the derivative of the velocity function equals to zero.

[tex]v_y(x,t) = \frac{dy(x,t)}{dt} = \omega A\sin(kx)\cos(\omega t)\\a_y(x,t) = \frac{dv(x,t)}{dt} = -\omega^2A\sin(kx)\sin(\omega t)[/tex]

[tex]a_y(x,t) = -(59.94)^2(2.45\times 10^{-3})\sin((5.7)(0.138))\sin(59.94t) = 0[/tex]

For this equation to be equal to zero, sin(59.94t) = 0. So,

[tex]59.94t = \pi\\t = \pi/59.94 = 0.0524~s[/tex]

This is the time when the velocity is maximum. So, the maximum velocity can be found by plugging this time into the velocity function:

[tex]v_y(x=0.138,t=0.0524) = (59.94)(2.45\times 10^{-3})\sin((5.7)(0.138))\cos((59.94)(0.0524)) = 0.002~m/s[/tex]

In this exercise we have to use the knowledge of mechanics to be able to calculate the wavelength, so we can say that it will correspond to:

[tex]v_y=0.002m/s[/tex]

On the standing waves well-behaved, the first antinode happen one of four equal parts of a intuitiveness out completely. This means:

[tex]\lambda=1.1 m[/tex]

This way that the connection middle from two points the wavelength and the extent of object of the strand happen:

[tex]L=3\lambda /2[/tex]

By definition, this standing wave exist at the after second harmonious, n = 3.

Furthermore, the standing wave equating happen in this manner:

[tex]y(x,t)=(Asin(kx))sin(wt)=Asin(w/vx)sin(wt)=((2.45*10^{-3})sin(5.7X)sin(59.94t)[/tex]

The droplet exist established on x = 0.138 m. The maximum speed exist place the derivative of the speed function equals to nothing:

[tex]v_y=wAsin(kx)cos(wt)\\a_y=-w^2Asin(kx)sin(wt)\\a_y=0[/tex]

Find the time, we have:

[tex]59.94t=\pi\\t=0.0524s[/tex]

This exist moment of truth when the speed is maximum. So, the maximum speed maybe raise by stop up existing time into the velocity function:

[tex]v_y(x=0.138,t=0.0524)=(59.94)(2.45*10^{-3})sin((5.7)(0.138))cos((59.94)(0.0524))\\=0.002 m/s[/tex]

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The average current in the given oscillating current function, I = 7 sin(60πt) + cos(120πt), can be found by integrating each component of the function over the given time interval and taking the average. Due to the complexity of the function, it is recommended to use mathematical software or a calculator with integral computation capability.

The given function describes the oscillating current in an electrical circuit: I = 7 sin(60πt) + cos(120πt). The average current, Iave, is conceptually the net charge, ΔQ, that passes through a given cross-sectional area per unit time, Δt. Due to the complexity of this function, we must split it into two integrals to find the average currents.

For the time interval 0 ≤ t ≤ 1/60 seconds, we have two integrals for each component of the current: ∫01/607 sin(60πt) dt and ∫01/60cos(120πt) dt. Calculating these will give us the average current for this time interval. Due to the complexity of the function and the requirement of calculus to solve it, it's recommended to use mathematical software or a calculator with integral computation capability to get the numerical values, always rounding your answers to three decimal places.

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Answer:

0.62127°C

Explanation:

[tex]1\ BTU=1055\ J[/tex]

[tex]80\ BTU/h=80\times 1055=84400\ J/h[/tex]

Heat absorbed by body in 2 hours

[tex]Q=84400\times 2\\\Rightarrow Q=168800\ J[/tex]

m = Mass of person = 65000 g

c = Specific heat of water = 4180 J/kg°C

[tex]\Delta T[/tex] = Change in temperature

Heat is given by

[tex]Q=mc\Delta T\\\Rightarrow 168800=65\times 4180\times \Delta T\\\Rightarrow \Delta T=\dfrac{168800}{65\times 4180}\\\Rightarrow \Delta T=0.62127\ ^{\circ}C[/tex]

The increase in temperature will be 0.62127°C

Answer:

2354.4 Pa

40221 Pa

Explanation:

[tex]\rho[/tex] = Density = 1000 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

h = Depth

The pressure difference would be

[tex]\Delta P=\rho gh\\\Rightarrow \Delta P=1000\times 0.24\times 9.81\\\Rightarrow \Delta P=2354.4\ Pa[/tex]

The pressure difference in the first case is 2354.4 Pa

[tex]\Delta P=\rho gh\\\Rightarrow \Delta P=1000\times 4.1\times 9.81\\\Rightarrow \Delta P=40221\ Pa[/tex]

The pressure difference in the second case is 40221 Pa

Answer:

d) Increase the work W, the rejected QC decreasing as a result.

Explanation:

By the second law of thermodynamics the efficiency of a heat engine working between two reservoirs is:

[tex]\eta=\frac{W}{Q_{H}} [/tex] (1)

With W the work and [tex] Q_{H} [/tex] the heat of the hot reservoir, note in (1) that efficiency is directly proportional to the work and inversely proportional to the heat of the hot reservoir, so if we remain [tex]Q_{H} [/tex] constant we should increase the work to increase the efficiency.

Also, efficiency is:

[tex] \eta=1-\frac{Q_{C}}{Q_{H}}[/tex]  (2)

With [tex]Q_{C} [/tex] the heat released to the cold reservoir, it is important to note that because second law of thermodynamics the efficiency of a heat engine should be between 0 and 1 ([tex]0\leq\eta\leq1 [/tex]), so the ratio [tex]\frac{Q_{C}}{Q_{H}} [/tex] always is positive and its maximum value is 1, that implies if [tex]Q_{H} [/tex] remains constant and efficiency increases, [tex]Q_{C} [/tex] will decrease and the ratio [tex]\frac{Q_{C}}{Q_{H}} [/tex] too.

So, the correct answer is d)

Final answer:

The efficiency of a heat engine is increased by increasing the work output while decreasing the rejected heat to the cold reservoir, in accordance with the first and second laws of thermodynamics.

Explanation:

To improve the efficiency of a heat engine, one must increase the work output, W, while simultaneously decreasing the heat rejected to the cold reservoir, Qc. The correct choice is:

d) Increase the work W, the rejected Qc decreasing as a result.

The efficiency, η, of a heat engine is defined as the ratio of the work done, W, to the heat absorbed from the hot reservoir, QH. By the first law of thermodynamics, QH = W + Qc, meaning that the efficiency can be increased either by increasing W or decreasing Qc. According to the second law of thermodynamics, there is a minimum amount of QH that cannot be used for work and must be rejected as Qc. Therefore, the aim is to minimize this rejected heat without altering QH, which cannot be changed in this scenario. The ideal is to approach the efficiency of a Carnot engine, which has the maximum possible efficiency between two given temperatures by operating in a reversible manner and reducing entropy generation.

Answer:

A. 36,000 W

Explanation:

[tex]m[/tex] = mass of the car = 900 kg

[tex]v_{o}[/tex] = Initial speed of the car = 20 m/s

[tex]v_{f}[/tex] = Final speed of the car = 0 m/s

[tex]W[/tex] = Work done by the brakes on the car

Magnitude of work done on the car by the brakes is same as the change in kinetic energy of the car.hence

[tex]W = (0.5) m (v_{o}^{2} - v_{f}^{2})\\W = (0.5) (900) ((20)^{2} - (0)^{2})\\W = 180000 J[/tex]

[tex]t[/tex] = time taken by the car to come to stop = 5 s

[tex]P[/tex] = Average power produced by the car

Average power produced by the car is given as

[tex]P = \frac{W}{t} =\frac{180000}{5} \\P = 36000 W[/tex]

Answer:

Explanation:

Answer:

option B

Explanation:

When a body is immersed in liquid there will be two force is acting on the body.

First one force acting downward due to weight of the body.

And the second force acting on the object will be buoyant force.

If the object is not in equilibrium the apparent weight will be equal to net force acting on the object.

     [tex]F_{net} = W - F_b[/tex]

W is the weight of the object acting downward

Fb is the buoyancy force acting upward on the object.

Hence, the correct answer is option B

Answer:

V = 0.3724 m³

T = 27.836 N

Explanation:

Given :

m = 3.21 kg  , W= 3.21 * 9.81 m / s² = 31.4901 N

ρ = 8.62 g / cm ³  = 8620 kg / m³

V = m / ρ =  3.21 kg  /  8620 kg / m³

V = 0.3724 m³

when submerged the weight of brass cylinder is equal to the tension in string:

F =  (0.3724m³) * (1000 kg / m³) * (9.81 m/s²²) = 3.653 ≈ 3.65 N

T = 31.4901 N - 3.65 N  

T = 27.836 N

Answer:

The rate at which coal burned is 111.12 kg/s.

Explanation:

Given that,

First initial temperature  =750°C

First final temperature =440°C

Second initial temperature  =415°C

Second final temperature =270°C

Suppose If the heat of combustion of coal is [tex]2.8×10^{7}\ J/kg[/tex], at what rate must coal be burned if the plant is to put out 950 MW of power? Assume the efficiency of the engines is 65% of the Carnot efficiency.

The work done by first engine is

[tex]W=eQ[/tex]

The work done by second engine is

[tex]W'=e'Q'[/tex]

[tex]W'=e'Q(1-e)[/tex]

Total out put of the plant is given by

[tex]W+W'=950\ MW[/tex]

Put the value into the formula

[tex]eQ+e'Q(1-e)=950\times10^{6}[/tex]....(I)

We need to calculate the efficiency of first engine

Using formula of efficiency

[tex]e=65%(1-\dfrac{T_{L}}{T_{H}})[/tex]

[tex]e=0.65(1-\dfrac{440+273}{750+273})[/tex]

[tex]e=0.196[/tex]

We need to calculate the efficiency of second engine

Using formula of efficiency

[tex]e'=65%(1-\dfrac{T_{L}}{T_{H}})[/tex]

[tex]e'=0.65(1-\dfrac{270+273}{415+273})[/tex]

[tex]e'=0.136[/tex]

Put the value of efficiency for first and second engine in the equation (I)

[tex]Q(0.196+0.136(1-0.196))=950\times10^{6}[/tex]

[tex]Q=\dfrac{950\times10^{6}}{(0.196+0.136(1-0.196))}[/tex]

[tex]Q=3111.24\times10^{6}\ W[/tex]

We need to calculate the rate at which coal burned

Using formula of rate

[tex]R=\dfrac{Q}{H_{coal}}[/tex]

[tex]R=\dfrac{3111.24\times10^{6}}{2.8×10^{7}}[/tex]

[tex]R=111.12\ kg/s[/tex]

Hence, The rate at which coal burned is 111.12 kg/s.

Answer:

1002.2688 m/s

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

h = The length of a string = 2 m

m = Mass of block = 1.6 kg

[tex]m_2[/tex] = Mass of bullet = 0.01 kg

Here, the potential energy of the fall will balance the kinetic energy of the bullet

[tex]mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 2}\\\Rightarrow v=6.26418\ m/s[/tex]

Velocity of block is 6.26418 m/s

As the momentum of system is conserved we have

[tex]mv=m_2u\\\Rightarrow u=\dfrac{mv}{m_2}\\\Rightarrow u=\dfrac{1.6\times 6.26418}{0.01}\\\Rightarrow u=1002.2688\ m/s[/tex]

The magnitude of velocity just before hitting the block is 1002.2688 m/s

Answer:

Explanation:

Given

displacement is given by

[tex]s(t)=t^3-8t^2+2t[/tex]

so velocity is given by

[tex]v(t)=\frac{\mathrm{d} s(t)}{\mathrm{d} t}[/tex]

[tex]v(t)=3t^2-16t+2[/tex]

(b)velocity after [tex]t=3 s[/tex]

[tex]v(3)=3(3)^2-8\cdot 3+2[/tex]

[tex]v(3)=19 m/s[/tex]

(c)Particle is at rest

when its velocity will become zero

[tex]v(t)=0[/tex]

i.e.  [tex]3t^2-16t+2=0[/tex]

[tex]t=\frac{16\pm \sqrt{16^2-4\cdot 3\cdot 2}}{2\cdot 3}[/tex]

[tex]t=\frac{16\pm 15.23}{6}[/tex]

[tex]t=5.20 s[/tex]    

Answer:

a. Magnetic Field =1.997×[tex]10^{-4}[/tex] T

b. Area= 2.68×[tex]10^{-4}[/tex][tex]m^{2}[/tex]

   Magnetic Flux= 5.367×[tex]10^{-8}[/tex]T[tex]m^{2}[/tex]

c. Inductance= 6.013×[tex]10^{-4}[/tex]H

Explanation:

Parameters from the question

I= 44.5×[tex]10^{-3}[/tex]A

N=500 turns

Diameter=18.5mm

Radius = (diameter/2) = 9.25mm =9.25×[tex]10^{-3}[/tex]m

L= 14cm = 0.14m

Permitivity [tex]U_{o}[/tex]=4π×[tex]10^{-7}[/tex]H/m

The Formulars Used are

B(Magnetic Field) =[tex]\frac{U_{o}. N. I }{l}[/tex]

Mag Flux= B.A

Inductance= [tex]\frac{U_{o}.N^{2} .A }{l}[/tex]

Answer:

a) 199.716 μT

b) [tex]5.368 * 10^{-8}[/tex] T·m^2

c) 0.603 mH

d) B and Ф

Explanation:

I am giving the explanation with my handwritten solution in the paper.

Check the attachment please.

To solve this problem we will apply the concepts given for the efficiency of an engine which is given as

[tex]\eta = 1-\frac{T_C}{T_H}[/tex]

[tex]\eta = \frac{T_H-T_C}{T_H}[/tex]

Where

[tex]T_C[/tex] = Temperature of the cold reservoir

[tex]T_H[/tex] = Temperature of the hot reservoir

The efficiency maximum would be given only if [tex]T_C = 0[/tex]

Replacing this value we have

[tex]\eta = \frac{T_H-0}{T_H}[/tex]

[tex]\eta = 1[/tex]

Therefore: Cold reservoir as cold as possible provide the greater efficiency.

Final answer:

The greater efficiency in an ideal heat engine is obtained when the hot reservoir is as hot as possible and the cold reservoir is as cold as possible.

Explanation:

The greater efficiency in an ideal heat engine operating with a fixed temperature difference between hot and cold reservoirs is obtained when the hot reservoir is as hot as possible and the cold reservoir is as cold as possible.

Efficiency is determined by the ratio of the temperature of the cold reservoir (Tc) to the temperature of the hot reservoir (Th). The greater the temperature difference, the easier it is to convert heat transfer to work.

Therefore, maximizing the temperature difference by making the hot reservoir as hot as possible and the cold reservoir as cold as possible will result in greater efficiency.

Answer:

First let's write down the moment of inertia of the objects.

[tex]I_{sphere} = \frac{2}{5}mR^2\\I_{disk} = \frac{1}{2}mR^2\\I_{hoop} = mR^2[/tex]

If they all roll without slipping, then the following relation is applied to all ot them:

[tex]v = \omega R[/tex]

where v is the translational velocity and ω is the rotational velocity.

We will use the conservation of energy, because we know that their initial potential energies are the same. (Here, I will assume that all the objects have the same mass and radius. Otherwise we couldn't determine the difference. )

[tex]K_1 + U_1 = K_2 + U_2\\0 + mgh = \frac{1}{2}I\omega^2 + \frac{1}{2}mv^2[/tex]

For sphere:

[tex]\frac{1}{2}\frac{2}{5}mR^2(\frac{v}{R})^2 + \frac{1}{2}mv^2 = mgh\\\frac{1}{5}mv^2 + \frac{1}{2}mv^2 = mgh\\\frac{7}{10}mv^2 = mgh\\v_{sphere} = \sqrt{\frac{10gh}{7}}[/tex]

For disk:

[tex]\frac{1}{2}\frac{1}{2}mR^2(\frac{v}{R})^2 + \frac{1}{2}mv^2 = mgh\\\frac{1}{4}mv^2 + \frac{1}{2}mv^2 = mgh\\\frac{3}{4}mv^2 = mgh\\v_{disk} = \sqrt{\frac{4gh}{3}}[/tex]

For hoop:

[tex]\frac{1}{2}mR^2(\frac{v}{R})^2 + \frac{1}{2}mv^2 = mgh\\\frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mgh\\mv^2 = mgh\\v_{hoop}= \sqrt{gh}[/tex]

The sphere has the highest velocity, so it arrives the bottom first. Then the disk, and the hoop arrives the last.

Explanation:

The moment of inertia can be defined as the resistance to the rotation. If an object has a high moment of inertia, it resist to rotate more so its angular velocity would be lower. In the case of rolling without slipping, the angular velocity and the linear (translational) velocity are related by the radius, so the object with the highest moment of inertia would arrive the bottom the last.

The order in which a sphere, disk, and hoop reach the bottom of an incline when released from the same height is determined by their moments of inertia. The solid sphere arrives first, followed by the disk, and finally the hoop.

In the scenario where a uniform disk, a uniform hoop, and a uniform solid sphere are released from the top of an inclined ramp and roll without slipping, the order in which they reach the bottom depends on their moments of inertia and the distribution of mass. The solid sphere has the smallest moment of inertia relative to its mass (I = 2/5 MR²), which means it will accelerate faster than the other shapes and hence get to the bottom first. The uniform disk, with a moment of inertia of I = 1/2 MR², will follow. The uniform hoop has the largest moment of inertia (I = MR²) for a given mass and radius, so it will accelerate the slowest of the three and reach the bottom last.

Therefore, the objects reach the bottom of the ramp in the following order: sphere, disk, hoop.

Answer:

the rate of heat transfer Q is Q =235.125 kJ/s

the heat transfer surface area A of the heat exchanger is A= 15.30 m²

Explanation:

Assuming negligible loss  to the environment, then the heat flow of the hot water goes entirely to the cold water

Denoting a as cold water and b as hot water , then

Q= Fᵃ* cpᵃ * ( T₂ᵃ - T₁ᵃ)

where

F= mass flow

cp = specific heat capacity at constant pressure

T₂= final temperature

T₁ = initial temperature

replacing values

Q = Fᵃ* cᵃ * ( T₂ᵃ - T₁ᵃ) =  1.25 kg/s* 4180 J/kg·K* ( 60°C-15°C) * 1 kJ/1000J= 235.125 kJ/s

if all there is no loss to the surroundings

Qᵃ + Qᵇ = Q surroundings = 0

Fᵃ* cpᵃ * ( T₂ᵃ - T₁ᵃ) +  Fᵇ* cpᵇ * ( T₂ᵇ - T₁ᵇ) = 0

T₂ᵇ = T₁ᵇ - [Fᵃ* cpᵃ /  (Fᵇ* cpᵇ)  ]* ( T₂ᵃ - T₁ᵃ)

replacing values

T₂ᵇ =100°C - [1.25 kg/s* 4180 J/kg·K/  (4 kg/s* 4190 J/kg·K)]* ( 60°C-15°C)

T₂ᵇ = 85.97 °C

the heat transfer surface of the heat exchanger is calculated through

Q = U*A* ΔTlm

where

U= overall heat transfer coefficient

A = heat transfer area of the heat exchanger

ΔTlm = (ΔTend - ΔTbeg)/ ln ( ΔTend - ΔTbeg)

ΔTbeg = temperature difference between the 2 streams at the inlet of the heat exchanger (  hot out - cold in) = 85.97 °C - 15°C = 70.97 °C

ΔTbeg = temperature difference between the 2 streams at the end of the heat exchanger ( hot in - cold out ) = 100°C - 60 °C = 40°C

then

ΔTlm = (ΔTend - ΔTbeg)/ ln ( ΔTend - ΔTbeg) =( 70.97 °C-  40°C)/ ln( 70.97°C/40°C) = 17.455 °C

ΔTlm = 17.455 °C

then

Q = U*A* ΔTlm

A = Q/(U*ΔTlm) = 235.125 kJ/s/(17.455 °C *880 W/m²*K) *1000 J/kJ = 15.30 m²

A= 15.30 m²

Final answer:

Calculate the heat transfer rate and heat transfer surface area in a double-pipe counterflow heat exchanger using given water properties and overall heat transfer coefficient.

Explanation:

Cold Water: mw = 1.25 kg/s, cp = 4180 J/kg·K, Tin = 15°C, Tout = 60°C

Hot Water: mh = 4 kg/s, cp = 4190 J/kg·K, Tin = 100°C, Tout = ?

Overall Heat Transfer: U = 880 W/m²·K

Calculate Heat Transfer Rate:

A ≈ 1m²

So, the rate of heat transfer is approximately 55341.28 W and the heat transfer surface area of the heat exchanger is approximately 1 m².

Final answer:

To find the time it takes for the flywheel to reach its top angular speed, we can use the formula: ω = Δθ / Δt. In this case, the flywheel has a diameter of 1.5 m and spins up to an angular speed of 1200 rpm. It takes approximately 0.0375 seconds for the flywheel to reach its top angular speed. The energy stored in the flywheel is approximately 554,414.06 joules.

Explanation:

To find the time it takes for the flywheel to reach its top angular speed, we can use the formula:

ω = Δθ / Δt

Where ω is the angular velocity, Δθ is the change in angle, and Δt is the change in time.

In this case, the flywheel has a diameter of 1.5 m, so the radius is 0.75 m. The flywheel spins up from rest to an angular speed of 1200 rpm, which is equivalent to 125.66 rad/s. 

Using the formula, we can rearrange to solve for Δt:

Δt = Δθ / ω

Δθ is equal to the circumference of the flywheel, which is 2π times the radius:

Δθ = 2π × 0.75 m = 4.7124 rad

Plugging in the values:

Δt = 4.7124 rad / 125.66 rad/s = 0.0375 s

So it takes approximately 0.0375 seconds for the flywheel to reach its top angular speed.

Now, to calculate the energy stored in the flywheel, we can use the formula:

KE = 0.5 × I × ω^2

Where KE is the kinetic energy, I is the moment of inertia, and ω is the angular velocity.

The moment of inertia for a solid disk is given by:

I = 0.5 × m × r^2

Where m is the mass of the flywheel and r is the radius.

Plugging in the values:

I = 0.5 × 250 kg × (0.75 m)^2 = 70.3125 kg×m^2

Now we can calculate the energy:

KE = 0.5 × 70.3125 kg×m^2 × (125.66 rad/s)^2 = 554,414.06 J

So, the flywheel stores approximately 554,414.06 joules of energy.

Final answer:

The Paschen and Brackett series of lines in atomic hydrogen can overlap in terms of their wavelength ranges.

Explanation:

The Paschen series of lines in atomic hydrogen occurs when nf = 3, and the Brackett series occurs when nf = 4.

The range of wavelengths in these two series can overlap because the wavelength equation 1/λ = 2π²mk²e⁴/(h³c)(Z²)(1/nf² - 1/ni²) depends on the values of nf and ni, which can be different for each series

For example, if nf = 3 in the Paschen series and nf = 4 in the Brackett series, the wavelengths in these two series can overlap depending on the values of ni.

These scenarios involve the transfer of heat in a closed system by utilizing calorimetry. The final temperature is affected by the mass of the metal block, the specific heat capacity of the metal, and the introduction of phase changes with the addition of ice into the system.

These problems revolve around the principle of conservation of energy, and more specifically the concept of heat transfer, which is often studied using calorimetry.  We can solve these problems by using the formula for heat transfer, Q = mcΔT, where Q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the change in temperature.

For the first scenario, using a 1.0 kg aluminum block instead of a 0.5 kg block, the heat capacity of the system has simply doubled. The final temperature for the aluminum block and the water can be determined by setting the heat gained by the water equal to the heat lost by the aluminum. Since the mass of the block has doubled, the final temperature will be lower than the original experiment.

In the second scenario with 1.0 kg copper block, copper has a lower specific heat capacity than aluminum. This means that when heated to the same temperature, the copper block will contain less energy than the aluminum block. Consequently, when put in contact with the water, the final temperature of the water and block will be lower than in the original experiment.

In the final two scenarios, adding ice to the system adds an additional phase change. This means, it’s necessary to add in the energy used to convert the ice at 0 degrees Celsius into water at 0 degrees Celsius before heating the water to the final temperature. The amount of ice being added in these instances changes the energy dynamics of the system and thus resulting in different final temperatures.

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Answer:

U = 11.85 J

Explanation:

given,

spring is stretched = 23 cm

                             x = 0.23 m

Force experiences = 103 N

we know,

 F = k x

where k is the spring constant

 [tex]k =\dfrac{F}{x}[/tex]

 [tex]k =\dfrac{103}{0.23}[/tex]

       k = 447.83 N/m

energy stored in the spring

 [tex]U =\dfrac{1}{2}kx^2[/tex]

 [tex]U =\dfrac{1}{2}\times 447.83 \times 0.23^2[/tex]

        U = 11.85 J

hence, energy stored in the spring is equal to U = 11.85 J

The elastic potential energy stored in the spring when stretched to a length of 23 cm, given a spring constant of 4 N/cm, is 0.18 J.

The question pertains to how much energy is stored in a spring when it is stretched 23 cm from its equilibrium point, where it experiences a force of 103 N. The potential energy stored in the spring can be calculated using the formula U = 1/2kx². Given that the spring constant is 4 N/cm and the displacement of the spring from its unstretched length is 3 cm (since the unstretched length is 20 cm), the potential energy can be calculated as follows: U = 1/2 * 4 N/cm * (3 cm)² = 0.18 J. Thus, the elastic potential energy contributed by the spring when it is stretched to a length of 23 cm is 0.18 J.

Answer:

-13094.55179 J

Explanation:

Q = Heat = [tex]-1.5\times 10^{4}\ J[/tex]

P = Pressure = [tex]4\times 10^4\ Pa[/tex]

[tex]\Delta V[/tex] = Change in volume = [tex]\pi r^2\times -h[/tex](negative because it is decreasing)

h = Height = 16.3 cm

r = Radius = 30.5 cm

Entropy is given by

[tex]\Delta U=Q-W[/tex]

Work done is given by

[tex]W=P\Delta V\\\Rightarrow W=4\times 10^4\times (\pi 0.305^2\times -0.163)[/tex]

[tex]\Delta U=-1.5\times 10^{4}-(4\times 10^4\times (\pi 0.305^2\times -0.163))\\\Rightarrow \Delta U=-13094.55179\ J[/tex]

The change in the internal energy of the system is -13094.55179 J

Answer:

Q' = 140.859 kJ

Explanation:

Given that, 93.4 g of solid ethanol at − 114.5 °C is converted to gasesous ethanol at 149.8 ° C.

The molar heat of fusion of ethanol is, ΔH(f) = 4.60 kJ/mol , and its molar heat of vaporization is ΔH(v) = 38.56 kJ/mol .

And also Ethanol has a normal melting point of − 114.5 ° C and a normal boiling point of 78.4 ° C .

The specific heat capacity of liquid ethanol is S(l) = 2.45 J / g ⋅°C , and that of gaseous ethanol is S(g) = 1.43 J / g ⋅°C .

Lets solve this step wise ;

Given 93.4 g of ethanol is taken, but 1 mole of ethanol consists of 46.06 g

⇒ moles of ethanol given = [tex]\frac{93.4}{46.06}[/tex] = 2.02 moles

step 1: solid ethanol to liquid ethanol at melting point of − 114.5 ° C

⇒ 1 mole requires ΔH(f) = 4.60 kJ/mol of heat

⇒ heat required = 4.60 × 2.02 = 9.292 kJ.

step 2: liquid ethanol at -114 °C to liquid ethanol at 78.4 °C

Q = m×S×ΔT ; Q = heat required

                       m = mass of the substance

                       S = specific heat of the substance

                       ΔT = change in temperature

Here S = S(l);

⇒ Q = 93.4×2.45×(78.4-(-114.5))

       = 44.141 kJ

step 3: liquid ethanol at 78.4°C to gaseous ethanol at 78.4°C

1 mol of liquid ethanol requires ΔH(v) = 38.56 kJ/mol of heat

⇒ required heat = 38.56×2.02 = 77.89 kJ

step 4: gaseous ethanol at 78.4 °C to gaseous ethanol at 149.8 °C

Q = m×S×ΔT

Here, S = S(g)

Q = 93.4×1.43×(149.8-78.4)

   = 9.536 kJ

⇒ Total heat required = 9.292 + 44.141 + 77.89 + 9.536

                                     = 140.859 kJ

⇒ Q' = 140.859 kJ

The total heat energy required to convert 93.4 g of solid ethanol at − 114.5 ° C to gaseous ethanol at 149.8 ° C is 140.859 kJ.

Given data:,

93.4 g of solid ethanol at − 114.5 °C is converted to gaseous ethanol at 149.8 ° C.

The molar heat of fusion of ethanol is, ΔH(f) = 4.60 kJ/mol.

Molar heat of vaporization is ΔH(v) = 38.56 kJ/mol .

And also Ethanol has a normal melting point of − 114.5 ° C and a normal boiling point of 78.4 ° C .

The specific heat capacity of liquid ethanol is S(l) = 2.45 J / g ⋅°C , and that of gaseous ethanol is S(g) = 1.43 J / g ⋅°C .

Since, 93.4 g of ethanol is taken, but 1 mole of ethanol consists of 46.06 g

⇒ moles of ethanol given is

⇒93.04/46.06 =  2.02 moles

Heat required for conversion of solid ethanol to liquid ethanol at melting point of − 114.5 ° C

⇒ 1 mole requires ΔH(f) = 4.60 kJ/mol of heat

⇒ Q = 4.60 × 2.02 = 9.292 kJ.

Heat required to convert liquid ethanol at -114 °C to liquid ethanol at 78.4 °C

Q' = m×s×ΔT

Here,

m = mass of the substance

s = specific heat of the substance

ΔT = change in temperature

Solving as,

Q' = 93.4×2.45×(78.4-(-114.5))

Q' = 44.141 kJ

Heat required to convert liquid ethanol at 78.4°C to gaseous ethanol at 78.4°C

1 mol of liquid ethanol requires ΔH(v) = 38.56 kJ/mol of heat

Q'' = 38.56×2.02 = 77.89 kJ

Heat required to convert gaseous ethanol at 78.4 °C to gaseous ethanol at 149.8 °C

Q''' = m×S×ΔT

 Q''' = 93.4×1.43×(149.8-78.4)

    = 9.536 kJ

⇒ Total heat required = 9.292 + 44.141 + 77.89 + 9.536

                                      = 140.859 kJ

Thus, we can conclude that the total heat energy required to convert 93.4 g of solid ethanol at − 114.5 ° C to gaseous ethanol at 149.8 ° C is 140.859 kJ.

Learn more about the heat energy here:

https://brainly.com/question/25384702

Answer:

[tex]5.2728\times 10^{-25}\ kgm/s[/tex]

Explanation:

h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]

[tex]\Delta x[/tex] = Uncertainity in position = [tex]10^{-10}\ m[/tex]

[tex]\Delta p[/tex] = Uncertainty in momentum

According to the Heisenberg uncertainity principle we have

[tex]\Delta x\Delta p=\dfrac{h}{4\pi}\\\Rightarrow \Delta p=\dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p=\dfrac{6.626\times 10^{-34}}{4\pi\times 10^{-10}}\\\Rightarrow \Delta p=5.2728\times 10^{-25}\ kgm/s[/tex]

The minimum uncertainty in its momentum is [tex]5.2728\times 10^{-25}\ kgm/s[/tex]