PHYSICS COLLEGE

A piston-cylinder device initially contains 1.4 kg saturated liquid water at 200oC. Now heat is transferred to the water until the volume quadruples and the cylinder contains saturated vapor only. Determine (a) the volume of the cylinder, (b) the final temperature.

Answers

Answer 1

Answer:

Answer:

Explanation:

Given

mass of saturated liquid water $m=1.4\ kg$

at $200^(\circ)$ specific volume is $\nu =0.001157\ m^3\kg$ (From Table A-4,Saturated water Temperature table)

$V_1=m\nu _1$

$V_1=1.4* 0.001157$

$V_1=1.6198* 10^(-3)\ m^3$

Final Volume $V_2=4V_1$

$V_2=4* (1.6198* 10^(-3))$

$V_2=6.4792* 10^(-3)\ m^3$

Specific volume at this stage

$\nu _2=(V_2)/(m)$

$\nu _2=(6.4792* 10^(-3))/(1.4)$

$\nu _2=0.004628\ m^3/kg$

Now we see the value and find the temperature it corresponds to specific volume at vapor stage in the table.

$T_2=T_1^(*)+(T_2^(*)-T_1^(*))/(\alpha _2^(*)-\alpha _1^(*))* (\alpha _2-\alpha _1^(*))$

$T_2=370^(\circ)+(373.95-370)/(0.003106-0.004953)* (0.004628-0.004953)$

$T_2=370.7^(\circ) C$

Answer 2

Answer:

Final answer:

The problem in the question is solved using the principles of thermodynamics. The volume of the device after the heat transfer is 6311.2 cm³. The final temperature inside the cylinder, when the water reached the state of saturated vapor, is approximately 240°C.

Explanation:

The subject question is a thermodynamics problem; more specifically dealing with changes of state, volume, and temperature in a system under certain conditions.

For solving part (a), one would first need to find the specific volume (v) at the initial state, which is saturated liquid at 200°C. Looking up in the property tables, we see that v = 1.127 cm³/g for saturated water at 200°C. Then, the initial volume (V) is mass times specific volume, so V = 1.4 kg x 1.127cm³/g x 1000g/kg = 1577.8 cm³. Because volume quadrupled, the final volume is 4 x 1577.8 cm³ = 6311.2 cm³.

For part (b), at the final state, the water is a saturated vapor. The specific volume at the final state is the final volume divided by the mass, which equals to 6311.2 cm³ / 1.4kg / 1000g/kg = 4.507 cm³/g. Look this value up in the property table to find the corresponding temperature. We get a final temperature of about 240°C.

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A pressure antinode in a sound wave is a region of high pressure, while a pressure node is a region of low pressure.True
False

Answers

A pressure antinode in a sound wave is not a region of high pressure, while a pressure node is not a region of low pressure.

The answer is false

Final answer:

A pressure antinode in a sound wave is indeed a region of high pressure, while a pressure node is a region of low pressure. These definitions hold true for all types of waves.

Explanation:

That's true. In terms of sound waves, a pressure antinode is a region of high pressure, while a pressure node is a region of low pressure. This is true for all types of waves, not only sound waves. In essence, a wave moves through a medium (in case of a sound wave, that medium is typically air) by creating areas of high and low pressure - the high pressure areas are called antinodes, and the low pressure areas are called nodes.

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A ball with a mass of 4 kg is initially traveling at 2 m/s and has a 5 N force applied for 3 s. What is the initial momentum of the ball?

Answers

Answer:

The initial momentum of the ball is 8 kg-m/s.

Explanation:

Given that,

Mass of the ball is 4 kg

Initial speed of the ball is 2 m/s

Force applied to the ball is 5 N for 3 seconds

It is required to find the initial momentum of the ball. Initial momentum means that the product of mass and initial velocity of the ball. It is given as :

$p_i=mu\n\np_i=4\ kg* 2\ m/s\n\np_i=8\ kg-m/s$

So, the initial momentum of the ball is 8 kg-m/s.

Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg), who is sitting at the very bottom end, and whom he holds onto when he arrives. Laughing, John & William leave the slide horizontally and land in the muddy ground near the foot of the slide. (A) If John starts out 1.8 m above William, and the slide is essentially frictionless, how fast are they going when they leave the slide? (B) Thanks to the mud he acquired, John will now experience an average frictional force of 105 N as he slides down. How much slower is he going when he reaches the bottom than when friction was absent?

Answers

Answer:

$v=3.564\ m.s^(-1)$

$\Delta v =2.16\ m.s^(-1)$

Explanation:

Given:

mass of John, $m_J=30\ kg$

mass of William, $m_W=30\ kg$

length of slide, $l=3\ m$

(A)

height between John and William, $h=1.8\ m$

Using the equation of motion:

$v_J^2=u_J^2+2 (g.sin\theta).l$

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

$v_J^2=0^2+2* (9.8* (1.8)/(3))* 3$

$v_J=5.94\ m.s^(-1)$

Now using the law of conservation of momentum at the bottom of the slide:

Sum of initial momentum of kids before & after collision must be equal.

$m_J.v_J+m_w.v_w=(m_J+m_w).v$

where: v = velocity with which they move together after collision

$30* 5.94+0=(30+20)v$

$v=3.564\ m.s^(-1)$ is the velocity with which they leave the slide.

(B)

frictional force due to mud, $f=105\ N$

Now we find the force along the slide due to the body weight:

$F=m_J.g.sin\theta$

$F=30* 9.8* (1.8)/(3)$

$F=176.4\ N$

Hence the net force along the slide:

$F_R=71.4\ N$

Now the acceleration of John:

$a_j=(F_R)/(m_J)$

$a_j=(71.4)/(30)$

$a_j=2.38\ m.s^(-2)$

Now the new velocity:

$v_J_n^2=u_J^2+2.(a_j).l$

$v_J_n^2=0^2+2* 2.38* 3$

$v_J_n=3.78\ m.s^(-1)$

Hence the new velocity is slower by

$\Delta v =(v_J-v_J_n)$

$\Delta v =5.94-3.78= 2.16\ m.s^(-1)$

The magnetic flux that passes through one turn of a 11-turn coil of wire changes to 5.60 from 9.69 Wb in a time of 0.0657 s. The average induced current in the coil is 297 A. What is the resistance of the wire?

Answers

Answer:

2.31 Ω

Explanation:

According to the Faraday's law of electromagnetic induction,

Induced emf = - N (dΦ/dt)

Emf = -N (ΔΦ/t)

where N = number of turns = 11

Φ = magnetic flux

ΔΦ = change in magnetic flux = 9.69 - 5.60 = 4.09 Wb

t = time taken for the change = 0.0657 s

Emf = 11(4.09/0.0657)

Emf = - 684.78 V (the minus sign indicates that the direction of the induced emf is opposite to the direction of change of magnetic flux)

From Ohm's law,

Emf = IR

R = (Emf)/I

I = current = 297 A

R = (684.78)/297

R = 2.31 Ω

Hope this Helps!!

Explanation:

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The tensile strength (the maximum tensile stress it can support without breaking) for a certain steel wire is 3000 MN/m2. What is the maximum load that can be applied to a wire with a diameter of 3.0 mm made of this steel without breaking the wire?

Answers

Answer:

The correct answer is "21195 N".

Explanation:

The given values are:

Tensile strength,

= 3000 MN/m²

Diameter,

= 3.0 mm

i.e.,

= 3×10⁻³ m

Now,

The maximum load will be:

= $Tensile \ strength* Area$

On substituting the values, we get

= $(3000* 10^6)((\pi)/(4) (3* 10^(-3))^2)$

= $(3000* 10^6)((3.14)/(4) (3* 10^(-3))^2)$

= $21195 \ N$

Final answer:

The maximum load that can be applied to a 3.0 mm diameter steel wire with a tensile strength of 3000 MN/m2 without breaking it is 21,200 Newtons.

Explanation:

The subject of this question revolves around the concept of tensile strength in the field of Physics. The maximum load that can be applied to a wire without it breaking depends on the wire's tensile strength and its cross-sectional area. For a steel wire with a tensile strength of 3000 MN/m2 and a diameter of 3.0 mm, we first need to calculate the cross-sectional area, which can be found using the formula for the area of a circle, A = πr^2, where r is the radius of the wire. Given the diameter is 3.0 mm, the radius will be 1.5 mm or 1.5 x 10^-3 m. So, A = π(1.5 x 10^-3 m)^2 ≈ 7.07 x 10^-6 m^2.

We can then use the tensile strength (σ) to find the maximum load (F) using the equation F = σA. Substituting the given values, we get F = 3000 MN/m^2 * 7.07 x 10^-6 m^2 = 21.2 kN, which is equivalent to 21,200 N. Therefore, the maximum load that can be applied to the wire without breaking it is 21,200 Newtons.

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1. On a force vs. mass graph, what would be the slope of the line?2. On a Free Body Diagram, if the forces are all balanced, what do you know about the
object? Can it be moving?

Answers

1. By Newton's second law,

F = ma

so the slope of the line would represent the mass of the object.

2. If all the forces are balanced, then the object is in equilibrium with zero net force, which in turn means the object is not accelerating. So the object is either motionless or moving at a constant speed.

Final answer:

The slope on a Force vs. Mass graph represents acceleration. In a Free Body Diagram, if all the forces are balanced, the object could be either at rest or moving at a constant velocity.

Explanation:

1. On a Force vs. Mass graph, the slope of the line represents acceleration, according to Newton's second law of motion, which is force equals mass times acceleration (F=ma). The slope of the line is calculated as the change in force divided by the change in mass, which results in acceleration.

2. In a Free Body Diagram, if all the forces are balanced, it means the net force acting on the object is zero. This does not necessarily mean that the object is stationary. The object could be at rest, or it could be moving at a constant velocity. If an object is moving at a constant velocity, it is said to be in equilibrium because the forces are balanced.

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The acceleration of a particle moving along a straight line is given by a = −kt2 m/s2 where k is a constant and time t is in seconds. The initial velocity of the particle at t = 0 is v0 = 12 m/s and the particle reverses it direction of motion at t = 6 s. Determine the constant k and the displacement of the particle over the same 6-second interval of motion. Ans: k = 1/6 m/s4, Δs = 54 m

A proton (charge e), traveling perpendicular to a magnetic field, experiences the same force as an alpha particle (charge 2e) which is also traveling perpendicular to the same field. The ratio of their speeds, vproton/valpha is:

a painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer. How far from the wall should the observer stand to get the best view?