A frictionless piston-cylinder device contains 10 kg of superheated vapor at 550 kPa and 340oC. Steam is then cooled at constant pressure until 60 percent of it, by mass, condenses. Determine (a) the work (W) done during the process. (b) What-if Scenario: What would the work done be if steam were cooled at constant pressure until 80 percent of it, by mass, condenses

Answers

Answer 1

Answer:

a) the work (W) done during the process is -2043.25 kJ

b) the work (W) done during the process is -2418.96 kJ

Explanation:

Given the data in the question;

mass of water vapor m = 10 kg

initial pressure P₁ = 550 kPa

Initial temperature T₁ = 340 °C

steam cooled at constant pressure until 60 percent of it, by mass, condenses; x = 100% - 60% = 40% = 0.4

from superheated steam table

specific volume v₁ = 0.5092 m³/kg

so the properties of steam at p₂ = 550 kPa, and dryness fraction

x = 0.4

specific volume v₂ = v[tex]_f[/tex] + xv[tex]_{fg[/tex]

v₂ = 0.001097 + 0.4( 0.34261 - 0.001097 )

v₂ = 0.1377 m³/kg

Now, work done during the process;

W = mP₁( v₂ - v₁ )

W = 10 × 550( 0.1377 - 0.5092 )

W = 5500 × -0.3715

W = -2043.25 kJ

Therefore, the work (W) done during the process is -2043.25 kJ

( The negative, indicates work is done on the system )

b)

What would the work done be if steam were cooled at constant pressure until 80 percent of it, by mass, condenses

x₂ = 100% - 80% = 20% = 0.2

specific volume v₂ = v[tex]_f[/tex] + x₂v[tex]_{fg[/tex]

v₂ = 0.001097 + 0.2( 0.34261 - 0.001097 )

v₂ = 0.06939 m³/kg

Now, work done during the process will be;

W = mP₁( v₂ - v₁ )

W = 10 × 550( 0.06939 - 0.5092 )

W = 5500 × -0.43981

W = -2418.96 kJ

Therefore, the work (W) done during the process is -2418.96 kJ


Related Questions

You walk into a room and you see 4 chickens on a bed 2 cows on the floor and 2 cats in a chair. How many legs are on the ground? (I know this answer just a riddle to see who knows it) (:

Answers

Answer:

18

Explanation:

I'm pretty sure I got it right

After de Broglie proposed the wave nature of matter, Davisson and Germer demonstrated the wavelike behavior of electrons by observing an interference pattern from electrons scattering off what

Answers

Answer:

Scattering is an interaction that can happen when a given particle or wave, like an electron, impacts a target or material. Then the electron changes it's original path and leaves some energy in the process. (This is a really simplified explanation of scattering, this is a really complex phenomenon, but let's not dive into that path)

Particularly, Davisson and Germer used a beam of electrons against a target of nickel, and these scattered electrons were detected by a detector. All of that in a vacuum chamber.

Then the correct answer is a nickel target.

"After de Broglie proposed the wave nature of matter, Davisson and Germer demonstrated the wavelike behavior of electrons by observing an interference pattern from electrons scattering off a nickel target"

The value of mass remains constant but weight changes place to place why​

Answers

Explanation:

No matter where you are in the universe, your mass is always the same: mass is a measure of the amount of matter which makes up an object. Weight, however, changes because it is a measure of the force between an object and body on which an object resides (whether that body is the Earth, the Moon, Mars, et cetera).

Explanation:

Hence, weight of a body will change from one place to another place because the value of g is different in different places. For example, the value of g on moon is 1/6 times of the value of g on earth. As mass is independent of g , so it will not change from place to place.

Mary needs to row her boat across a 160 m-wide river that is flowing to the east at a speed of 1.5 m/s. Mary can row with a speed of 3.6 m/s. If Mary points her boat due north, how far from her intended landing spot will she be when she reaches the opposite shore? What is her speed with respect to the shore?

Answers

Answer: 66.67 m, 44.44 s

Explanation:

Given

Velocity of flow is [tex]u=1.5\ m/s[/tex]

Mary can row with speed [tex]v=3.6\ m/s[/tex]

Width of the river [tex]y=160\ m[/tex]

Flow will drift the Mary towards east, while Mary boat will cause it to travel in North direction

time taken to cross river

[tex]\Rightarrow t=\dfrac{160}{3.6}\\\\\Rightarrow t=\dfrac{400}{9}\ s[/tex]

Flow will drift Mary by

[tex]\Rightarrow x=ut\\\\\Rightarrow x=1.5\times \dfrac{400}{9}\\\\\Rightarrow x=66.67\ m[/tex]

Velocity w.r.t shore is

[tex]\Rightarrow v_{net}=\sqrt{3.6^2+1.5^2}\\\Rightarrow v_{net}=\sqrt{15.21}\\\Rightarrow v_{net}=3.9\ m/s[/tex]

1. A 20.0 N force directed 20.0° above the horizontal is applied to a 6.00 kg crate that is traveling on a horizontal
surface. What is the magnitude of the normal force exerted by the surface on the crate?

Answers

N = 52.0 N

Explanation:

Given: [tex]F_a= 20.0\:\text{N}=\:\text{applied\:force}[/tex]

[tex]m=6.00\:\text{kg}[/tex]

[tex]N = \text{normal force}[/tex]

The net force [tex]F_{net}[/tex] is given by

[tex]F_{net} = N + F_a\sin 20 - mg=0[/tex]

Solving for N, we get

[tex]N = mg - F_a\sin 20[/tex]

[tex]\:\:\:\:\:\:= (6.00\:\text{kg})(9.8\:\text{m/s}^2) - (20.0\:\text{N}\sin 20)[/tex]

[tex]\:\:\:\:\:\:= 52.0\:\text{N}[/tex]

A car hurtles off a cliff and crashes on the canyon floor below. Identify the system in which the net momentum is zero during the crash.

Answers

Solution :

It is given that a car ran off from a cliff and it crashes on canyon floor. Now the system of a car as well as the earth together have a  [tex]\text{ net momentum of zero}[/tex] when the car crashes on the canyon floor, thus reducing the momentum of the car to zero. The earth also stops its upward motion and it also reduces the momentum to zero.

The slope of a d vs t graph represents velocity. Describe 3 ways you know this to be true.

Answers

Answer:

Look at explanation

Explanation:

I only know 1 way, there is another way you can rephrase this using derivatives but that's pretty much the same thing.

The slope is calculated by Δy/Δx so the slope of distance vs time graph is Δd/Δt which is the velocity

explain why sound wave travel faster in liquid than gas​

Answers

Answer:

Because gas contains free molecules but not liquid.

Please mark as brainliast

A cylindrical disk of wood weighing 45.0 N and having a diameter of 30.0 cm floats on a cylinder of oil of density 0.850 g>cm3 (Fig. E12.19). The cylinder of oil is 75.0 cm deep and has a diameter the same as that of the wood. (a) What is the gauge pressure at the top of the oil column

Answers

Answer:

665.25 Pa

Explanation:

Given data :

Weight of the disk, w = 45 N

Diameter, d = 30 cm

                    = 0.30 m

Therefore, radius of the disk,

[tex]$r=\frac{d}{2}$[/tex]

[tex]$r=\frac{0.30}{2}$[/tex]

   = 0.15 m

Now, area of the cylindrical disk,

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

[tex]$A=3.14 \times (0.15)^2$[/tex]

   [tex]$=0.07065 \ m^2$[/tex]

∴ The gauge pressure at the top of the oil column is :

   [tex]$p=\frac{w}{A}$[/tex]

   [tex]$p=\frac{47}{0.07065}$[/tex]

      = 665.25 Pa

Therefore, the gauge pressure is 665.25 Pa.

The definition of pressure allows to find the result for the pressure at the top of the oil cylinder is:

The pressure is: P = 636.6 Pa

The pressure is defined by the relationship between perpendicular force and area.

          [tex]P = \frac{F}{A}[/tex]

where P is pressure, F is force, and A is area.

They indicate that the wooden cylinder weighs W = 45.0 N and has a diameter of d = 30 cm = 0.30 m.

The area is:

        A = π r² = [tex]\pi \frac{d^2}{4}[/tex]  

In the attachment we see a diagram of the forces, where the weight of the cylinder and the thrust are equal.

         B-W = 0

          B = W

The force applied to the liquid is the weights of the cylinder. Let's replace.

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

Let's calculate.

          [tex]P = \frac{45 \ 4 }{\pi \ 0.30^2 }[/tex] P = 45 4 / pi 0.30²

          P = 636.6 Pa

In conclusion using the definition of pressure we can find the result for the pressure at the top of the oil cylinder is:

The pressure is: P = 636.6 Pa.

Learn more about pressure here: brainly.com/question/17467912

If a bale of hay behind the target exerts a constant friction force, how much farther will your arrow burry itself into the hay than the arrow from the younger shooter

Answers

Answer:

The arrow will bury itself farther by 3S₁

Explanation:

lets assume; the Arrow shot by me has a speed twice the speed of the arrow fired by the younger shooter

Given that ; acceleration is constant , Frictional force is constant

                    A₂ =   A₁

Vf²₂ - Vi²₂ / 2s₂  = Vf₁² - Vi₁² / 2s₁ ---- ( 1 )

final velocities = 0

Initial velocities : Vi₂ = 2(Vi₁ )

Back to equation 1

0 - (2Vi₁ )² / 2s₂ =  0 - Vi₁² / 2s₁

hence :

s₂ = 4s₁

hence the Arrow shot by me will burry itself farther by :

s₂ - s₁ = 3s₁

Note :  S1 = distance travelled by the arrow shot by the younger shooter

a student weighs 1200N they are standing in an elevator that is moving downwards at a constant speed of

Answers

Answer:

Elevator That Is Moving Downwards At A Constant Speed Of 4.9 M/S. What Is The Magnitude Of The Net Force Acing On The Student?

This problem has been solved!

This problem has been solved!See the answer

This problem has been solved!See the answerA student weighs 1200N. They are standing in an elevator that is moving downwards at a constant speed of 4.9 m/s. What is the magnitude of the net force acing on the student?

Explanation:

use this R= m(g-a), where R = reaction = weight, m= mass, a= acceleration and g= acceleration due to gravity

An astronaut on a distant planet wants to determine its acceleration due to gravity. The astronaut throws a rock straight up with a velocity of 19 m/s and measures a time of 24.4 s before the rock returns to his hand. What is the acceleration (magnitude and direction) due to gravity on this planet

Answers

Answer:

1.56 m/s²

Explanation:

Projectile motion is a form of motion where an object moves in parabolic path (trajectory). Projectile motion only occurs when there is one force applied at the beginning on the trajectory, after which the only interference is from gravity.

The total time (time of flight) of an object is given by:

T = 2usinθ / g

where u is the initial velocity, θ is the angle with horizontal and g is the acceleration due to gravity

Since the astronaut throws a rock straight up, hence θ = 90°, u = 19 m/s, T = 24.4 s.

T = 2usinθ / g

Substituting:

24.4 = 2(19)(sin90)/g

g = 2(19)(sin90) / 24.4

g = 1.56 m/s²

2.
Select the correct answer.
Erica is working in the lab. She wants to remove the fine dust particles suspended in a sample of oil. Which method is she most likely to use?

Answers

Answer:

Reverse Osmosis

Explanation:

Reverse osmosis is a type of filtration that involves passing a solvent through a semipermeable membrane in the opposite direction that natural osmosis does. Separation is always enforced through the use of pressure in this process. Ions, fine dust particles, molecules, and larger particles are typically removed from solvents using this method. The technique is particularly popular in the treatment and purification of water.

Answer:

filtration is used to separate

A wave moves in a rope with a certain wavelength. A second wave is made to move in the same rope with twice the wavelength of the first wave. The frequency of the second wave is _______________ the frequency of the first wave.

Answers

Answer:

The frequency of the second wave is half of the frequency of first one.

Explanation:

The wavelength of the second wave is double is the first wave.

As we know that the frequency is inversely proportional to the wavelength of the velocity is same.

velocity = frequency x wavelength

So, the ratio of frequency of second wave to the first wave is

[tex]\frac{f_2}{f_1} =\frac{\lambda _1}{\lambda _2}\\\\\frac{f_2}{f_1} =\frac{\lambda _1}{2\lambda _1}\\\\\frac{f_2}{f_1} =\frac{1}{2}\\\\[/tex]

The frequency of the second wave is half of the frequency of first one.

Two pendulums have the same dimensions (length {L}) and total mass (m). Pendulum A is a very small ball swinging at the end of a uniform massless bar. In pendulum B, half the mass is in the ball and half is in the uniform bar.
1. Find the period of pendulum A for small oscillations.
2. Find the period of pendulum B for small oscillations.

Answers

Answer:

1) [tex]T_{A} = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex], 2) [tex]T_{B} \approx 1.137\cdot T_{A}[/tex], where [tex]T_{A} = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex].

Explanation:

1) Pendulum A is a simple pendulum, whose period ([tex]T_{A}[/tex]) is determined by the following formula:

[tex]T_{A} = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex] (1)

Where:

[tex]l[/tex] - Length of the massless bar.

[tex]g[/tex] - Gravitational acceleration.

2) Pendulum B is a physical pendulum, whose period ([tex]T_{B}[/tex]) is determined by the following formula:

[tex]T_{B} = 2\pi \cdot \sqrt{\frac{I_{O}}{m\cdot g\cdot l} }[/tex] (2)

Where:

[tex]m[/tex] - Total mass of the pendulum.

[tex]g[/tex] - Gravitational acceleration.

[tex]l[/tex] - Length of the uniform bar.

[tex]I_{O}[/tex] - Moment of inertia of the pendulum with respect to its suspension axis.

The moment of inertia can be found by applying the formulae of the moment of inertia for a particle and the uniform bar and Steiner's Theorem:

[tex]I_{O} = \frac{1}{2} \cdot m\cdot l^{2}+\frac{1}{24}\cdot m\cdot l^{2} + \frac{3}{4}\cdot m\cdot l^{2}[/tex]

[tex]I_{O} = \frac{31}{24}\cdot m\cdot l^{2}[/tex] (3)

By applying (3) in (2) we get the following expression:

[tex]T_{B} = 2\pi \cdot \sqrt{\frac{\frac{31}{24}\cdot m \cdot l^{2} }{m\cdot g \cdot l} }[/tex]

[tex]T_{B} = 2\pi \cdot \sqrt{\frac{31\cdot l}{24\cdot g} }[/tex]

[tex]T_{B} = \sqrt{\frac{31}{24} } \cdot \left(2\pi \cdot \sqrt{\frac{l}{g} }\right)[/tex]

[tex]T_{B} \approx 1.137\cdot T_{A}[/tex]

1. The period of pendulum A for small oscillations is  

[tex]T_A=2\pi\sqrt{\dfrac{L}{g}}[/tex]

2. The period of pendulum B for small oscillations.

[tex]T_B=1.137.T_A[/tex]

What is simple harmonic motion?

Simple harmonic motion is the periodic motion or back and forth motion of any object with respect to its equilibrium or mean position. The restoring force is always acting on the object which try to bring it to the equilibrium.

1) Pendulum A is a simple pendulum, whose period () is determined by the following formula:

[tex]T_A=2\pi\sqrt{\dfrac{L}{g}}[/tex]

Where:

l - Length of the massless bar.

g - Gravitational acceleration.

2) Pendulum B is a physical pendulum, whose period () is determined by the following formula:

[tex]T_A=2\pi\sqrt{\dfrac{I_o}{mgl}}[/tex] .............................2

Where:

m - Total mass of the pendulum.

g - Gravitational acceleration.

l - Length of the uniform bar.

Io- Moment of inertia of the pendulum with respect to its suspension axis.

The moment of inertia can be found by applying the formulae of the moment of inertia for a particle and the uniform bar and Steiner's Theorem:

[tex]I_o=\dfrac{1}{2}ml^2+\dfrac{1}{24}ml^2+\dfrac{3}{4}ml^2[/tex]

[tex]I_o=\dfrac{31}{24}ml^2[/tex]..................................3

By applying (3) in (2) we get the following expression:

[tex]T_B=2\pi\sqrt{\dfrac{\frac{31}{24}ml^2}{mgl}[/tex]

[tex]T_B=2\pi\sqrt{\dfrac{31l}{24g}}[/tex]

[tex]T_B=\sqrt{\dfrac{31}{24}}. (2\pi\sqrt{\dfrac{l}{g}})[/tex]

[tex]TB=1.137.T_A[/tex]

Thus to know more about Simple harmomnic motion follow

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Question 7 of 10
A railroad freight car with a mass of 32,000 kg is moving at 2.0 m/s when it
runs into an at-rest freight car with a mass of 28,000 kg. The cars lock
together. What is their final velocity?
A.1.1 m/s
B. 2.2 m/s
C. 60,000 kg•m/s
D. 0.5 m/s

Answers

Answer:

a

Explanation:

you take 32,000kg ÷2.0m

A 36.0 kg child slides down a long slide in a playground. She starts from rest at a height h1 of 24.00 m. When she is partway down the slide, at a height h2 of 11.00 m, she is moving at a speed of 7.80 m/s.
Calculate the mechanical energy lost due to friction (as heat, etc.).

Answers

Answer:

E = 3495.96 J

Explanation:

From the law of conservation of energy:

Total Mechanical Energy at h1 = Total Mechanical Energy at h2

Kinetic energy at h1 + potential energy at h1 = Kinetic energy at h2 + potential energy at h2 + Mechanical Energy Lost due to Friction

[tex]K.E_{h1}+P.E_{h1} = K.E_{h2}+P.E_{h2} + E\\\\\frac{1}{2}mv_1^2\ J + mgh_1 = \frac{1}{2}mv_2^2 + mgh_2 + E\\\\\frac{1}{2}(36\ kg)(0\ m/s)_1^2\ J + (36\ kg)(9.81\ m/s^2)(24\ m)_1 = \frac{1}{2}(36\ kg)(7.8\ m/s)_2^2 + (36\ kg)(9.81\ m/s^2)(11\ m)_2 + E\\\\0\ J + 8475.84\ J = 1095.12\ J + 3884.76\ J + E\\E = 8475.84\ J - 1095.12\ J - 3884.76\ J\\[/tex]

E = 3495.96 J

A uniform horizontal bar of mass m1 and length L is supported by two identical massless strings. String A Both strings are vertical. String A is attached at a distance d

Answers

Answer:

a)  T_A = [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex] ,  b) T_B = g [m₂ ( [tex]\frac{x}{d} -1[/tex]) + m₁ ( [tex]\frac{L}{ 2d} -1[/tex]) ]

c)  x = [tex]d - \frac{m_1}{m_2} \ \frac{L}{2d}[/tex],  d)  m₂ = m₁  ( [tex]\frac{ L}{2d} -1[/tex])

Explanation:

After carefully reading your long sentence, I understand your exercise. In the attachment is a diagram of the assembly described. This is a balancing act

a) The tension of string A is requested

The expression for the rotational equilibrium taking the ends of the bar as the turning point, the counterclockwise rotations are positive

      ∑ τ = 0

      T_A d - W₂ x -W₁ L/2 = 0

      T_A = [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex]

b) the tension in string B

we write the expression of the translational equilibrium

       ∑ F = 0

       T_A - W₂ - W₁ - T_B = 0

       T_B = T_A -W₂ - W₁

       T_ B =   [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex]  - g m₂ - g m₁

       T_B = g [m₂ ( [tex]\frac{x}{d} -1[/tex]) + m₁ ( [tex]\frac{L}{ 2d} -1[/tex]) ]

c) The minimum value of x for the system to remain stable, we use the expression for the endowment equilibrium, for this case the axis of rotation is the support point of the chord A, for which we will write the equation for this system

         T_A 0 + W₂ (d-x) - W₁ (L / 2-d) - T_B d = 0

at the point that begins to rotate T_B = 0

          g m₂ (d -x) -  g m₁  (0.5 L -d) + 0 = 0

          m₂ (d-x) = m₁ (0.5 L- d)

          m₂ x = m₂ d - m₁ (0.5 L- d)

          x = [tex]d - \frac{m_1}{m_2} \ \frac{L}{2d}[/tex]

 

d) The mass of the block for which it is always in equilibrium

this is the mass for which x = 0

           0 = d - \frac{m_1}{m_2} \  \frac{L}{2d}

         [tex]\frac{m_1}{m_2} \ (0.5L -d) = d[/tex]

          [tex]\frac{m_1}{m_2} = \frac{ d}{0.5L-d}[/tex]

          m₂ = m₁  [tex]\frac{0.5 L -d}{d}[/tex]

          m₂ = m₁  ( [tex]\frac{ L}{2d} -1[/tex])

During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race

Answers

Answer:

The maximum speed of Bolt for the 100 m race is 14.66 m/s

Explanation:

Given;

initial distance covered by Bolt, d = 200 m

time of this motion, t = 19.3 s

The second distance covered by Bolt, = 100 m

Assuming Bolt maintained the same acceleration for both races.

His acceleration can be determined from the 200 m race.

d = ut + ¹/₂at²

where;

u is his initial velocity = 0

d =  ¹/₂at²

[tex]at^2 = 2d\\\\a = \frac{2d}{t^2} \\\\a = \frac{2\times 200}{19.3^2} \\\\a = 1.074 \ m/s^2[/tex]

Let the final or maximum velocity for the 100 m race = v

v² = u² + 2ad₂

v² =  2 x 1.074 x 100

v² = 214.8

v = √214.8

v = 14.66 m/s

The maximum speed of Bolt for the 100 m race is 14.66 m/s

3. Calculate the force it would take to accelerate a 50 ka bike at a rate of 3 m/s2 (6 points)

Answers

Answer:

150 N

Explanation:

Given that,

Acceleration (a) = 3 m/s²Mass of the bike (m) = 50 kg

We are asked to calculate force required.

[tex]\longrightarrow[/tex] F = ma

[tex]\longrightarrow[/tex] F = (50 × 3) N

[tex]\longrightarrow[/tex] F = 150 N

uppose that 3 J of work is needed to stretch a spring from its natural length of 32 cm to a length of 49 cm. (a) How much work (in J) is needed to stretch the spring from 37 cm to 45 cm

Answers

Answer:

0.113 J

Explanation:

Applying,

w = ke²/2................. Equation 1

Where w = workdone in stretching the spring, k = spring constant, e = extension

make k the subject of the equation

k = 2w/e²................ Equation 2

From the question,

Given: w = 3 J, e = 49-32 = 17 cm = 0.17 m

Substitute these values into equation 2

k = (2×3)/0.17²

k = 6/0.17

k = 35.29 N/m

(a) if the spring from 37 cm to 45 cm,

Then,

w = ke²/2

Given: e = 45-37 = 8 cm = 0.08

w = 35.29(0.08²)/2

w = 0.113 J

Traveling waves propagate with a fixed speed usually denoted as v (but sometimes c). The waves are called __________ if their waveform repeats every time interval T.

a. transverse
b. longitudinal
c. periodic
d. sinusoidal

Answers

Answer:

periodic

Explanation:

ASK YOUR TEACHER A 2.0-kg mass swings at the end of a light string with the length of 3.0 m. Its speed at the lowest point on its circular path is 6.0 m/s. What is its kinetic energy at an instant when the string makes an angle of 50 degree with the vertical

Answers

Answer:

  K_b = 78 J

Explanation:

For this exercise we can use the conservation of energy relations

starting point. Lowest of the trajectory

        Em₀ = K = ½ mv²

final point. When it is at tea = 50º

        Em_f = K + U

        Em_f = ½ m v_b² + m g h

where h is the height from the lowest point

        h = L - L cos 50

        Em_f = ½ m v_b² + mg L (1 - cos50)

energy be conserve

        Em₀ = Em_f

         ½ mv² = ½ m v_b² + mg L (1 - cos50)

         K_b = ½ m v_b² + mg L (1 - cos50)

let's calculate

          K_b = ½ 2.0 6.0² + 2.0 9.8 6.0 (1 - cos50)

          K_b = 36 +42.0

          K_b = 78 J

two identical eggs are dropped from the same height. The first eggs lands on a dish and breaks, while the second lands on a pillow and does not break. Which quantities are the same in both situations

Answers

Answer:

The height is the same

Explanation:

Because they were at the same height but they fell at different velocities

A 190 g glider on a horizontal, frictionless air track is attached to a fixed ideal spring with force constant 160 N/m. At the instant you make measurements on the glider, it is moving at 0.835 m/sm/s and is 4.00 cmcm from its equilibrium point.

Required:
a. Use energy conservation to find the amplitude of the motion.
b. Use energy conservation to find the maximum speed of the glider.
c. What is the angular frequency of the oscillations?

Answers

(a) Let x be the maximum elongation of the spring. At this point, the glider would have zero velocity and thus zero kinetic energy. The total work W done by the spring on the glider to get it from the given point (4.00 cm from equilibrium) to x is

W = - (1/2 kx ² - 1/2 k (0.0400 m)²)

(note that x > 4.00 cm, and the restoring force of the spring opposes its elongation, so the total work is negative)

By the work-energy theorem, the total work is equal to the change in the glider's kinetic energy as it moves from 4.00 cm from equilibrium to x, so

W = ∆K = 0 - 1/2 m (0.835 m/s)²

Solve for x :

- (1/2 (160 N/m) x ² - 1/2 (160 N/m) (0.0400 m)²) = -1/2 (0.190 kg) (0.835 m/s)²

==>   x ≈ 0.0493 m ≈ 4.93 cm

(b) The glider attains its maximum speed at the equilibrium point. The work done by the spring as it is stretched away from equilibrium to the 4.00 cm position is

W = - 1/2 k (0.0400 m)²

If v is the glider's maximum speed, then by the work-energy theorem,

W = ∆K = 1/2 m (0.835 m/s)² - 1/2 mv ²

Solve for v :

- 1/2 (160 N/m) (0.0400 m)² = 1/2 (0.190 kg) (0.835 m/s)² - 1/2 (0.190 kg) v ²

==>   v1.43 m/s

(c) The angular frequency of the glider's oscillation is

√(k/m) = √((160 N/m) / (0.190 kg)) ≈ 29.0 Hz

The amplitude of the motion is 0.049 cm. The maximum speed of the glider is 1.429 m/s. The angular frequency of the oscillation is 29.02 rad/s

From the given information;

the mass of the glider = 190 gForce constant k = 160 N/mthe horizontal speed of the glider [tex]v_x[/tex] = 0.835 m/sthe distance away from the equilibrium = 4.0 cm = 0.04 m

Using energy conservation E, the amplitude of the motion can be calculated by using the formula:

[tex]\mathbf{E = \dfrac{1}{2}mv^2 + \dfrac{1}{2}kx^2}[/tex]

[tex]\mathbf{E = \dfrac{1}{2}(0.19 \ kg )\times (0.835)^2 + \dfrac{1}{2}(160) (0.04)^2}[/tex]

[tex]\mathbf{E =0.194 \ J}[/tex]

Similarly, we know that:

[tex]\mathbf{E = \dfrac{1}{2}kA^2}[/tex]

Making amplitude A the subject, we have:

[tex]\mathbf{A = \sqrt{\dfrac{2E}{k}}}[/tex]

[tex]\mathbf{A = \sqrt{\dfrac{2(0.194)}{160}}}[/tex]

[tex]\mathbf{A =0.049 \ cm}[/tex]

Again, using the energy conservation, the maximum speed of the glider can be calculated by using the formula:

[tex]\mathbf{E =\dfrac{1}{2} mv^2 _{max}}[/tex]

[tex]\mathbf{v _{max} = \sqrt{\dfrac{2E}{m}}}[/tex]

[tex]\mathbf{v _{max} = \sqrt{\dfrac{2\times 0.194}{0.19}}}[/tex]

[tex]\mathbf{v _{max} = 1.429 \ m/s}[/tex]

The angular frequency of the oscillation can be computed by using the expression:

[tex]\mathbf{\omega = \sqrt{\dfrac{k}{m}}}[/tex]

[tex]\mathbf{\omega = \sqrt{\dfrac{160}{0.19}}}[/tex]

ω = 29.02 rad/s

Learn more about energy conservation here:

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No esporte coletivo, um dos principais fatores desenvolvidos é o desenvolvimento social. Qual desses não faz parte das virtudes ensinadas no esporte?

Companheirismo
Humildade
Ser justo (Fair Play)
Vencer independente do que precise ser feito

Answers

Answer:

fair palybtgshsisuehdh

State whether plastic is biodegradable or non-biodegradable ? Give reasons for your answer.​

Answers

Answer:

non biodegradable

Explanation:

It is non biodegradable because plastic cannot dispose off easily ..

If you change the motor in your vehicle you need to notify the DMV within ____,
days of this change.

-20
-25
-10
-15

Answers

when you change your motor on your vehicle you need to notify the DMV within 10 days

If you change the motor in your vehicle you need to notify the DMV within 10 days of this change.

An engine or motor

An engine or motor is a machine designed to convert one or more forms of energy into mechanical energy.

Available energy sources include potential energy (e.g. energy of the Earth's gravitational field as exploited in hydroelectric power generation), heat energy (e.g. geothermal), chemical energy, electric potential, and nuclear energy (from nuclear fission or nuclear fusion). Many of these processes generate heat as an intermediate energy form, so heat engines have special importance. Some natural processes, such as atmospheric convection cells convert environmental heat into motion (e.g. in the form of rising air currents). Mechanical energy is of particular importance in transportation but also plays a role in many industrial processes such as cutting, grinding, crushing, and mixing.

Mechanical heat engines convert heat into work via various thermodynamic processes. The internal combustion engine is perhaps the most common example of a mechanical heat engine, in which heat from the combustion of fuel causes rapid pressurization of the gaseous combustion products in the combustion chamber, causing them to expand and drive a piston, which turns a crankshaft. Unlike internal combustion engines, a reaction engine (such as a jet engine) produces thrust by expelling reaction mass, by Newton's third law of motion.

Learn more about motor

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The period of a simple pendulum is 3.5 s. The length of the pendulum is doubled. What is the period T of the longer pendulum?

Answers

Explanation:

The period T of a simple pendulum is given by

[tex]T = 2 \pi \sqrt{\dfrac{l}{g}}[/tex]

Doubling the length of the pendulum gives us a new period T'

[tex]T' = 2 \pi \sqrt{\dfrac{l'}{g}} = 2 \pi \sqrt{\dfrac{2l}{g}}[/tex]

[tex]\:\:\:\:\:\:\:= \sqrt{2} \left(2 \pi \sqrt{\dfrac{l}{g}} \right)[/tex]

[tex]\:\:\:\:\:\:\:= \sqrt{2}\:T = \sqrt{2}(3.5\:\text{s})= 4.95\:\text{s}[/tex]

The free-fall acceleration at the surface of planet 1 is 22 m/s^2. The radius and the mass of planet 2 are twice those of planet 1. What is the free-fall acceleration on planet 2?

Answers

Answer:

g₂ = 11 m/s²

Explanation:

The value of free-fall acceleration on the surface of a planet is given by the following formula:

[tex]g = \frac{Gm}{r^2}[/tex]

where,

g = free-fall acceleration

G = Universal Gravitational Constant

m = mass of the planet

r = radius of planet

FOR PLANET 1:

[tex]g_1 = \frac{Gm_1}{r_1^2}\\\\\frac{Gm_1}{r_1^2} = 22 m/s^2[/tex] --------------------- equation (1)

FOR PLANET 2:

[tex]g_2 = \frac{Gm_2}{r_2^2}\\\\g_2 = \frac{G(2m_1)}{(2r_1)^2}\\\\g_2 = \frac{1}{2}\frac{Gm_1}{r_1^2}\\\\[/tex]

using equation (1):

[tex]g_2 = \frac{g_1}{2}\\\\g_2 = \frac{22\ m/s^2}{2}[/tex]

g₂ = 11 m/s²

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