Two positive point charges q are placed on the x-axis, one at x = a and one at x = -a.a. Derive an expression for the electric field at points on the x-axis, where ?a < x < a.

Answers

Answer 1

Th electric field at a point between 2 points is given by E =   [tex]\frac{4kQax}{(a^2 - x^2)^2}[/tex] (-i^)

Formula for electric field at a point on x-axis is given by:

E = [tex]\frac{kq}{r^2}[/tex]  ----(i)

where k is constant, q is magnitude of charge and r is the distance from charge where electric field is to be found.

                                         

   Q (-a,0)____________________(x,0)_____________________Q(a,0)

so we have to fing electric field at point (x,0) due to the both charges shown in the figure.

due to point charge at pont (-a,0) the electric field is in the positive x direction and due to charge at point (a,0) the electric field is in negative x direction.

let [tex]E_1[/tex] be the electric field due to charge at point (-a,0)

let [tex]E_2[/tex] be the electric field due to charge at point (a,0)

[tex]E_1[/tex] = [tex]\frac{kQ}{(a+x)^2}[/tex] (i^)

where i^ is unit vector along x- axis

[tex]E_2[/tex] = [tex]\frac{kQ}{(a-x)^2}[/tex] (-i^)

[tex]E_1+E_2 = \frac{kQ}{(x+a)^2}[/tex] i^  [tex]+ \frac{kQ}{(a-x)^2}[/tex] (-i^)

=>   [tex]\frac{-4kQax}{(a^2-x^2)^2}[/tex] i^

so [tex]E_1 + E_2 =[/tex] [tex]\frac{4kQax}{(a^2 - x^2)^2}[/tex] (-i^)

so for any point a < x < a .

Th electric field is given by E =   [tex]\frac{4kQax}{(a^2 - x^2)^2}[/tex] (-i^)

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Related Questions

What is the function of the muscles around the lens in the human eye?

They focus light rays onto light-sensor cells at the back of the eye.
They control the amount of light that passes through the lens.
They send signals to the brain.
They change the shape of the lens.

Answers

Answer: They change the shape of the lens

Explanation: It enables the lens to change its shape to focus on objects at various distances

They change the shape of the lens

a 1.90-m-diameter vat of liquid is 2.10 m deep. the pressure at the bottom of the vat is 1.30 atm. what is the mass of the liquid in the vat?

Answers

A liquid vat with a 1.90 m diameter is 2.10 m deep. 1.30 atm of pressure is present at the vat's bottom. The amount of liquid in the container weighs 2788kg

Suppose we define d as the liquid's density;

The pressure at the bottom of the vat, which has a depth (h), can then be calculated as P = P0+Dgh, where P0 is atmospheric pressure.

P-P0 = Dgh

D= P-P0/gh

We are aware that density (D) = mass/volume m/v = P-P0/gh.

Volume equals area times height, hence the formula for mass is: m = P-P0/gh x area (a) x height (h) = (P-P0)/gh x π(d/2)^2

Given that P = 1.3 atm and P0 = 1 atm, the diameter of the vat (d) is 1.9 atm, which is

(1.3-1)/9.8 x (1.9/2)2 x 1.013 x 105 Pa, = 2788 kg.

As a result, we can say that the mass of the liquid in the vat is 2788kg

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how many different orientations can this largest orbital angular momentum vector have with respect to the vertical axis

Answers

For l = 3, m = -3, -2, -1, 0, 1, 2, 3, and L2 = m th, the angular momentum can be projected in 7 distinct directions.

As the rotating equivalent of linear momentum in physics, angular momentum (rarely moment of momentum or rotational momentum) is used. It is a conserved quantity, meaning that in a closed system, the total angular momentum stays constant, making it a significant physical quantity. The magnitude and direction of angular momentum are both preserved. The conservation of angular momentum is responsible for the beneficial characteristics of bicycles, motorbikes, frisbees, rifled bullets, and gyroscopes. Hurricanes from spiral galaxies and neutron stars rotate rapidly because of the conservation of angular momentum. Conservation often sets a limit on a system's potential motion but does not determine it in any way.

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rank the light intensity, from largest to smallest, at the point p in the figures.

Answers

The light intensity from  largest to smallest ranking will be  :B > D > A=C > E. where the area is measured on the plane perpendicular to the direction .

In physics, the intensity is the amount of energy that is transmitted per unit area, and the area is measured on a plane perpendicular to the direction that the energy equation will propagate. I = P/ 4(d2), with P denoting power. Let power of 1 bulb equal = P where I = intensity, d = distance at which the intensity must be determined.

case A = I =  P / (1) (1) 1 = P case with 2 = P In the situation B = I = 2P/(0.5)2 = 8P C = I = 4P / (2) (2) ^2 = P case Case (1)2 = 3P: D = I = 3P E = I = 2P /(1.5) (1.5) ^2 = 0.8 P

B > D > A=C > E will be the order of light intensity, from greatest to least.

rank the light intensity, from largest to smallest, at the point p in the figures?

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lisa exposes an iron paper clip to a very strong magnet. she is now able to pick up other paper clips using her newly magnetized clip. what happens within the paper clip in order for it to become magnetized?

Answers

Lisa can now pick up other paper clips with her freshly magnetized clip after exposing an iron paper clip to a very powerful magnet, which aligns the magnetic domains within the clip.

When a small portion of the iron filings comes into contact with the poles, which are the ends of the magnet, all of the filings are drawn to them. An opposite pole and a similar pole make up a magnet's two poles. While similar poles repel, opposite poles attract. The magnetic force between two magnets increases as the distance between them decreases.

There are three different types of magnets, Permanent magnet, Temporary magnet ,Electromagnets.

When ferromagnetic substances like iron and nickel are exposed to a magnetic field, magnets are produced. When heated to a certain temperature, metals acquire permanent magnetic properties.

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which changes by the largest factor along the main sequence from spectral types o to m: mass or luminosity?

Answers

The mass varies from only about 100M suns to about 0.08M suns or by about a factor of 1000 along the main sequence, whereas the luminosity varies from 10^6 for the most luminous stars to less than 0.001 or factors of more than 10^9 for the least luminous.

Only the radii of main sequence stars are fairly similar to that of the sun, which implies that the radii change by a smaller factor than the luminosity. The validity of the conclusion can be shown by noting that the luminosity of a star is proportional to R^2T^4, but this is beyond what is covered.

A main-sequence star is a star that burns hydrogen just like the Sun. Massive main-sequence stars are hot and in the upper left of the strip. Low-mass stars have low temperatures and are located in the lower right corner. Massive stars rapidly burn up all their hydrogen in just a few million years. luminosity and distance.

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a mass on a spring in shm has amplitude a and period t. what is the total distance traveled by the mass after a time interval t?

Answers

The displacement is zero because the mass's starting and final positions are identical (it returns to its initial position).

What is amplitude example?

It alludes to the greatest departure from equilibrium that such an object in harmonic oscillator can exhibit. As an illustration, consider how a pendulum moves through its market equilibrium (straight down) and then moves outward to its maximum distance.

Do you mean height by amplitude?

The vertical separation of both the wave from average is measured by amplitude. The wave axis, which is typically thought of as zero, is the average size of the wave across one cycle. Positive and negative numbers are assigned to heights between and above the average, accordingly.

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what minimum horsepower must a motor have to be able to drag 310 kg box along a level floor at a speed of 1.20 m/s if the coefficient of friction is 0.45

Answers

The minimum horsepower must a motor have to be able to drag 310 kg box along a level floor at a speed of 1.20 m/s is 2.199 horse power.

What is called friction?

The resistance to motion of one object travelling in relation to another is referred to as friction. It is not regarded as a fundamental force like gravity or electromagnetic, according to the International Journal of Parallel, Emergent and Distributed Systems(opens in new tab).

First change the mass of 310kg by multiplying by 9.8(gravity), to get 3038 N which equals 'n'

f=μn  =  (.45)(3038)

        = 1367.1N

Use work formula W= fΔx

                              = (1367.1N)(1.2m/s) = 1640.52 Joules/sec

                               (1 Watt = 1 Joules / sec) so you get 1640.52 watts

1 horsepower = 746 watts

1640.52 watt/ 746 watt

= 2.199 horse power

Therefore, the minimum horsepower must a motor have to be able to drag 310 kg box along a level floor at a speed of 1.20 m/s is 2.199 horse power.

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block is sliding down on a frictionless incline. is its linear momentum conserved in any directions?

Answers

No, the momentum of the block is not conserved.

The block is not isolated—that is, free from outside influences. In actuality, a part of the block's weight force operates in a direction perpendicular to the incline.

Momentum

Momentum is the result of a particle's mass and velocity. Being a vector quantity, momentum possesses both magnitude and direction. According to Isaac Newton's second equation of motion, the force applied on a particle is equal to the time rate of change of momentum.

According to Newton's second law, if a particle is subjected to a constant force for a specific amount of time, the result of the force and time is equal to the change in momentum. On the other hand, a particle's momentum represents the length of time needed for a consistent force to bring it to rest.

Any grouping of particles has a momentum equal to the vector sum of its individual momenta. A change in one particle's momentum is precisely balanced by an equal and opposite change in the momentum of another particle, according to Newton's third law, which states that particles exert equal and opposite forces on one another. The law of conservation of momentum states that when no net external force is acting on a group of particles, their total momentum remains constant.

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a ball rolls down an incline and off a horizontal ramp. ignoring air resistance, what force or forces act on the ball as it moves through the air just after leaving the horizontal ramp?

Answers

The weight of the ball acting vertically downward is the only force acting on the ball as it moves through the air just leaving the horizontal ramp.

A moving body's force is equal to the product of mass and acceleration. This is the relation stated from Newton's second law of motion. As the mass of the object increases the force that has to be applied on it to change its state will be greater.

The ball which is moving down the horizontal ramp has weight acting downwards. As it is said to ignore air resistance, it can be neglected.

Weight of the ball is nothing but the product of the mass of the ball and acceleration due to gravity.

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bert and ernie are pushing on opposite sides of the box. if they can both push with the same amount of force, what is true about the motion of the box?

Answers

The box will be in rest due to net force applied on it is zero.

What is force?

The definition of force in physics is: The push or pull on a massed object changes its velocity.

An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.

Given that Bert and Ernie are pushing on opposite sides of the box with force of same magnitude, let it be F.

So, net force acting om the force = F-F = 0.

As that net force acting on the box is zero, the box will be in rest.

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would the magnitude of the acceleration due to gravity near earth's surface increase more if earth's mass were doubled and

Answers

Yes, the magnitude of the acceleration due to gravity near Earth surface increases more if Earth's mass were doubled.

The magnitude of the acceleration due to gravity on the surface of the Earth is given by the relation,

g = GM/r²

Where,

G is the universal gravitational constant,

M the mass of the earth and,

R is the radius of the earth.

Now if the radius of the earth is kept constant and mass of the earth is doubled.

The value of the new acceleration due to gravity will become,

g' = G2M/r²

Putting GM/r² = g,

g' = 2g

So, we can see that the new acceleration due to gravity because of the increase in the mass of the earth is two times of the earlier acceleration due to gravity.

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016 (part 1 of 2) 10.0 points
A neutron in a reactor makes an elastic headon collision with the nucleus of an atom initially at rest.
Assume: The mass of the atomic nucleus is
about 11.1 the mass of the neutron.
What fraction of the neutron’s kinetic energy is transferred to the atomic nucleus?
017 (part 2 of 2) 10.0 points
If the initial kinetic energy of the neutron is
4.2 × 10−13 J, find its final kinetic energy.
Answer in units of J.
No Rounding

Answers

(a) The fraction of neutron's kinetic energy transferred to the atomic nucleus is 0.09.

(b) The final kinetic energy of the neutron is 3.82 x 10⁻¹³ J.

What is the final velocity of the atom?

The final velocity of the atom is calculated by applying the principle of conservation of linear momentum as follows;

initial momentum of the neutron = final momentum of the atom

m₁u₁  = m₂u₂

where;

m₁ is the mass of the neutronu₁ is the initial velocity of the neutronm₂ is the mass of the atomic nucleus u₂ is the final velocity of the atomic nucleus

The mass of the atomic nucleus = 11.1 m₁

u₂  = m₁u₁ / m₂

u₂  = m₁u₁ / (11.1 m₁)

u₂  = 0.09u₁

The initial kinetic energy of the neutron is calculated as;

K.Ei = ¹/₂m₁u₁²

The final kinetic energy of the atomic nucleus is calculated as;

K.Ef =  ¹/₂m₂u₂²

K.Ef =  ¹/₂(11.1 m₁)(0.09u₁)²

K.Ef = 0.09 (¹/₂m₁u₁²)

K.Ef = 0.09 (K.Ei)

The fraction of neutron's kinetic energy transferred to the atomic nucleus is calculated as;

= 0.09 (K.Ei) / K.Ei

= 0.09

= 9 %

The final kinetic energy of the neutron is calculated as follows;

K.E.f (neutron) = (1 - 0.09) x (4.2 x 10⁻¹³ J)

K.E.f (neutron) = 3.82 x 10⁻¹³ J

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Calculate the potential difference between two points of a uniform field located along one line of tension and 30 cm apart. It is known that the electric field strength is 30 V/m.

1) 0.01 V
2) 9v
3) 100V
4) 900V

Answers

Potential difference is the work done in moving a unit charge to a reference point. The potential difference is 9 volts. Thus option B.

Potential difference is a term that is used to describe the work done in transferring a unit positive charge from infinity to a reference point. While the electric field strength is the number of electric field passing through a unit area in a period.

Tension is an example of contact force which is applicable to extended strings, lines, ropes etc.

Given:

electric field strength = 30 V/m

distance between the two points = 30 cm

                                         = 0.3 m

Potential difference between the points = 0.3 x 30

                                          = 9.0 V

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angelina jumps off a stool. as she is falling, the earth’s gravitational force on her is larger in magnitude than the gravitational force she is exerting on the earth.
a. true
b. false

Answers

The answer to that question is true

which type of wave is characterized by particle motion that is both perpendicular and parallel to the wave? responses longitudinal longitudinal electromagnetic electromagnetic surface surface transverse

Answers

Mechanical waves come in two different varieties: longitudinal waves, in which the medium is moved in the same direction as the wave.

What kind of wave is distinguished by the motion of a particle that moves perpendicular to the wave?

The motion of the wave and the particles is always parallel in longitudinal waves. A classic illustration of a longitudinal wave is a sound wave that is moving through air.

What kind of wave best explains situations where particle motion is opposed to energy motion?

Particles are displaced perpendicular to the wave's direction in a transverse wave. Transverse waves, such as those that cause ripples on the water's surface and vibrations in a string, are examples.

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hrough what potential difference would the charged particles have to be accelerated to aquire the velocity of v0

Answers

Charged particles would need to be accelerated by a potential difference of qmv0^2/2 in order to reach the velocity of v0.

The rate at which the charge between two places is exerting work is known as the electric potential difference.

V=W/q is the formula for the electric potential difference.

According to the Work-Energy Theorem, an object's work is equal to the change in kinetic energy.

Relation between Work-Energy Theorem and

Work completed is equal to 1/2MV^2

Considering that the particle's mass is m

Charge of the particle equals q, and particle velocity equals V0.

Work performed on the particle (W) = 0.5 m v 0 2.

Difference in electric potential (V) = W/q

V = 1/2mv0^2 /q

in order for V = qmv0^2/2.

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A baseball is thrown from the roof of 20.8m-tall building with an initial velocity of magnitude 10.6m/sand directed at an angle of 57.3^\circabove the horizontal.A. What is the speed of the ball just before it strikes the ground? Use energy methods and ignore air resistance.B. What is the answer for part (A) if the initial velocity is at an angle of 57.3^\circbelow the horizontal?C. If the effects of air resistance are included, will part (A) or (B) give the higher speed?- part A will give the higher speed- part B will give the higher speed

Answers

a) The speed with which it strikes the ground is 22.212 m/s.

b) It is not influenced by angle and simply relies on mechanical energy conservation.

c) None of the situation would give the higher speed.

Given that,

The height from which the baseball is thrown = 20.8 m

Initial velocity v₀ = 10.6 m/s

Angle = 57.3°

Initially, the potential energy of the ball = m × g × h = m × 9.8 × 20.8 = 204.82 m J

Initial velocity contains two components, vx and vy.

vx = 10.6 cos 57.3° = 5.73

vy = 10.6 sin 57.3° = 8.92

The x component remains constant through the whole motion, so we need to consider the y component only, as the x component will get cancelled as it is same everywhere.

Initially the kinetic energy is,

K.E i = 1/2 m (vy)² = 1/2 m × 8.92² = 39.78 m

Final potential energy = m × g ×h = m × 9.8 ×h

At the height the ball reached, the final kinetic energy is zero.

The conservation of mechanical energy enables us to know that,

P.E i + K.E i = P.E f + K.E f

204.82 m + 39.78 m = m × 9.8 ×h

244.6 = 9.8 ×h

h = 24.96 metres

Now, the potential energy at the highest point is,

P.E f = m × 9.8 × 24.96 = 244.61 m J

Initial kinetic energy was zero, because of the zero speed at the top.

Final kinetic energy will be,

K.E f = 1/2 m × v²

244.61 m = 1/2 m × v²

v² = 244.61 × 2

v = 22.12 m/s

b) If the ball would have been thrown at an angle of 57.3°, below the horizontal, then also the final velocity would be 22.12 m/s. It is not influenced by angle and simply relies on mechanical energy conservation.

c) The air resistance will produce a decrease in the final speed.

We know that friction is a non conservative force and our system will become a dissipative system as the total mechanical energy would not be conserved.

Air friction will cause a decrease in speed. None of the situation would give the higher speed.

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suppose a collection of five batteries is connected as shown in the figure, in which b1 = 4.9 v.

Answers

1. The equivalent emf of the collection is 15.6 V

2. The current through the resistor is 4.7 A

1. How do I determine the equivalent emf?

The equivalent emf of the collection can be obatined as illustrated below.

The following were obtained from the question:

Emf 1 (E₁) = 3.0 VEmf 2 (E₂) = 4.5 VEmf 3 (E₃) = 1.5 VEmf 4 (E₄) = 2.0 VEmf 5 (E₅) = B₁ = 4.6 VEquivalent emf (E) = ?

E = E₁ + E₂ + E₃ + E₄ + E₅

E = 3 + 4.5 + 1.5 + 2 + 4.6

E = 15.6 V

Thus, the equivalent emf is 15.6 V

2. How do I determine the current ?

The current through the resistor can be obtained as follow:

Resistance (R) = 3.3 ΩEquivalent emf (E) = 15.6 VCurrent (I) =?

Emf (E) = Current (I) × resistance (R)

15.6 = Current × 3.3

Divide both sides by 3.3

Current = 15.6 / 3.3

Current = 4.7 A

Thus, the current is 4.7 A

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how large a net force is required to accelerate a 1200-kg suv from rest to a speed of 25 m/s in a distance of 200 m?

Answers

A 1200 kg SUV will take 1875.6 N of net force to accelerate it across 200 meters at 25 m/s from a standstill.

Where:

a = acceleration (m/s²)

v = final velocity (m/s)

u = initial velocity (m/s)

s = distance (m)

v² - u² = 2as

a = v² - u²/2s

a = 25² - 0²/2(200)

a = 625/400

a = 1.563m/s²

Since a = 1.563m/s²

F = 1200 × 1.563

F = 1875.6N

Therefore, a 1200 kg SUV will need to accelerate with a net force of 1875.6 N over a 200 m distance to reach a speed of 25 m/s.

In mechanics, a force is any action that has the potential to change, maintain, or deform a body's motion. The three principles of motion outlined by Isaac Newton in his Principia Mathematica are frequently used to illustrate the concept of force (1687). Newton's first law states that a body at rest or moving uniformly in a straight line will stay in that state until a force is applied to it. According to the second law, a body will accelerate (change in velocity) in the direction of any external force acting on it. The amount of acceleration is directly related to the amount of external force.

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what happens to the sun as it transitions between a main sequence star and a white dwarf?

Answers

In the cores of main sequence stars, hydrogen atoms are fused to form helium atoms. The sun is one of the majority of main sequence stars, which make up about 90% of all stars in the universe.

What is Transition between Sun?

These stars can be as much as 200 times as massive than the sun or only a tenth as massive.

Stars begin their lives as gas and dust clouds. These clouds are brought together by gravity. The falling material acts as the fuel for a tiny protostar. Protostars can be difficult to find and frequently form in tightly packed gas clouds.

Mark Morris of the University of California, Los Angeles (UCLS), stated in a statement that "Nature doesn't generate stars in solitude" (opens in new tab). "It creates them in groups from natal clouds that burst under.

Therefore, In the cores of main sequence stars, hydrogen atoms are fused to form helium atoms. The sun is one of the majority of main sequence stars, which make up about 90% of all stars in the universe.

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some people with turner syndrome are 45,x/46,xy mosaics. how could this mosaicism arise?

Answers

This mosaicism in Turner's syndrome develops as a result of some cells in an XY embryo rapidly losing a Y chromosome after fertilization.

The zygote will have karyotype 44+XY if the sperm contains a Y chromosome. A gamete with the X chromosome will be produced by a mother. The resulting embryo would be a 44+XY (total 46 chromosomes). This zygote begins to divide its cells, but if the embryo's cells somehow lose the Y chromosome, the karyotype will change to 44+X0 (totaling 45 chromosomes, which is the Turner's syndrome disease).

Thus, certain cells that were originally destined to create an XY embryo with 46 chromosomes now have 45 chromosomes and the Turner's syndrome X0 condition.

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The Sun reaches the zenith (directly overhead) at midday ___________. Select all the options that make the statement true.
A. everywhere in the Southern Hemisphere on the December solstice
B. on the equator only on the March and September equinoxes
C. on the Arctic circle (66.5°N) on the June solstice
D. on the equator every day
E. everywhere in the Northern Hemisphere on the June solstice
F. everywhere within the tropics (between latitudes 23.5°N and 23.5°S) on two dates each year

Answers

Answer: (option b) on the equator only on the March and September equinoxes

Explanation:One of the abecedarian data of life at Earth’s midlatitudes, where utmost of this book’s compendiums live, is that there are significant variations in the heat we admit from the Sun during the course of the time. We therefore divide the time into seasons, each with its different quantum of sun. The difference between seasons gets more pronounced the further north or south from the ambit we travel, and the seasons in the Southern Hemisphere are the contrary of what we find on the northern half of Earth. With these observed data in mind, let us ask what causes the seasons.

numerous people have believed that the seasons were the result of the changing distance between Earth and the Sun. This sounds reasonable at first it should be colder when Earth is further from the Sun. But the data do n’t bear out this thesis. Although Earth’s route around the Sun is an cirque, its distance from the Sun varies by only about 3. That’s not enough to beget significant variations in the Sun’s heating. To make matters worse for people in North America who hold this thesis, Earth is actually closest to the Sun in January, when the Northern Hemisphere is in the middle of downtime.



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The Sun reaches the zenith (directly overhead) at midday on the equator every day. Therefore, option D is correct.

The relationship between the Sun and the equator also affects the seasons. As the Earth orbits the Sun, its axial tilt causes different parts of the Earth to receive varying amounts of sunlight throughout the year. When the Sun is directly overhead at the Tropic of Cancer during the June solstice, it marks the start of summer in the Northern Hemisphere and winter in the Southern Hemisphere.

Conversely, when the Sun is directly overhead at the Tropic of Capricorn during the December solstice, it marks the start of summer in the Southern Hemisphere and winter in the Northern Hemisphere.

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in a two-slit interference pattern on a distant screen, are the bright fringes midway between the dark fringes? is this ever a good approximation?

Answers

Due to its dependence on L, the spacings between various fringes get smaller as the distance between the slits gets more. The space between various fringes grows as the light's wavelength rises because this is a wavelength-dependent property.

How far apart are the second bright fringes on either side of the main bright fringe?

Distance between the core bright fringe and two side-by-side dark fringes is 2 mm (1. 2 inches).

How do you calculate the separation between the dark and brilliant fringes?

As a result, "xn = (2n+1)D/2d" is how far the "n" brilliant fringe is from the "centre." Similar to the example above, the formula for the distance between the "n-1" dark fringe and the "centre" is: "x (n-1)= (2(n-1) +1)D/2d."

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To test this hypothesis, you would change the and observe the in the system. Therefore, the independent variable would be the and the dependent variable would be the .

Answers

You would alter the separation width and track the system's diffraction angle to verify this theory. Therefore, the Gap Width is the variable and the Diffraction Angle would serve as the dependent variable.

What does a physics hypothesis look like?

Salt will become more soluble as even the temperature rises. UV radiation exposure is associated to the risk of skin cancer. Many brands of light bulbs have the same lifespan. Cats don't care what color their cat chow is.

What makes a strong thesis statement?

A great hypothesis statement includes a prediction, is unambiguous and testable. While "testable" denotes verifiability or falsifiability, it also implies that you may conduct the required tests without going against any ethical norms.

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The complete question is-

The goal of this experiment was to answer the question, "How does diffraction occur?" Suppose you formulated this alternate hypothesis:

If the gap width is increased while keeping wavelength constant, then the observed diffraction will decrease because the less the ratio between wavelength and gap width is, the smaller the observed diffraction angle will be.

To test this hypothesis, you would change the and observe the in the system.

Therefore, the independent variable would be the and the dependent variable would be the .

1. is gap width

2. is diffraction angle

3. is gap width

4. is diffraction angle

Explanation:

A superball with mass m equal to 50 grams is dropped from a height of hi = 1.5 m. It collides with a table, then bounces up to a height of hf = 1.0 m. The duration of the collision (the time during which the superball is in contact with the table) is tc = 15 ms. In this problem, take the positive y direction to be upward, and use g = 9.8 m/s2 for the magnitude of the acceleration due to gravity. Neglect air resistance. Part A: Find the y component of the momentum of the ball immediately before the collision. Part B: Find the y component of the momentum of the ball immediately after the collision, that is, just as it is leaving the table. Part C: Find the y component of the impulse imparted to the ball during the collision. Part D: Find the y component of the time-averaged force Favg, in newtons, that the table exerts on the ball. Part E: Find Kafter-Kbefore, the change in the kinetic energy of the ball during the collision, in joules.

Answers

The different component of the momentum of the ball is as bellow,

As per the details share in the above question are s bellow,

The mass of the super ball m is =50gm

The super ball drop from the height of = 1.5 m

Consider the ball's speed be v right before it hits the table.

Decrease of Potential Energy = Increase of Kinetic Energy

=> m x 9.8 x 1.5 = [tex]0.5 m v^{2}[/tex]

=> Therefore velocity = 5.42 m/s (- y direction)

A) Prior to a collision, the momentum component Y = mv

= - 0.05 x 5.42 = - 0.271 kg m/s

Consider the ball's speed be u right before it hits the table.

Increase of PE = Decrease of KE

=> m x 9.8 x 1 = [tex]0.5 m u^{2}[/tex]

=> u = 4.427 m/s (+ y direction)

B) Following the impact, the ball's Y contribution of momentum = mu

= 0.05 x 4.427 = 0.221 kg m/s

C) Now when Impulse imparted,

Consider Jy = change in momentum

= 0.221 - (-0.271) = 0.492 kg m/s

D) The ball time of contact is = 15 ms

Force = change inimpulse / time

= 0.492/0.015 = 32.82 N

E) change in Kinetic energy during collision

= KE after - KE before

[tex]= 0.5 \times 0.05 \times 4.427^{2} - 0.5\times0.05\times5.42^{2}[/tex]

= - 0.24445 J

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Note: The correct correct question would be as bellow,

As shown in the figure (Figure 1) , a superball with mass m equal to 50 grams is dropped from a height of hi=1.5m . It collides with a table, then bounces up to a height of hf=1.0m . The duration of the collision (the time during which the superball is in contact with the table) is tc=15ms . In this problem, take the positive y direction to be upward, and use g=9.8m/s2 for the magnitude of the acceleration due to gravity. Neglect air resistance.

A) Find the y component of the momentum Pbefore,y of the ball immediately before the collision (in kg x m/s)

B) Find the y component of the momentum of the ball immediately after the collision, that is, just as it is leaving the table (in kg x m/s)

C) Find Jy, the component of the impulse imparted to the ball during collision (in kg x m/s)

D) Find the y component of the time-averaged force Favg,y in Newtons, that the table exerts on the ball

E)Find Kafter - Kbefore, the change in the kinetic energy of the ball during the collision, in Joules.

thin-straight uniform rod of length 0.85 m hangs from a pivot at the end of the rod. what is the period of oscillation? (include units with answer) a: 1.51 s c: 1.510 s 1 hint available

Answers

The time period of oscillation is 1.84 s.

Simple pendulum's time period is nothing but the time taken to complete one oscillation. It is measured in seconds and minutes.

Given that, length of the uniform rod = 0.85 m

We know the expression for time period as,

T = 2π√(l/g)

where, T is time period

l is length

g is acceleration due to gravity

Substituting the values in the above equation, we have

T = 2π√(l/g) = 2π* √(0.85/10) = 2π* 0.292 = 1.84 s

Thus, the time period of oscillation is found to be 1.84 s.

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a tire is filled with air at 15oc to a gauge pressure of 2.2x105 pa. if the tire reaches a temperature of 38oc, what will the new gauge pressure be inside it?

Answers

Manometer-style pressure gauges have a small, liquid-filled U-shaped tube inside.

Reading the marks on this tube will tell you how pressurized your system is. When pressure is applied to either side of the gauge, the water in the tube rises in one direction or the other. The most important kinds of pressure measurements use atmospheric pressure as their standard. When a tire is inflated to 30 psi, for instance, a standard tire pressure gauge will display this number as 30 psig (the "g" stands for gauge pressure). The liquid fill makes pressure measurements simpler by attenuating vibration, mechanical stress, and pressure pulsations. Due to the ambient atmosphere, the liquid fill prevents corrosion, moisture infiltration, and ice.

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water weighs about 10 n per liter on earth. a 4.0 liter ball is pushed underwater on the space shuttle, where gravity was 1/10 that on earth. what is the buoyant force on the ball?

Answers

the buoyant force on the ball is 4 N

buoyant force formula,

Fb=-ρgV

To calculate the buoyant force, the formula can be used: Fb=ρVg Fb = ρVg where Fb is the buoyant force in Newtons, ρ is the density of the liquid in kilograms per cubic meter, and V is the displaced volume of liquid in cubic metres. ), g is the gravitational acceleration

Buoyancy is the upward force on an object that is wholly or partially immersed in liquid. This upward force is also called thrust. Buoyancy causes a body that is wholly or partially immersed in liquid to appear weighed down. H. Easier

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a waterskier skis over the ramp shown in the figure at a speed of 30 ftys. how fast is she rising as she leaves the ramp?

Answers

A waterskier skis over the ramp shown in the figure at a speed of 30 ft/s, 7.73 ft/sec is rising as she leaves the ramp.

We assume the hypotenues = h

From the given figure (attached) ;

h² = x² + y²

Put the  values of x and y from the figure

h² = 15² + 4²

h² = 225 + 16

h² = 241

Take under root on both sides

h = √241

From the figure the value of sin∅ = 4/h

Put value of h

sin∅ = 4 / √241

The rate at which she is rising = verticle velocity

The rate at which she is rising = 30sin∅

Put the value of sign∅

The rate at which she is rising = (30 * 4) / √241

The rate at which she is rising = 7.73 ft/sec

The missing figure is attached.

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