the temperature of the photosphere is about the temperature of the photosphere is about 3,200 k. one million k. ten million k. 11,000 k. 5,800 k.

The rms emf of the AC generator  is 297 Volt. Hence, option (C) is correct.

Energy is transformed from one form to another using a generator or a battery. One terminal in these gadgets charges up positively, whereas the other charges up negatively. Therefore, work performed on a unit of electric charge constitutes an electromotive force.

Given that: the  maximum output emf of  the AC generator:

V₀ =4.20 x 10^2 V.

Hence, r.m.s  output emf of  the AC generator:

[tex]V_{rms[/tex] = V₀/√2

= 4.20 x 10^2/√2 Volt

=297 Volt.

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rises or reductions in electromagnetic wave frequency Decide what happens to each as O stays constant as the wave travels away from the source.

The two are interconnected; a shifting magnetic field will result in a shifting electric field, and the opposite is true. These fluctuating fields emit electromagnetic waves. In order to travel, mechanical waves require a medium, but electromagnetic waves do not.

In other words, oscillating magnetic and electric fields are what make up electromagnetic radiation (EM). Electromagnetic waves are produced when a magnetic field interacts with an electric field. It is referred to as "electromagnetic waves" to describe them.

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The object requires the MOST power to lift  a 2 kg box 2 m in 1 s. Hence, option (B) is correct.

The quantity of energy moved or converted per unit of time is known as power in physics. The watt, or one joule per second, is the unit of power in the International System of Units. Power is also referred to as activity in ancient writings. A scalar quantity is power.

Power required for lifting a 2 kg box 1 m in 1 s = (2×9.8×1)/1 watt = 19.6 watt.

Power required for lifting a 2 kg box 2 m in 1 s = (2×9.8×2)/1 watt = 39.2 watt.

Power required for lifting a 2 kg box 1 m in 2 s = (2×9.8×1)/2watt =9.8 watt.

Power required for lifting a 2 kg box 2 m in 2 s = (2×9.8×2)/2watt =19.6 watt.

So, The most power required in  lifting a 2 kg box 2 m in 1 s.

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Excavated and brought to a construction site, this soil will be used in a compact The volume fill is 8680.73 m³.

Given,

         The specific gravity of soil solids = 2.75

     The situ moisture content of the soil =15%

                                       Saturation level = 70%

                                                 void ratio = 0.9.

                                               Dry weight = 103.5

We know that,

                             Yd = [tex]\frac{16.5}{1+\frac{18}{100} }[/tex]

                            Yd  = 14.34KN/m³

The volume of soil to be excavated to produce 7651m³ is

                               [tex]V=\frac{Yd (compacted)}{Yd(insite)} Compacted Fill Volume[/tex]

                               [tex]V=\frac{16.27}{14.34}7651[/tex]

                               [tex]V=8680.73m^{3}[/tex]

             No of trucks = [tex]\frac{V*Y(insity)}{1.78}[/tex] = 826 trucks

The most conventional approach involves using the human body, such as pinching or measuring the volume of an object using the size of your hand. On the other hand, because of its variations, the human body is very unpredictable. Utilizing reasonably consistent and durable natural things, such as gourds, sheep or pig stomachs, and bladders allows for a more accurate method of measuring volume. Since metallurgy and glass manufacture have evolved throughout time, small amounts are routinely measured using standardized man-made containers today. For granular goods, the container is shaken or leveled off to produce a virtually smooth surface. This method is not the most accurate way to measure.

Hence,

          Excavated and brought to a construction site, this soil will be used in a compact The volume fill is 8680.73 m³.

And

               No of trucks = [tex]\frac{V*Y(insity)}{1.78}[/tex] = 826 trucks.

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The following Seyfart spiral galaxy is not one of Hubble's types of galaxies.

It has hundreds of billions of stars, enough gas and dust to create billions more stars, and at least ten times as much dark matter as all the stars and gas combined, making our galaxy, the Milky Way, typical. Gravity also keeps everything in place.

The Milky Way is spiral-shaped, just like more than two-thirds of all known galaxies. There is a lot of energy and occasionally bright flares being produced in the spiral's center. Astronomers come to the conclusion that the Milky Way's core is a supermassive black hole based on the enormous gravitational force that would be necessary to account for star motion and energy released.

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The duct enlarges until it takes on a square cross-section with sides of 15 inches. Considering that the air can be regarded as an incompressible fluid V₂ is 0.94 ft/sec.

By the equation of continuity

(Rate of volumetric flow rate is constant)

A₁v₁ = A₂v₂

(6×6)×5.9(inch²ft/sec) = (15×15)×v₂

V₂ = 5.9(36/225) ft/sec

V₂ = 0.944 ft/sec

In fluid dynamics, or more broadly in continuum mechanics, an incompressible fluid  (also known as an isochoric flow) is a flow in which the material density within a fluid parcel—an infinitesimal volume that moves with the flow velocity—remains constant. The idea that the flow velocity has zero divergences is comparable to the idea that incompressibility exists (see the derivation below, which illustrates why these conditions are equivalent).

The fact that fluid flows incompressibly does not mean that it cannot be compressed. In the derivation that follows, it is demonstrated that even compressible fluids can be roughly described as incompressible flow under the appropriate circumstances. If a fluid parcel flows with the flow velocity, it is said to have incompressible flow, meaning the density stays constant.

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The arrow that best represented the acceleration position is the arrow 4.

Arrow 4 represents the direction of the acceleration vector at the position of the arrow.

In a velocity-time graph, the acceleration corresponds to the slope of the curve.

In fact, acceleration is defined as the ratio between the change in velocity and the time interval.

A arrow indicates the location of an acceleration at a certain location. The acceleration line will radiate outward in the case of a positive direction. The direction of the acceleration at a certain location is not, however, shown by the arrow. As a result, reading the values of positive signs on a field line is useless.

Since, direction of motion at I is nearly parallel to 4 so the arrow represented will be 4.

Therefore the arrow that best represented the acceleration position is the arrow 4.

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The average speed of the particle during the 5 second time interval isis 25m/s. Option C.

Calculation:-

Let acceleration = a

vi = 0

v1 = 50m /s

f = 5sec

v_{1} = v_{i} + at

a = 10 m/s^2

d = 1/2 a t ^ 2 +vi f

d = 1/2 * 10 *s ^ 2

d = 125m

Vavg = d/t = 125/5 .

Vavg = 25 m/s.

Acceleration is the rate of change of velocity over time, both in terms of velocity and direction. A point or object moving in a straight line accelerates as it accelerates or decelerates. Circumferential motion is accelerated because it always changes direction even at constant velocity.

Acceleration is the change in velocity over time. Whenever a change in velocity or acceleration in the direction of movement occurs. At constant velocity, acceleration is zero. Deceleration is called negative acceleration or deceleration.

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The work done by the steam when the piston moves 0.800 m is 1.75×10⁵ J and the amount of work by calculating the force exerted times the distance traveled is 1.85×10⁵ J. From this, the work done by the second part is more than the initial one. So they are not the same.

The work produced by the compression or expansion of fluid is known as pressure-volume work. Gas molecules gain kinetic energy when energy is added, which causes the gas to expand and affect its surroundings. The work is calculated using the formula W=pΔV where p is pressure, W is work done and ΔV is the change in the volume.

Given the gauge pressure (Pg) of the steam locomotive is  1.75×10⁶ N/m², the radium of piston r is 0.200 m, and the distance moved by the piston l is 0.800 m.

a) First, calculate the ΔV.

ΔV = area×distance

ΔV = πr²×l

ΔV = π×(0.20)²×0.80

ΔV = 0.1 m³

Then, the work done will be,

W = 1.75×10⁶× 0.1

W = 1.75×10⁵ J

The required answer is 1.75×10⁵ J.

b) The absolute pressure is given by the sum of gauge pressure and the atmospheric pressure, i.e., P=P₀+Pg. Then,

P = 1.01×10⁵ +  1.75×10⁶

P = 18.5×10⁵ Pa.

The force is calculated as,

F = P×A

F = 18.5×10⁵×π×(0.20)²

F = 232477 N

The work done by the piston is calculated as,

W = Force×distance

W = 232477×0.80

W = 1.85×10⁵ J

The required answer is 1.85×10⁵ J.

The answer for the first and second parts are not the same.

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The center of Milky Way Galaxy allows us to determine (d) mass of the Milky Way Galaxy that lies within 40,000 light-years of the galactic center.

What is Kepler's Third Law (Orbital Velocity Law) ?

The Orbital Velocity Law states that the orbital velocity is directly proportional to the body's mass for which it is being calculated and inversely proportional to the body's radius.

orbital velocity is the speed at which one body orbits the other body .

The term "orbit" refers to an object's consistent circular motion around the Earth.

From the above data , we can conclude that ,

A star orbiting 40,000 light-years from the center of the Milky Way Galaxy helps us to determine the mass of the Milky Way Galaxy that lies within 40,000 light-years of the galactic center.

Therefore , the correct option is (d) .

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The upper part of a thunderstorm cloud is normally positively charged, and the middle and lower parts are negatively charged.

A thunderstorm cloud is a type of dense, vertically towering cloud. They usually produce torrential rain and may cause flash flooding and straight-line winds, but they also may produce bolts of lightning.

Generally, positive charges are accumulating at higher altitudes. The ground (lower parts) normally has a slight negative charge, but when the cloud is directly overhead, the negative charge in the middle of the storm cloud repels the ground underneath, causing the ground to become positively charged.

Lightning happens when the accumulated electric charge in a thunderstorm becomes sufficiently large.

Attached below is an image of a thunderstorm cloud by the NOAA/AOML/Hurricane Research Division.

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The brain (the pressure is enough to perfuse the brain with blood is 33330.5 Pascal and Blood pressure at the brain is 311.666  torr.

Given quantities are

Blood pressure at the heart Ph=250  torr = 250∗133.322 = 33330.5 Pascal   as    [ 1torr= 133.322 Pa]

Height to the head from its heart hh =2  m

Height to the heart  from feet hf =2  m

Density of blood ρ=1.06×103  kg/m3

Blood pressure at the brain

Pb = Ph − hhρg

=   33330.5 + 2*1.06*103*9.8 = 54106.5Pa

Increase in blood pressure

P = Pf − P

b = 405.833 − 94.167 = 311.666  torr

a pressure measurement in the meter-kilogram-second system that is equal to one newton per square meter. The computer programming language Pascal, often known as PASCAL, was created from Algol and is built to handle both textual and numerical data. In the International System of Units, the pascal (Pa) is the unit of pressure or stress (SI). The mathematician and physicist Blaise Pascal is honored by the name of this object. Over a surface area of one square meter, one pascal is equal to one newton (N) of force (m2). French mathematician, philosopher, and physicist Blaise Pascal lived in the 17th century. The syringe, the hydraulic press, the roulette wheel, and an early digital calculator were all inventions credited to him. He made significant advances in atmospheric pressure and established the vacuum above the atmosphere.

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Given that the flower pot's height is 45 meters, the speed at which it will hit the ground is v = 29.69 m/s.

We are aware that kinetic energy is equivalent to KE=1/2mv2, where v is the pot's velocity.

Utilizing the law of conservation of energy, potential energy PE = mgh with v = 29.69 m/s and KE = PE 1/2mv2=mgh

V = 29.69 m/s is the speed at which the pot impacted the earth.

The definition of energy is the "capacity to accomplish work, which is the ability to apply a force producing the displacement of an item." Despite this unclear description, energy simply refers to the power that propels motion. There are two categories of energy: potential and actual.

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The force constant is 827.4 N/m, the distance is 1.48 cm and the amount of work done is 3.42 J.

To find the force constant of the spring, we use the equation F = kx, where F is the force applied to the spring, k is the force constant, and x is the displacement of the spring. We are given that the mass of the object is 2.30 kg, so the force applied to the spring is equal to the weight of the object, which is given as:

F = mg = (2.30 kg)(9.8 m/s^2) = 22.54 N.

We are also given that the spring stretches 2.74 cm, or 0.0274 m. Plugging these values into the equation above, we get:

22.54 N = k * 0.0274 m

Solving for k, we find that the force constant of the spring is k = 827.4 N/m.

To find the distance the spring stretches when a 1.25-kg object is suspended from it, we use the same equation F = k*x, but with the new values for F and x. The force applied to the spring is now:

F = mg = (1.25 kg)(9.8 m/s^2) = 12.25 N.

Solving for x, we find that the spring stretches:

x = F/k = (12.25 N)/(827.4 N/m) = 0.0148 m = 1.48 cm.

To find the amount of work an external agent must do to stretch the spring 6.40 cm from its un-stretched position, we use the following equation:

W = F×d

where W is the work done, F is the force applied to the spring, and d is the displacement of the spring.

In this case, the force applied to the spring is given as:

F = k*x = (827.4 N/m)(0.0640 m) = 53.29 N,

and the displacement of the spring is d = 0.0640 m. Plugging these values into the equation above, we find that the work done is

W = (53.29 N)(0.0640 m) = 3.42 J.

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The speed of the plane relative to the ground is obtained as  192.9 m/s.

We know that the relative speed can be obtained as the resultant of the speed of the airplane and that of the wind. We have to recall that the relative speed is the single speed that has the same effect in magnitude and direction as two or more speeds acting together.

Now we can write;

R^2 = 176^2 + 79^2

We would now have to take the square root of the both sides so as to obtain the resultant velocity and we now have;

R = √ 176^2 + 79^2

R = 192.9 m/s

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A block of mass 3 kg slides down an inclined plane at an angle of p with a massless tether attached to a pulley with mass 1 kg and radius 0.5 m at the top of the incline. The acceleration of the block is a = [m1g(sinΘ - μcosΘ)]/(m1 + 1/2m2).

Formula of acceleration = a bar = (v - v0)/t = delta v / delta t

a bar = average acceleration

v = final velocity

v0 = starting velocity

t = elapsed time

After solving by the formula we get,  a = [m1g(sinΘ - μcosΘ)]/(m1 + 1/2m2).

Acceleration refers to the rate at which an object's velocity with respect to time varies. Acceleration is a vector quantity (in that they have magnitude and direction). The direction of an object's acceleration is determined by the direction of the net force acting on it. According to Newton's Second Law, the amount of an object's acceleration is the combined result of two causes: According to the materials used to make the object, the magnitude of the net balance of all external forces acting on it is inversely proportional to its mass, but the magnitude of the net resulting force is directly proportional to the net force.

We are aware that the angle = 45°, the block's mass m1 is 3 kg, and the pulley's mass m2 is 1 kg.

Pulley radius is 0.5 meters, and its coefficient of friction is 0.4.

Four forces are at work on the mass-m1 block:

Weight is represented by m1g, normal response force by N, tension by T, and frictional force by f.

After solving we get acceleration = a = [m1g(sinΘ - μcosΘ)]/(m1 + 1/2m2)

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The speed of sound in air can be found using the formula is  v = √γRTM.

The equation for the rate of sound in air v = √γRTM may be simplified to present the equation for the speed of sound in air as a characteristic of absolute temperature.

Calculation:-

v=√γRTM

=√γRTM(273K273K)

=√(273K)γRM√T273K

⇒ 331m.

The relationship of the speed of sound, its frequency, and wavelength is the same as for all waves: vw=fλ, where vw is the rate of sound, f is its frequency, andλ is its wavelength.

The rate of that wind can be measured the use of a device known as an anemometer. An anemometer looks like a weather vane, but as opposed to measuring which route the wind is blowing with hints, it has 4 cups so that it may more accurately degree wind speed.

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(a) The flux through the triangle formed by the rails and bar at t 3.00 s is calculated to be 85.2 Wb.

(b) The EMF around the triangle at that time is 56.8 V.

(c) If the EMF is at n, where a and n are constants, the value of n is 1.

(a) The height of the triangular area enclosed by the rails and bar is the same as the distance traveled in time t. So, we have d = v * t

where, v = 5.2 m/s

We also note that the base of the triangle is the distance between the intersection points of the bar with the rails which is 2 d. The area of the triangle is

A = 1/2 * b * h = 1/2 * 2 v t * v t = v² t²

As the field is said to be uniform B = 0.35 T

The magnitude of flux is given by,

Ф = B * A = 0.35 * v² t² = 0.35 * 5.2² * t² = 9.46 v² t²

At t = 3 s, we get Ф = 85.2 Wb

(b) The magnitude of EMF is

ε = dФ/dt = 9.46 dt²/dt = 18.9 t

At t = 3 s, ε = 18.9 t = 56.8 V

(c) Our solution part of (b) shows, n = 1

The question is incomplete. The attachment shows the figure.

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The string's angle of deflection  is useful for quantitatively comparing the amount of charge on each sphere.

The angles of trajectory that all of your strings take across and behind the nut and saddles of your instrument are, in a nutshell, what is meant by this term. The string's angle, which are located above the headstock, are the downward angles your strings make when they travel through the nut slot, emerge over the fingerboard, and exit the rear edge of the nut to the string wraps on the machine head posts. It is the angle that the string makes at the bridge between your saddle and the spot where the ball end of your string is located.

The string's ability to transfer downward stresses into the guitar is governed by break angles.

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The force that the pedal exerts on the water causes the boat to move in the forward direction.

According to the third law of motion proposed by Newton, for every action force there is an equal and opposite reaction force generated.

When the man paddles it exerts a force on the water, in return the water also exerts a force on the boat of the man.

Because the mass of the man is very very less in comparison to the water body.

The water body does not move backward but the force applied by The Man on the water body makes the boat to move forward.

So we can conclude here the force that the pedal exerts on the water body causes the boat to move in the forward direction.

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The observable aspect which most useful to astronomers when measuring the dark matter content of galaxies is stellar velocities.

Dark matter is composed of particles which do not absorb, reflect, or emit light, so they cannot be detected by observing electromagnetic radiation. Dark matter is material which cannot be seen directly. We know that dark matter exists because of the effect it has on objects that we can observe directly.

Dark matter's existence was first inferred by Fritz Zwicky in 1933, a Swiss American astronomer, who discovered that the mass of all the stars in the Coma cluster of galaxies provided only about 1 percent of the mass required to keep the galaxies from escaping the cluster's gravitational pull.

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The MSE for the given situation is 14.8 if SSE for variance problem is 399.6

In insights, the mean squared error (MSE) or mean squared deviation (MSD) of an assessor (of a system for assessing an unseen amount) gauges the normal of the squares of the mistakes — that is, the typical squared contrast between the assessed values and the real worth. MSE is a gamble capability, relating to the normal worth of the squared mistake misfortune

We know very well that MSE=SSE / r×(c-1)

where r is defined as the number of treatments

and c is defined as the total number of treatments

MSE=Variance within the sample

SSE=Sum of squares of the error

We have values of SSE=399.6,r=3 and value of c=10.

According  to the question, so on putting the values, we get

=>MSE=399.6/3(10-1)

=>MSE=399.6/(3×9)

=>MSE=399.6/27

=>MSE=14.8

Hence, the  MSE for the analysis would be 14.8

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The magnitude of the electric field along the positive z axis is kqz/√(z^2+a^2)^3/2 , the angular frequency of these oscillations is √(kqq0/ma^3)

1. Given that a uniformly charged ring in the xy plane, centered at the origin.

Radius of the ring(r) = a

positive charge distributed evenly along its circumference = q

k= 1/4piepsion

The electric field can be described as, E = kq/r^2

The force exerted on an electric charge by a charge is F = q0E

Any vector can be projected along the z = A(z)=Acosθ

We know that ∑F =ma

Acceleration along z axis = a =d^2z/dt^2

The equation of motion for a simple harmonic oscillation along z is d^2z/dt^2 = -ω^2z

Here,  ω is the cosine function in a right-angled triangle is cosθ = z/√(z^2+a^2)

dE(z)=dEcosθ we know that E = kq/z^2+a^2

To calculate the magnitude of the electric field caused by complete rings, integrate dE(z)=dEcosθ

E(z) = Ecosθ = kq/(z^2+a^2) x  z/√(z^2+a^2)

E(z) = kqz/(z^2+a^2)^3/2

2. We know that F= q0E = kqq0z/(z^2+a^2)^3/2

Given The ball will oscillate along the z axis between z=d and z=-d in simple harmonic motion. Let z =d

F = kqq0d/(d^2+a^2)^3/2 = kqq0d/a^3

Force at any point along z = - kqq0z/a^3

We know F = ma =  

- kqq0z/a^3 = md^2z/dt^2

d^2z/dt^2= - kqq0z/ma^3

This equation of motion can be compared with the equation of motion for simple harmonic motions to find the angular frequency.

ω =√(kqq0/ma^3)

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The volume is decreasing by 80cm³/min.

Boyle's Law is a fundamental chemical principle that describes how a gas behaves when it is kept at a fixed temperature. The law asserts that at a certain temperature, the volume of a gas is inversely proportional to the pressure it exerts (Robert A. Boyle found this rule in 1662).

Given, V = 600 cm³

Pressure P = 150 KPa

dP/dT = 20 KPa/min

now, V × dP/dT + P × dV/dT = 0

    600×20 + 150 × dV/dT = 0

    dV/dT = -1200/150

     dV/dT = -80 cm³/min

The -ve sign indicates that volume is decreasing.

Therefore, volume is decreasing at a rate of 80 cm³/sec.

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The complete question is as follows;

Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and the volume V satisfy the equation PV=C, where C is a constant. Suppose that at a certain instant the volume is 600cm3, the pressure is 150 kPa, and the pressure is increasing at a rate of 20 kPa/min. At what rate is the volume decreasing at this instant?

Surface tension is a property of a liquid that allows it to resist an external force. It is caused by the cohesive forces between molecules in the liquid, which tend to draw the molecules together and make the surface of the liquid behave as if it were a stretched membrane or elastic skin.

Examples of surface tension include:

The ability of a liquid to form a droplet or a bead when it is placed on a surface. The surface tension of the liquid helps to hold the droplet together and prevent it from spreading out.

The ability of a liquid to flow through a small hole or capillary tube. The surface tension of the liquid helps to pull it through the small opening.

The ability of insects such as water striders to walk on water. The surface tension of the water helps to support their weight.

The ability of soap bubbles to hold their shape. The surface tension of the soap solution helps to keep the bubble walls stretched and maintain the bubble's round shape.

The ability of mercury in a thermometer to form a thin, stable column. The surface tension of the mercury helps to keep it together and prevent it from spreading out.

Betelgeuse is a very large/bright star located about 500 light years away. if a similar star is located at 2000 light years away, its relative brightness is 1/16, option (a) is correct, compared to Betelgeuse.

The brightness of a star, as seen from Earth, is determined by its luminosity (the amount of energy it emits) and its distance from Earth. As a star gets further away, its apparent brightness (the amount of light it appears to emit) decreases. To compare the relative brightness of two stars, you can use the inverse square law, which states that the intensity of a point source of light (such as a star) is inversely proportional to the square of the distance from the source. In other words, if a star is located twice as far away as another star, its apparent brightness will be 1/4 of the other star's apparent brightness. In this case, the star located at 2000 light years away is four times further away from Earth than Betelgeuse, which is located 500 light years away. Therefore, according to the inverse square law, the relative brightness of the star located at 2000 light years away compared to Betelgeuse would be 1/4 x 1/4 = 1/16. In other words, the star located at 2000 light years away would appear to be 16 times dimmer than Betelgeuse.

So, the correct answer is "1/16".

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Question - Betelgeuse is a very large/bright star located about 500 light years away. if a similar star is located at 2000 light years away, its relative brightness, compared to Betelgeuse, would be which of the following?

(a) 1/16, (b) 1/2, (c) 1/35, (d) 1/15

Answer:

7.068m

Explanation:

We have the physics formula  [tex]x=x_{0}+v_{0}t+\frac{1}{2}at^{2}[/tex]

[tex]v_{0}=[/tex] 1.1m/s

[tex]x_0=[/tex] 0m

[tex]a=[/tex] 0.4m/s²

[tex]t=[/tex] 3.8sec

Now we plug those values into the formula.

[tex]x=0+1.1(3.8)+\frac{1}{2}0.4 (3.8)^{2}[/tex]

[tex]x=4.18+0.2(14.44)[/tex]

[tex]x=4.18+2.888[/tex]

[tex]x=7.068[/tex]m

Blossom traveled 7.068m.

Hope this helps!

Answer:

Below

Explanation:

Adding salt to the water will make it CONDUCTIVE and the bulb will then have a complete circuit and possibly glow

Steven sets up an experiment as per the diagram and the bulb is not glowing, so he should mix a few teaspoons of salt into the water. Hence, option B is correct.

Charged material experiences a force when exposed to an electromagnetic field because of the fundamental characteristic of electric charge. It's conceivable for electrical ions to be positive or negative. Two charges that are opposed to one another repel one another.

An object with no net charge is said to be "neutral." Classical electrodynamics refers to the older theory of how charged particles interact, and it is still valid for problems that do not necessitate accounting for quantum phenomena.

The net charge of an isolated system, which is the total of the negative and positive charges, cannot change since electrical current is a conserved property. Electric charge is carried by subatomic particles.

The bulb is not glowing because the conduction of electricity does not take place through pure water, so he needs to put some teaspoons of salt into the water due to which the conduction of electric will take place and the bulb be glowed.

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When a parachutist opens her parachute after accomplishing terminal speed, in what path does she accelerate.

When a parachutist opens her chute, she slows down. That method she hurries up upward.Once the parachute is opened, the air resistance overwhelms the downward pressure of gravity. The internet pressure and the acceleration at the falling skydiver is upward.

A 500-N parachutist opens his chute and reports an air resistance pressure of 800 N. The internet pressure at the parachutist is. three hundred N upward (Air resistance exerts a pressure withinside the path contrary the path of the movement of the person. So, the internet pressure is 500 N down plus 800 N up, that is three hundred N up

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Option D; As the material changes, you end up with coal, and the name scientists give to the material changes. In order, from coolest (first) to warmest (last) the names given are: Peat, lignite, bituminous, anthracite.

A form of coal known as bituminous coal, sometimes known as black coal, contains bitumen, also known as asphalt. It can be black or occasionally dark brown in hue, and the seams frequently contain distinct bands of shiny and dull material. Usually, it is rigid but hard. Its quality is graded lower than that of anthracite but higher than that of lignite and sub-bituminous coal. It is the most prevalent type of coal, with deposits found all across the world, frequently in Carboniferous-era strata. Sub-bituminous coal that has been sufficiently buried to reach temperatures of at least 85 °C (185 °F) is converted into bituminous coal.

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