The mass of a certain man is 250g..
I. What is the weight of the man on earth
ii. What is the weight of the man on the moon
I promise to mark u the brainliest, please help me

Answer:

105N

Explanation:

force = mass x acceleration

force without the 25N = 40 x 2

= 80N

On top of this, he has to counteract the 25N so the actual force is 80N + 25N which is 105N

From the analysis, it can be concluded that planet A has the strongest gravitational field, followed by planet C, and planets B and D have the same gravitational field strength.

The gravitational force acting on various masses is measured on different planets. The table shows the measured values for the forces acting on the corresponding masses:Planet Force (N) Mass (kg)A 8.0 0.50B 30 3.0C 45 3.0D 60 6.0

Method for comparing the gravitational field strengths on the different planets:First, we can use the formula for calculating gravitational force: \(`F = G (m_1m_2 / r^2)`\)where G is the universal gravitational constant `\(6.67 * 10^{-11 }Nm^2/kg^2\), m1 and m2 are the masses of the two objects in kg, and r is the distance between the centers of the objects in meters.

We know that the force is proportional to mass (F = ma). So we can calculate the acceleration due to gravity (g) on each planet by dividing the force by the mass. Therefore, we can use the formula: `g = F / m`.

Comparing the gravitational field strengths on the different planets:We will calculate the acceleration due to gravity (g) on each planet.

For planet A: `

g = F / m

= 8.0 N / 0.50 kg

= 16 \(m/s^2\)`

For planet B: `g = F / m

= 30 N / 3.0 kg

= 10 \(m/s^2\)

For planet C: `g = F / m

= 45 N / 3.0 kg

= 15 \(m/s^2\)

For planet D: `g = F / m

= 60 N / 6.0 kg

= 10 \(m/s^2\)

`So we see that planet A has the strongest gravitational field, followed by planet C, then planet B and planet D have the same gravitational field strength.

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In this kind of movement, the pace of progress of speed as such speed increase is steady. Under the influence of gravity, a ball falls when dropped from a height at rest.

A stone dropped from a tower is a free-falling body moving at a constant speed—gravity-induced acceleration—of 9.8 meters per second.

We observe that three distinct but constant acceleration motions, one after the other, constitute this issue.

Given that alt) = got3+15 +2 +10 t +5 m/s2 =

= fact) dt + C = [(20t3 + 1st 2 + 10 ++5) dt + C]

fact) dt + C = [(20t3 + 1st 2 + 10 ++5) dt + C V(f) = 4 +15 +10 +2 =5+"+5+3 +5 +2 +5 + + (Given vehicle starts from rest) V(0)=0 =) 0=5(0) 4 +5103) +5(0)2 +5(0) +(  (=0 = [(st +st +St) dt +C = 5

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1. Volcanoes occur at specific locations on the Earth's surface, mainly along tectonic plate boundaries.

2. Volcanism is the result of geological processes that occur inside the Earth.

3. Scientists use various methods to monitor volcanoes and predict eruptions. They use seismometers to detect earthquakes, which can indicate the movement of magma beneath the surface.

4. Volcanoes can also occur in other tectonic settings, such as within continental plates or on mid-ocean ridges.

Where volcanoes occur:

Volcanoes occur at specific locations on the Earth's surface, mainly along tectonic plate boundaries. There are three main types of plate boundaries: divergent, convergent, and transform. Volcanoes typically occur at divergent and convergent plate boundaries. At divergent boundaries, magma rises up to the surface as the plates move away from each other, forming fissures or vents that release lava and ash. At convergent boundaries, one plate subducts or goes beneath the other, and the magma rises up through the crust and erupts on the surface. There are also volcanic hotspots, which occur when a plume of hot magma rises from deep within the Earth's mantle and breaks through the crust to form a volcano.

Why volcanoes occur:

Volcanism is the result of geological processes that occur inside the Earth. The Earth's crust is made up of tectonic plates that move and interact with each other. When plates move apart at divergent boundaries, the underlying mantle is partially melted, and magma rises up to the surface. When plates collide at convergent boundaries, the denser oceanic plate is forced beneath the lighter continental plate in a process known as subduction. As the subducting plate sinks deeper into the Earth, it heats up and melts, creating magma that rises to the surface to form volcanoes. Volcanic hotspots are thought to be caused by a plume of hot magma rising from deep within the Earth's mantle and melting the crust above it.

How scientists can predict volcanic eruptions:

Scientists use various methods to monitor volcanoes and predict eruptions. They use seismometers to detect earthquakes, which can indicate the movement of magma beneath the surface. They also monitor gas emissions from the volcano, as changes in the types and amounts of gas can indicate an impending eruption. Ground deformation, measured using GPS and other instruments, can also provide clues about the movement of magma beneath the surface. Scientists use this data to create models of the volcano's behavior and predict when an eruption is likely to occur. They can then issue warnings and evacuation orders to minimize the impact of the eruption.

Outline the function of volcanoes in other tectonic settings:

Volcanoes can also occur in other tectonic settings, such as within continental plates or on mid-ocean ridges. In these settings, volcanoes can create new land, add nutrients to the soil, and provide geothermal energy. Volcanic eruptions can also release gases that affect the Earth's climate, such as carbon dioxide and sulfur dioxide. Over time, volcanic activity can change the landscape and shape the Earth's surface. For example, the Hawaiian Islands were formed by a volcanic hotspot that created a series of volcanic islands as the Pacific plate moved over it.

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Explanation:

Use half life equation:

A = A₀ (½)^(t / T)

where A is the final amount,

A₀ is the initial amount,

t is time,

and T is half life.

0.375 = 3 (½)^(1 / T)

0.125 = (½)^(1 / T)

(½)^3 = (½)^(1 / T)

3 = 1 / T

T = 1/3 hours

T = 20 minutes

Question 1.

A stereo with a resistance of 65 Ω is connected across a potential difference of 140 V. What is the current in this device?

Answer :-

Resistance = 65 Ω

Potential difference = 140V

where, I denotes Current, V denotes potential difference and R denotes the resistance

substituting the values,

\(\: :\implies \) I = 140/65

\(\: :\implies \) I = 2.15 Ampere

Therefore, 2.15 ampere is the required current in the device.

Question 2

A 1150 W electric toaster operates on a household circuit of 100 V. What is the resistance of the wire that makes up the heating element of the toaster?

Answer :-

We know that,

where, P denotes Power , I denotes Current and V denotes the potential difference.

Substituting the values,

\(\: :\implies \) 1150 = I × 110

\(\: :\implies \) 1150/100 = I

\(\: :\implies \) 11.5 = I

Required current = 11.5 A

Now,

where, R denotes resistance, V denotes potential difference

and I denotes the Current.

Substituting the values,

\(\: :\implies \) R = 110/1.5

\(\: :\implies \) R = 8.69 Ω

Therefore, 8.69Ω is the required resistance of the wire that makes up the heating element of the toaster

The force on the window due to the pressure of the water and atmosphere is 86.46 kN.

The pressure on the window of the submersible is calculated using the formula for calculating pressure due to fluid.

The formula for calculating pressure due to fluid is given below:

Pressure  = 454 * 1000 * 9.8

Pressure = 4449.2 kPa

Total pressure = 4449.2 kPa + 101.3 kPa

Total pressure = 4550.5 kPa

Force on the window = pressure * area

Force = 4550.5 kPa * 0.1 m * 0.19 m

Force = 86.46 kN

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The answer to the multiple questions asked on friction, Humidity,  triboelectric effect and static electricity is given below.

14. A-tape (or B-tape) is attracted to a neutral object because the tape acquires a static charge through friction, and like charges repel each other while opposite charges attract. When the charged tape comes near a neutral object, the neutral object acquires an electric charge of the opposite sign, causing an attraction between the charged tape and the neutral object.

15. Humid air makes it difficult to perform experiments with static electricity because the humidity in the air provides a path for the static charge to flow to the ground, thus reducing the charge on the object. This is because water is a good conductor of electricity, allowing the static charge to dissipate more quickly than in dry air.

16. When you rub your fingers on the non-sticky side of A-tape, you transfer some electrons from your fingers to the tape, effectively neutralizing its charge. The triboelectric effect, which is the transfer of electrons from one material to another through friction, can result in a buildup of static charge on the tape. By rubbing your fingers on the tape, you equalize the distribution of electrons between the tape and your fingers, neutralizing the charge on the tape.

17. Clothes taken out of the dryer tend to have a lot of static electricity because of the triboelectric effect. As the clothes rub against each other and the dryer drum during the tumbling and spinning action, they transfer electrons, resulting in a buildup of static charge. Dryer sheets work to eliminate this static charge by releasing positively charged ions into the dryer that neutralize the static charge on the clothes. The positively charged ions are attracted to the negatively charged clothes, effectively reducing the static charge and making the clothes less likely to cling to each other.

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If an object of mass 10.0 kg hangs from two ropes attached to the ceiling, and the tension in one rope is 150 N, the tension in the other rope will also be 150 N.

When an object is in equilibrium, the sum of the forces acting on it must be zero. In this case, the downward force due to gravity (weight) is balanced by the upward forces exerted by the two ropes. Since the object is not accelerating vertically, the tension in both ropes must be equal in magnitude. Therefore, the tension in the other rope is also 150 N.

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--The complete Question is, An object of mass 10.0 kg hangs from two ropes attached to the ceiling. If the tension in one rope is 150 N, what is the tension in the other rope?--

Answer:

There is direct evidence that the weak and electromagnetic forces were once unified as a single electroweak force from the study of subatomic particles.

Explanation:

The discovery of the W and Z bosons, which are particles that mediate the weak nuclear force, provided strong evidence that the weak force was once part of a larger electroweak force. The fact that the W and Z bosons have electric charge, and that their masses are related to the masses of the photon (the particle that mediates the electromagnetic force), supports the idea that the weak and electromagnetic forces were once part of a single electroweak force. Additionally, the mathematical symmetry of the electroweak theory, which unifies the weak and electromagnetic forces, provides further evidence for this theory.

Answer:

I'm not positive but I think the answer is Yes, because both scientific laws and theories are based on evidence.

Explanation:

Sorry I don't have notes on this, hope this helps! :)

The car will be 96.0 m from the starting line after the additional 4.0 s. The car starts from rest and travels 6.0 m in 1.0 s.

The car starts from rest and travels 6.0 m in 1.0 s. This means the car has an initial acceleration. To find the acceleration, we can use the formula: acceleration = (final velocity - initial velocity) / time. Since the car starts from rest, the initial velocity is 0 m/s.

The final velocity can be found using the formula: final velocity = initial velocity + (acceleration x time). Plugging in the values, we have: final velocity = 0 + (acceleration x 1.0). Since we know the distance traveled, we can use the formula: distance

= (initial velocity x time) + (0.5 x acceleration x time²)).

Plugging in the values, we have:

6.0 = (0 x 1.0) + (0.5 x acceleration x 1.0²).

Simplifying, we get: 6.0 = 0 + (0.5 x acceleration x 1.0).

Solving for acceleration, we find: acceleration = 12 m/s²)

Now, we can find the final velocity after the additional 4.0 s. Using the formula: final velocity = initial velocity + (acceleration x time), we have: final velocity = 0 + (12 x 4.0). Simplifying, we find: final velocity = 48 m/s.

To find the distance traveled during the additional 4.0 s, we can use the formula: distance = (initial velocity x time) + (0.5 x acceleration x time²)). Plugging in the values, we have: distance = (0 x 4.0) + (0.5 x 12 x 4.0²)). Simplifying, we find: distance = 0 + (0.5 x 12 x 16.0). Solving for distance, we find: distance = 96.0 m.

Therefore, the car will be 96.0 m from the starting line after the additional 4.0 s.

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Answer:

The net force on the skater is zero. (\(F_{net} = 0\,N\))

Explanation:

According to Newton's First Law, an object is at equilibrium when either it is at rest or moves at constant velocity, which means a net force of zero. Based on the given statement, there are no external forces acting on skate and, therefore, the net force on the skater is zero. (\(F_{net} = 0\,N\))

The light from the green glow stick held by the Olympic sprinter will appear redshifted.

The phenomenon of redshift occurs when the source of light is moving away from the observer. In this case, as the sprinter is running towards you, the distance between you and the glow stick is decreasing over time. This decrease in distance causes a Doppler shift in the frequency of the light emitted by the glow stick.

Since the light is redshifted, its wavelength increases and the frequency decreases compared to its original emitted frequency. As a result, the light that reaches your eyes appears more towards the red end of the visible spectrum.

It is important to note that the color of the glow stick itself remains the same, but due to the relative motion between the source (the sprinter) and the observer (you), the light undergoes a change in frequency and appears redshifted.

This phenomenon is similar to the redshift observed in cosmology, where the light from distant galaxies appears to be redshifted due to the expansion of the universe.

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Answer:

trigonometry (guessing)

Explanation:

ellipse: is the shape of an orbit : looks like an oval

periapsis : shortest distance between something like the moon and the planet its orbiting around like the earth

parallax is triangulation. like how gps works. looking at a star one day and then looking at it again 6 months later, an astronomer can see a difference in the viewing angle for the star. With trigonometry, the different angles yield a distance. This technique works for stars within about 400 light years of earth

https://science.howstuffworks.com/question224.htm

By comparing the intrinsic brightness to the star's apparent brightness we can calculate the distance of stars

1/r^2 rule states that the apparent brightness of a light source is proportional to the square of its distance.Jan 11, 2022

https://www.space.com/30417-parallax.html

alternative distance measurement for stars used by most astronomers is the parsec. A star with a parallax angle of 1 arcsecond has a distance of 1 parsec, or 1 parsec per arcsecond of parallax, which is about 3.26 light years

blossoms.mit.edu

.

Answer:

Explanation:

We shall apply law of conservation of momentum at the time of collision

Momentum before collision of both the pins = 1.8 x 3.6 + 0 = 6.48 kg m/s

Momentum after the collision of both the pins = 1.8 x .54 + 1.8 x v where v is velocity of second pin

So , according to law

6.48 = 1.8 x .54 + 1.8 x v

6.48 = .972  + 1.8 x v

v = 3.06 m /s to the right .

Answer:

A=4

B=3

c=2

D=1

Explanation:

since the person moving in opposite direction so displacement =distance in east/west - distance in west/east

Answer:

39850.785 J

Explanation:

0.5 x m x v^2

0.5x13x78.3^2

Answer:

KE =

\( \frac{1}{2} m {v}^{2} \\ \)

Explanation:

\(ke = \frac{1}{2} m {v}^{2} \\ = \frac{1}{2} \times 13 \times {78 . 3}^{2} \\ ke = 39850.785j \\ \)

If the cosmic microwave background peaked at a wavelength of 10 micrometers (10×10⁻⁶ meters), its temperature would be approximately 300 K.

The peak wavelength of the cosmic microwave background radiation is determined by the temperature of the universe. This relationship is described by Wien's displacement law, which states that the product of the peak wavelength and the temperature is a constant. Mathematically, this can be written as λmaxT = constant.

Given that the cosmic microwave background peaks at a wavelength of about 1 mm (1×10⁻³ meters) and has a temperature of about 3 K, we can use this information to find the constant. Rearranging the equation, we have:

constant = λmaxT

Now we can use the constant to find the temperature when the peak wavelength is 10 micrometers (10×10⁻⁶ meters). Substituting the new peak wavelength and solving for T, we get:

T = constant / λmax

Plugging in the values, we find:

T = constant / (10×10⁻⁶ meters)

Since we already know the constant from the previous information, we can calculate the temperature T, which is approximately 300 K.

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The correct option about light waves is "observed from a distance similar to or smaller than the separation between the sources".

In terms of wavelength the difference of phase is zero so the sources would experience "constructive" interphase at that point.

If the light from the sources differed by 1/2, 3/2, 5/2 wavelengths  etc, then the sources would be out of phase at that point.

The light waves coming from LED lights are the monochromatic as they are same color. Since the observer is far away from the two lights, the spacing between the two lights will be less than the distance of the observer but the waves are coherent.

Therefore, the waves are coherent because in the two-source interference waves are observed from a distance similar to or smaller than the separation between the sources.

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Answer:

The free-body diagram of the cannonball is found in the attachment below

Note The question is incomplete. The complete question is as follows:

A cannonball has just been shot out of a cannon aimed 45∘ above the horizontal rightward direction. Drag forces cannot be neglected.

Draw the free-body diagram of the cannonball.

Explanation:

Free-body diagrams are diagrams used to show the relative magnitude and direction of all forces acting upon an object in a given situation.

In order to construct free-body diagrams, it is important to know the various types of forces acting on the object in that situation. Then, the direction in which each of the forces is acting is determined. Finally the given object is drawn using any given representation, usually a box, and the direction of action of the forces are represented using arrows.

In the given situation of a cannonball which has just been shot out of a cannon aimed 45∘ above the horizontal rightward direction., the forces acting on it are:

F = force exerted by the cannon acting in the direction of angle of projection

Fdrag = drag force. The drag force acts in a direction opposite to the force exerted by the cannon

Fw = weight of the cannonball acting in a downward direction

The free body diagram is as shown in the attachment below.

When a light wave hits a medium and is not seen again, the wave has been absorbed. The correct answer is option c).

When a light wave hits a medium, it can either be transmitted through the medium, reflected off the medium, or absorbed by the medium. If the wave is absorbed, it is not seen again because the energy of the wave is converted into another form of energy, such as heat. This is why some materials appear opaque, as they absorb most of the light that hits them.

The absorbed light wave can also excite electrons in the medium, causing them to move to a higher energy level. This can result in the emission of light from the medium, such as in fluorescence or phosphorescence. Encoding, on the other hand, refers to the process of converting information into a different format or language, and is not related to the behavior of light waves in a medium.

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Answer:

M g - T = NB         vertical force on box B

T - m g = NA         vertical force on box A

T must be uniform (same) throughout length of string

(M - m) g = NB + NA = (M + m) a        since F accelerates M and m

Note: one can enclose the entire system with no external forces

No net vertical force or horizontal force is present - the net vertical force is A + B + C - masses of these objects * g

One would write a = (M - m) * g / (M + m)

The masses and tensions influence the accelerations of the boxes while considering friction and normal forces on block C.

When box A and B are released, their accelerations can be calculated using Newton's second law, (F = ma), where (F) is the net force and (a) is the acceleration.

The net force is the difference between the gravitational force (mg) and the tension in the rope (T).

For box A:

Net force: \(\rm \(F_{\text{net,A}} = T - mg\)\)

Using \(\rm \(T = m \cdot a_A\)\), where \(\rm \(a_A\)\) is the acceleration of box A:

\(\rm \(ma_A = T - mg\)\\\rm \(a_A = \frac{T}{m} - g\)\)

For box B:

Net force: \(\rm \(F_{\text{net,B}} = mg - T\)\)

Using \(\rm \(T = M \cdot a_B\)\), where \(\rm \(a_B\)\) is the acceleration of box B:

\(\rm \(Ma_B = mg - T\)\\\rm \(a_B = g - \frac{T}{M}\)\)

Now, considering friction and the normal force between block C and the ground:

The friction force \(\rm (\(F_{\text{friction}}\))\) can be calculated using the equation \(\rm \(F_{\text{friction}} = \mu N\)\), where \(\rm \(\mu\)\) is the coefficient of friction and (N) is the normal force.

The normal force (N) is equal in magnitude to the weight of block C \((mg_{C})\) since it's not accelerating vertically:

\(\rm \(N = mg_C\)\)

To summarize:

Acceleration of box A: \(\rm \(a_A = \frac{T}{m} - g\)\)

Acceleration of box B: \(\rm \(a_B = g - \frac{T}{M}\)\)

Friction force on block C: \(\rm \(F_{\text{friction}} = \mu mg_C\)\)

Normal force on block C: \(\rm \(N = mg_C\)\)

The masses and tensions influence the accelerations of the boxes while considering friction and normal forces on block C.

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Answer:

10 joules

Explanation:

Voltage= U(Potential energy)/qo

Q0 is the test charge

so,

250-50=w/5*10^-2

since potential energy = work done

w=10 joules

The tension in the inclined cable is  1,416.07 N and the tension in the horizontal cable is 1,132.85 N

Resolving Tension into its rectangular components,

we get,

The vertical component is Tsinθ

The horizontal component is Tcosθ

The vertical component is equal to the weight of the cat burglar,

It is given that θ = 34.6° and Weight (W) = 793 N

Now, Tsinθ = 793

Tsin34.6° = 793

T*0.56 = 793

T = 793/0.56 = 1,416.07 N

Hence, the tension in the inclined cable is  1,416.07 N

Now, the tension in the horizontal cable is  Tcosθ = 1,416.07cos34.6 = 1,416.07*0.8 = 1,132.85 N

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The Blackbody Spectrum simulation allows us to explore how the temperature of incandescent light sources affects their emitted light.

First, discuss the relationship between temperature, radiant energy, and lighting efficiency.
Incandescent light sources emit light due to their nonzero temperature. As the temperature of the source increases, the amount of radiant energy it gives off also increases. The Blackbody Spectrum simulation helps us visualize how the temperature of a substance affects the way light is emitted.
In the simulation, a curve represents the radiation intensity and energy emitted with respect to the wavelength at a given temperature. As the temperature rises, the curve's peak shifts towards shorter wavelengths, and the area under the curve increases. This shift indicates that the emitted light becomes more energetic and intense.
However, not all of this emitted energy is in the visible spectrum; a significant portion can be in the form of infrared radiation (heat). Incandescent light sources are not very energy-efficient, as a large portion of their energy output is wasted as heat rather than visible light. The lighting efficiency of an incandescent source is determined by the percentage of radiant energy that falls within the visible spectrum.
To summarize, the Blackbody Spectrum simulation allows us to explore how the temperature of incandescent light sources affects their emitted light. As the temperature increases, the emitted radiant energy also increases. However, a considerable amount of energy is lost as heat, making incandescent sources less energy-efficient compared to other lighting technologies.

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Answer:

the distance between the particles and the amount of electric charge that they carried.

Explanation:

hope this helps

Answer:  charge and distance between them

Explanation:

a species of bat navigates by emitting short bursts of sound waves that have a frequency range that peaks at 58 khz. if a bat is flying at speed  4.0 m/s toward a stationary object, The frequency of the sound waves reaching the object is 58.77 kHz.

f' = f ((v + v(O))/(v + v(bat))

where v is the speed of sound in air (about 343 m/s), f is the frequency of the bat's sound (58 kHz), v_obs is the observer's speed (zero in this case because the object is immobile), and v_bat is the bat's speed (4.0 m/s, in the direction of the object).

Putting all the values, we get frequency as

f' = 58 kHz ((343 + 0)/(343 - 4.0))

f' = 58 kHz (1.013)

f' = 58.77 kHz

Therefore, the frequency of the sound waves reaching the stationary object will be approximately 58.77 kHz.

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(a)Path Difference -  1.93 × 10^-6 m; (b) In terms of lambda -  3.00 ;(c) P corresponds to maximum condition

d=0.150mm, L=140cm, λ=643nm, and y=1.80cm.

(a) Path difference δ for the rays from the two slits arriving at

P :δ = dsinθ = d/sin(θ) = d/L * yδ = (0.150 mm / (1400 mm)) * 1.8 cmδ = 1.93 × 10^-6 m

(b) The path difference in terms of λ :δ/λ = (1.93 × 10^-6 m) / (643 × 10^-9 m)δ/λ = 3.00

(c) P corresponds to a maximum, a minimum, or an intermediate condition.

In this experiment, the distance between the two slits and the screen is enormous compared to the wavelength of light, and so the light waves can be thought of as travelling in parallel lines (θ ≈ tanθ = y/L). Since the paths are in phase, constructive interference occurs when δ = mλ, where m is an integer.

Thus, P corresponds to a maximum as constructive interference occurs.

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