The second law of thermodynamics states that spontaneous processestend to be accompanied by entropy increase. Consider, however, thefollowing spontaneous processes:
the growth of plants from simple seeds to well-organizedsystems
the growth of a fertilized egg from a single cell to a complexadult organism
the formation of snowflakes from molecules of liquid water withrandom motion to a highly ordered crystal
the growth of organized knowledge over time
In all these cases, systems evolve to a state of less disorder andlower entropy, apparently violating the second law ofthermodynamics. Could we, then, consider them as processesoccurring in systems that are not isolated?

True or False?

Answer:

3.258 m/s

Explanation:

k = Spring constant = 263 N/m (Assumed, as it is not given)

x = Displacement of spring = 0.7 m (Assumed, as it is not given)

[tex]\mu[/tex] = Coefficient of friction = 0.4

Energy stored in spring is given by

[tex]U=\dfrac{1}{2}kx^2\\\Rightarrow U=\dfrac{1}{2}\times 263\times 0.7^2\\\Rightarrow U=64.435\ J[/tex]

As the energy in the system is conserved we have

[tex]\dfrac{1}{2}mv^2=U-\mu mgx\\\Rightarrow v=\sqrt{2\dfrac{U-\mu mgx}{m}}\\\Rightarrow v=\sqrt{2\dfrac{64.435-0.4\times 8\times 9.81\times 0.7}{8}}\\\Rightarrow v=3.258\ m/s[/tex]

The speed of the 8 kg block just before collision is 3.258 m/s

The speed of the 8 kg block before the collision is found by considering the work done by friction and the initial potential energy in the spring and using the conservation of energy principle to solve for the speed.

To find the speed of the 8 kg block just before it collides with the 2 kg block, we must calculate the motion of the block under the influence of friction before the collision occurs.

First, we determine the force of friction using the coefficient of friction (μ) and the normal force (N), which in this case equals the weight of the block (m × g) since the surface is horizontal. So, the force of friction is given by Fₙ = μ × m × g = 0.4 × 8 kg × 9.8 m/s² = 31.36 N.

Since there is friction, work done by friction must be considered to find the velocity. The work done by friction (Wₙ) is the force of friction (Fₙ) multiplied by the distance the block moves before colliding with the second block (d), which is not given in this problem. But if we had the distance, we could use the work-energy principle, where the net work done on an object is equal to the change in its kinetic energy (KE).

Assuming no other forces act on the block, then the work done by friction is equal to the initial potential energy stored in the spring. Using Hooke's Law, the potential energy (PE) in the spring is given by PE = ½ kx², where k is the spring constant and x is the compression distance. This energy gets converted entirely into the kinetic energy of the 8 kg block minus the losses due to friction.

Using the conservation of energy:

PE - Wₙ = ½ m v²

½ kx² - Fₙ × d = ½ m v²

We solve this equation for v, the speed of the block, if we know the distance (d).

Answer:

21.73 cm

Explanation:

We have given parameters:

Mass of block, m = 2.0 kg

Force constant of spring, k = 264 N/m

Length of rough area,  L  = 10 cm = 0.1 m

Co-efficient of kinetic friction , [tex]\mu_{k}[/tex] = 0.54

Block's speed after crossing rough area, v = 2.7 m/s

Block's initial speed ( when it was released from compressed spring), [tex]v_{0}[/tex] = 0 m/s

We need to find the distance that the spring was initially compressed, x = ?

Hence, we well apply Work-Energy principle which indicates that,

Work done by the friction = Change in the total energy of block

[tex]- \mu_{k} * mg * L = (\frac{1}{2} * mv^{2} - \frac{1}{2} * mv_{0} ^{2} ) + (0 - \frac{1}{2} * kx^{2})[/tex]

-0.54 * 2 * 9.8 * 0.1 = (1/2 * 2 * [tex]2.7^{2}[/tex] - 1/2 * 2 * 0) + (0 - 1/2 * 264 * [tex]x^{2}[/tex])

x = 0.2173 m = 21.73 cm

Answer:

14009. 72 kgms^-1

Explanation:

Momentum is the product of an objects mass and velocity

The calculated value of  the magnitude of the car's momentum be  14006.72 kg.m/s or  68739 lb.mph.

Momentum is a property of a body moving in linear motion, It is product of mass and velocity. As velocity is a vector quantity, momentum is also a vector quantity with magnitude and direction. SI unit of momentum be kg.m/s.

Given parameters:

Mass of the car: m =  946.4 kg. = 2083 lb

Speed of the car: v = 14.8 m/s. = 33.0 mph

So, momentum of the car: p = mv = 946.4 × 14.8 kg.m/s = 14006.72 kg.m/s.

or p = mv = 2083 lb×33.0 mph  = 68739 lb.mph.

Hence,  the magnitude of its momentum be  14006.72 kg.m/s or  68739 lb.mph.

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Answer:

I = 0.0674 kg.m²

Explanation:

given,

mass = 13.5 Kg

torsion constant = k = 0.618 N.m

number of cycle = 28

time = 58.1 s

Time of one cycle

[tex]T = \dfrac{58.1}{28}[/tex]

[tex]T =2.075\ s[/tex]

we know,

[tex]T = 2\pi\sqrt{\dfrac{I}{k}}[/tex]

[tex]I = k (\dfrac{T}{2\pi})^2[/tex]

[tex]I =0.618\times \dfrac{T^2}{4\pi^2}[/tex]

[tex]I =0.618\times \dfrac{2.075^2}{4\pi^2}[/tex]

      I = 0.0674 kg.m²

the rotational inertia of the object is equal to  I = 0.0674 kg.m²

Answer:

B. Green

Explanation:

The change in wavelength that is caused when one object is moving towards another object from the perspective of the viewer is called the Doppler effect.

When objects move close to one another then wavelength reduces which is called blue shift while the opposite case causes the wavelength to increase which is called red shift.

Here, the color of the traffic light is yellow and you are getting closer to it so the wavelength should blue shift and green should appear.

Answer:

a. v' = 24.22 x 10 ⁻³

b. β = 0.825 T

Explanation:

V = 150 V , E = 6.0 x 10 ⁶ N / C

a.

¹/₂ * m * v² = e * V

v = √ 2 * e / m * V

v = √ 2 * 1.76 x 10 ¹¹ * 150 v  = √ 5.28 x 10 ¹³

v = 7.26 x 10 ⁶ m /s

v' / c = 7.26 x 10 ⁶ m /s  / 3.0 x 10 ⁸ = 24.22 x 10 ⁻³

b.  

β = E / V

β = 6.0 x 10 ⁶ / 7.266 x 10 ⁶

β = 0.825 T

c.

When the increasing the accelerating potential speed it doesn't change the up wired the electric force and the magnetic force the electron be beat down more .

Answer:

Explanation:

Given

mass of bullet [tex]m=7 gm[/tex]

mass of gun and Puck [tex]M=150 gm[/tex]

speed of gun and Puck is [tex]v=0.53 m/s[/tex]

Let speed of bullet be u

conserving Momentum

initial momentum=Final Momentum

[tex]0=Mv+mu[/tex]

[tex]u=-\frac{M}{m}v[/tex]

[tex]u=-\frac{150}{7}\times 0.53=-11.35 m/s[/tex]          

negative sign indicates that bullet is moving in opposite direction        

Using the law of conservation of momentum, it's possible to calculate that the bullet moves in the opposite direction of the gun and puck at a speed of approximately 113.6 m/s.

This question represents a practical example of the conservation of momentum. According to the law of conservation of momentum, the total momentum of an isolated system remains constant if no external forces act on it. As the system (gun and bullet) is initially at rest, the total momentum is zero, and it remains zero even after the bullet is fired.

Let's denote the bullet's velocity as v. The momentum of the bullet after firing is its mass times its velocity, or 0.007kg * v. The momentum of the gun and puck is their combined mass times their velocity, or 0.150kg * 0.530m/s. As the total momentum should remain zero, these two quantities must be equal and opposite. Thus, 0.007kg * v = - 0.150kg * 0.530m/s. Solving for v, we find that the bullet's speed is approximately -113.6 m/s. The negative sign indicates that the bullet moves in the opposite direction of the gun and puck.

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Answer:

Explanation:

Given

Pivot is at h=0.76 L

Where L is the length of String

Conserving Energy at A and B

[tex]mgL=\frac{1}{2}mu^2[/tex]

where u=velocity at bottom

[tex]u=\sqrt{2gL}[/tex]

After coming at bottom the ball completes the circle with radius r=L-0.76 L

Suppose v is the velocity at the top

Conserving Energy at B and C

[tex]\frac{1}{2}mu^2=mg(2r)+\frac{1}{2}mv^2[/tex]

Eliminating m

[tex]u^2=4r+v^2[/tex]

[tex]v^2=u^2-4\cdot gr[/tex]

[tex]v^2=2gL-4g(L-0.76L)[/tex]

[tex]v^2=1.04gL[/tex]

[tex]v=\sqrt{1.04gL}[/tex]          

It takes 0.05 seconds for a part undergoing simple harmonic motion with a frequency of 5.00 Hz to move from x=0 to x=-1.80 cm.

The student is asking how long it takes for a machine part undergoing simple harmonic motion (SHM) with a frequency of 5.00 Hz and amplitude of 1.80 cm to move from its equilibrium position (x=0) to its maximum negative displacement (x=-1.80 cm).

In SHM, the period (T) is the time required for one complete cycle of the motion. The period can be calculated using the formula T = 1/f, where f is the frequency. For a frequency of 5.00 Hz, the period is T = 1/5.00 Hz = 0.20 s. In half of a period, the part moves from one extreme to the other, passing through the equilibrium position at the quarter-period. Thus, to go from x=0 to x=-1.80 cm, it takes a quarter of this period, or (1/4)T = 0.05 s.

Therefore, it takes 0.05 seconds for the part to travel from x=0 to x=-1.80 cm while undergoing SHM.

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To solve this problem it is necessary to apply the equations related to the conservation of momentum. Mathematically this can be expressed as

[tex]m_1v_1+m_2v_2 = (m_1+m_2)v_f[/tex]

Where,

[tex]m_{1,2}[/tex]= Mass of each object

[tex]v_{1,2}[/tex] = Initial velocity of each object

[tex]v_f[/tex]= Final Velocity

Since the receiver's body is static for the initial velocity we have that the equation would become

[tex]m_2v_2 = (m_1+m_2)v_f[/tex]

[tex](0.42)(21) = (90+0.42)v_f[/tex]

[tex]v_f = 0.0975m/s[/tex]

Therefore the velocity right after catching the ball is 0.0975m/s

The receiver's final velocity is approximately 0.098 m/s.

First, we need to find the initial momentum of the ball by multiplying its mass (0.42 kg) by its velocity (21 m/s). The initial momentum of the ball is therefore 8.82 kg·m/s. Next, we need to find the final momentum of the receiver by multiplying his mass (90 kg) by his final velocity. Since he catches the ball at the highest point in his jump, his final velocity is 0 m/s. So the final momentum of the receiver is 0 kg·m/s. According to the law of conservation of momentum, the initial momentum of the ball must be equal to the final momentum of the receiver. Therefore, the final velocity of the receiver is 8.82 kg·m/s divided by his mass (90 kg), which is approximately 0.098 m/s.

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Answer:

6.13428 rev/s

Explanation:

[tex]I_f[/tex] = Final moment of inertia = 4.2 kgm²

I = Moment of inertia with fists close to chest = 5.7 kgm²

[tex]\omega_i[/tex] = Initial angular speed = 3 rev/s

[tex]\omega_f[/tex] = Final angular speed

r = Radius = 76 cm

m = Mass = 2.5 kg

Moment of inertia of the skater is given by

[tex]I_i=I+2mr^2[/tex]

In this system the angular momentum is conserved

[tex]L_f=L_i\\\Rightarrow I_f\omega_f=I_i\omega_i\\\Rightarrow \omega_f=\dfrac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\dfrac{(5.7+2\times 2.5\times 0.76^2)3}{4.2}\\\Rightarrow \omega_f=6.13428\ rev/s[/tex]

The rotational speed will be 6.13428 rev/s

Final answer:

To solve this problem, we need to use the principle of conservation of angular momentum. The skater will be spinning at approximately 8.0 rev/s when his hands are brought to his chest.

Explanation:

To solve this problem, we need to use the principle of conservation of angular momentum. According to this principle, the initial angular momentum of the skater with his arms extended is equal to the final angular momentum when his hands are brought to his chest.

The initial angular momentum is given by the equation:

Li = Iiωi

where Li is the initial angular momentum, Ii is the initial moment of inertia, and ωi is the initial angular velocity. The final angular momentum is given by:

Lf = Ifωf

where Lf is the final angular momentum, If is the final moment of inertia, and ωf is the final angular velocity.

Since the skater is pulling his hands in to his chest, his moment of inertia decreases, and his angular velocity increases. We can set up the following equation:

Iiωi = Ifωf

Substituting the given values, we have:

(5.7 kg.m2)(2π(3.0 rev/s)) = (4.2 kg.m2)ωf

Solving for ωf, we find that the skater will be spinning at approximately 8.0 rev/s when his hands are brought to his chest.

Answer:

Approximately [tex]\rm 2.0\; V[/tex].

Approximately [tex]\rm 30 \; mA[/tex]. (assumption: the LED here is an Ohmic resistor.)

Explanation:

The two resistors here [tex]R_1= 10\; \Omega[/tex] and [tex]R_2= 100\; \Omega[/tex] are connected in parallel. Their effective resistance would be equal to

[tex]\displaystyle \frac{1}{\dfrac{1}{R_1} + \dfrac{1}{R_2}} = \frac{1}{\dfrac{1}{10} + \dfrac{1}{100}} = \frac{10}{11} \; \Omega[/tex].

The current in a serial circuit is supposed to be the same everywhere. In this case, the current through the LED should be [tex]20\; \rm mA = 0.020\; \rm A[/tex]. That should also be the current through the effective [tex]\displaystyle \rm \frac{10}{11} \; \Omega[/tex] resistor. Make sure all values are in standard units. The voltage drop across that resistor would be

[tex]V = I \cdot R = 0.020 \times \dfrac{10}{11} \approx 0.182\; \rm V[/tex].

The voltage drop across the entire circuit would equal to

In this case, that value would be equal to [tex]1.83 + 0.182 \approx 2.0\; \rm V[/tex]. That's the voltage that needs to be supplied to the circuit to achieve a current of [tex]20\; \rm mA[/tex] through the LED.

Assuming that the LED is an Ohmic resistor. In other words, assume that its resistance is the same for all currents. Calculate its resistance:

[tex]\displaystyle R(\text{LED}) = \frac{V(\text{LED})}{I(\text{LED})}= \frac{1.83}{0.020} \approx 91.5\; \Omega[/tex].

The resistance of a serial circuit is equal to the resistance of its parts. In this case,

[tex]\displaystyle R = R(\text{LED}) + R(\text{Resistors}) = 91.5 + \frac{10}{11} \approx 100\; \Omega[/tex].

Again, the current in a serial circuit is the same in all appliances.

[tex]\displaystyle I = \frac{V}{R} = \frac{3}{100} \approx 0.030\; \rm A = 30\; mA[/tex].

Answer:

45 N, 30N ,20N

Explanation

torque = perpendicular force× distance

perpendicular force = torque/ distance

1) perpendicular force = 18/0.400 = 45 N

2)  perpendicular force = 18/0.600 = 30N

3) perpendicular force = 18/0.900 = 20N

We have that for the Question " the perpendicular force that you would have to apply to the end of the wrench in order to produce the desired torque" it can be said that

From the question we are told

You want to apply 18 N ⋅ m of torque to tighten a nut, using a wrench with a length of either 0.400 m, 0.600 m, or 0.900 m. For each wrench, determine the perpendicular force that you would have to apply to the end of the wrench in order to produce the desired torque.

Generally the equation for the Force is mathematically given as

t=F*d

a)

[tex]t=F*d\\\\F=\frac{t}{d}\\\\F=\frac{18}{0.4}\\\\[/tex]

F=45N

b)

[tex]t=F*d\\\\F=\frac{t}{d}\\\\\F=\frac{18}{0.6}\\\\[/tex]

F=30N

c)

[tex]t=F*d\\\\F=\frac{t}{d}\\\\F=\frac{18}{0.6}\\\\[/tex]

F=20N

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Answer:

Explanation:

Here is the full question and answer,

The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This spitting ability is enabled by the presence of a groove in the roof of the mouth of the archerfish. The groove forms a long, narrow tube when the fish places its tongue against it and propels drops of water along the tube by compressing its gill covers.

When an archerfish is hunting, its body shape allows it to swim very close to the water surface and look upward without creating a disturbance. The fish can then bring the tip of its mouth close to the surface and shoot the drops of water at the insects resting on overhead vegetation or floating on the water surface.

Part A: At what speed v should an archerfish spit the water to shoot down a floating insect located at a distance 0.800 m from the fish? Assume that the fish is located very close to the surface of the pond and spits the water at an angle 60 degrees above the water surface.

Part B: Now assume that the insect, instead of floating on the surface, is resting on a leaf above the water surface at a horizontal distance 0.600 m away from the fish. The archerfish successfully shoots down the resting insect by spitting water drops at the same angle 60 degrees above the surface and with the same initial speed v as before. At what height h above the surface was the insect?

Answer

A.) The path of a projectile is horizontal and symmetrical ground. The time is taken to reach maximum height, the total time that the particle is in flight will be double that amount.

Calculate the speed of the archer fish.

The time of the flight of spitted water is,

[tex]t = \frac{{2v\sin \theta }}{g}[/tex]

Substitute [tex]9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}[/tex] for g and [tex]60^\circ[/tex]  for [tex]\theta[/tex] in above equation.

[tex]t = \frac{{2v\sin 60^\circ }}{{9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}}}\\\\ = \left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\\[/tex]  

Spitted water will travel [tex]0.80{\rm{ m}}[/tex] horizontally.

Displacement of water in this time period is

[tex]x = vt\cos \theta[/tex]

Substitute [tex]\left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}[/tex] for [tex]t\rm 60^\circ[tex] for [tex]\theta[/tex] and [tex]0.80{\rm{ m}}[/tex] for x in above equation.

[tex]\\0.80{\rm{ m}} = v\left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\left( {\cos 60^\circ } \right)\\\\0.80{\rm{ m}} = {v^2}\left( {0.1767{\rm{ }}} \right)\frac{1}{2}{{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\\\\v = \sqrt {\frac{{2\left( {0.80{\rm{ m}}} \right)}}{{0.1767\;{{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}}}} \\\\ = 3.01{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\[/tex]

B.) There are two component of velocity vertical and horizontal. Calculate vertical velocity and horizontal velocity when the angle is given than calculate the time of flight when the horizontal distance is given. Value of the horizontal distance, angle and velocity are given. Use the kinematic equation to solve the height of insect above the surface.

Calculate the height of insect above the surface.

Vertical component of the velocity is,

[tex]{v_v} = v\sin \theta[/tex]

Substitute [tex]3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}[/tex] for v and [tex]60^\circ[/tex]  for [tex]\theta[/tex] in above equation.

[tex]\\{v_v} = \left( {3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}} \right)\sin 60^\circ \\\\ = 2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\[/tex]

Horizontal component of the velocity is,

[tex]{v_h} = v\cos \theta[/tex]

Substitute [tex]3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}[/tex] for v and [tex]60^\circ[/tex]  for [tex]\theta[/tex] in above equation.

[tex]\\{v_h} = \left( {3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}} \right)\cos 60^\circ \\\\ = 1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\[/tex]

When horizontal [tex]({0.60\;{\rm{m}}}[/tex] distance away from the fish.  

The time of flight for distance (d) is ,

[tex]t = \frac{d}{{{v_h}}}[/tex]

Substitute [tex]0.60\;{\rm{m}}[/tex] for d and [tex]1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}[/tex] for [tex]{v_h}[/tex] in equation [tex]t = \frac{d}{{{v_h}}}[/tex]

[tex]\\t = \frac{{0.60\;{\rm{m}}}}{{1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}}}\\\\ = 0.3987{\rm{ s}}\\[/tex]

Distance of the insect above the surface is,

[tex]s = {v_v}t + \frac{1}{2}g{t^2}[/tex]

Substitute [tex]2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}[/tex] for [tex]{v_v}[/tex] and [tex]0.3987{\rm{ s}}[/tex] for t and [tex]- 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}[/tex] for g in above equation.

[tex]\\s = \left( {2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}} \right)\left( {0.3987{\rm{ s}}} \right) + \frac{1}{2}\left( { - 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right){\left( {0.3987{\rm{ s}}} \right)^2}\\\\ = 0.260{\rm{ m}}\\[/tex]

Answer:

Explanation:

According to Kepler's law a radius vector joining planet and sun swept equal area in equal interval of time thus it can be applied for comets.

when a comet is nearer  to sun it has to swept more area so its velocity is more nearer to the sun ,

The basics of this formula comes from conservation of angular momentum  thus comet moves faster when it approaches the sun.

This question is wrong because of not correct values.The Correct question is

A gas in a cylinder is held at a constant pressure of 1.80×10⁵Pa and is cooled and compressed from 1.70m³ to 1.20m³. The internal energy of the gas decreases by 1.40×10⁵J.

(a) Find the work done by the gas.

(b) Find the absolute value of the heat flow, |Q|, into or out of the gas, and state the direction of the heat flow.

(c) Does it matter whether the gas is ideal? Why or why not?

Answer:

(a) W= -9×10⁴J

(b) |Q|=2.3×10⁵

(c) It does not matter whether gas is ideal or non-ideal

Explanation:

Given

V₁=1.70m³

V₂=1.20m³

p=1.8×10⁵ pa

ΔU= -1.40×10⁵J

For (a) work done by gas

[tex]W=p(V_{2}-V_{1} )\\W=1.8*10^{5}(1.2-1.7)\\ W=-9*10^{4}J[/tex]

For (b) Heat flow |Q|

|Q|=ΔU+W

[tex]Q=(-1.4*10^{5}) +(-9*10^{4})\\Q=-2.3*10^{5}J\\So\\|Q|=|-2.3*10^{5}|\\|Q|=2.3*10^{5}J[/tex]

For (c) part

It does not matter whether the gas is ideal or not because the first law of thermodynamics which applied in our solution could applied to any material ideal or non ideal  

Answer:

It depends on the rate at which the current through it is changing.          

Explanation:

As per the Faraday's law, the induced emf is given by :

[tex]\epsilon=-L\dfrac{di}{dt}[/tex]

Where

L is the inductance of the inductor

[tex]\dfrac{di}{dt}[/tex] is the rate of change of current

So, the self-induced emf of an inductor depends on the rate at which the current through it is changing. Hence, the correct option is (d).

Answer:

[tex]W=-3.2\times 10^4\ J[/tex]

Explanation:

Given:

Process 1:

Process 2:

We know that the work done by an ideal gas is given as:

[tex]W=P\times (V_f-V_i)[/tex]

Now for process 1:

[tex]W_1=0\ J[/tex]

∵there is no change in volume in this process.

For process 2:

[tex]W_2=4\times 10^5\time (0.12-0.2)\ J[/tex]

[tex]W_2=-3.2\times 10^4\ J[/tex]

∵Negative , sign indicates that the work is being done on the gas here since the gas is being compressed.

Hence the total work done by the gas during this two step process is :

[tex]W=W_1+W_2[/tex]

[tex]W=0-3.2\times 10^4[/tex]

[tex]W=-3.2\times 10^4\ J[/tex] is the work done by the gas.

To solve this problem it is necessary to apply the concepts related to the principle of superposition, as well as to constructive interference. From the definition we know that this can be expressed mathematically as

[tex]sin\theta_m = \frac{m\lambda}{d}[/tex]

Where

m = Any integer which represent the number of repetition of spectrum

[tex]\lambda[/tex]= Wavelength

d = Distance between slits

From triangle (Watch image below)

[tex]tan\theta_1 = \frac{y_1}{L}[/tex]

[tex]tan\theta_1 = \frac{0.488}{1.64}[/tex]

[tex]\theta_1 = 16.57\°[/tex]

Replacing the angle at the first equation for m=1 we have

[tex]\lambda = d sin \theta_1[/tex]

Each of the distances (d) would be defined by

d = \frac{1}{5300} = 0.0001886

[tex]\lambda = (\frac{1}{5300}) sin(16.57)[/tex]

[tex]\lambda = 538nm[/tex]

Therefore the wavelength of the laser light is 538nm