tom and his group were making a poster for their classroom to depict the way that electricity and magnetism work together which of the following should the group to their poster

The 80 kg person needs to sit 0.5625 m from the center of the seesaw to balance it with the 50 kg child on it.

We can use the principle of moments to solve this problem. The principle of moments states that the sum of the moments acting on an object must be equal to zero for the object to be in equilibrium.

In this case, the seesaw is in equilibrium when the clockwise moments acting on it are balanced by the counterclockwise moments. We can express the moments as the product of the force and the distance from the fulcrum.

Let x be the distance from the center of the seesaw to where the 80 kg person sits. The clockwise moment is then:

M_cw = m₁ * g * (L/2)

where g is the acceleration due to gravity.

The counterclockwise moment is:

M_ccw = m₂ * g * (L/2 - x)

Since the seesaw is in equilibrium, we have:

M_cw = M_ccw

Substituting the expressions for M_cw and M_ccw, we get:

m₁ * g * (L/2) = m₂ * g * (L/2 - x)

Canceling out the common factors, we get:

m₁ * (L/2) = m₂ * (L/2 - x)

Substituting the given values, we get:

50 kg * (4.5 m/2) = 80 kg * (4.5 m/2 - x)

Simplifying and solving for x, we get:

x = (80 kg * 4.5 m/2 - 50 kg * 4.5 m/2) / 80 kg

x = 0.5625 m

Therefore, the 80 kg person needs to sit 0.5625 m from the center of the seesaw to balance it with the 50 kg child on it.

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Answer:

The refractive index of a material is a measure of how much the speed of light is reduced when it passes through that material.

The formula for calculating the refractive index of a material is:

n = c/v

where n is the refractive index, c is the speed of light in a vacuum (approximately 3 x 10^8 m/s), and v is the speed of light in the material.

When light travels from one material to another and refracts at the boundary, the angle of refraction and the refractive index of the first material and the refractive index of the second material determine the way in which the light bends.

If light passes at an angle into a material, it will bend towards the normal (the perpendicular line) if the refractive index of the second material is greater than the refractive index of the first material. If the refractive index of the second material is less than the refractive index of the first material, the light will bend away from the normal.

The below diagram shows the path of light as it passes from air into a denser material (in this case, glass). The angle of refraction is determined by the refractive indices of the two materials and the angle of incidence.

The de Broglie wavelength of a particle is given by the equation:

λ = h/p

where λ is the de Broglie wavelength, h is Planck's constant (approximately 6.626 x 10^-34 J s), and p is the momentum of the particle.

To calculate the momentum of a particle, use the equation:

p = mv

where p is the momentum, m is the mass of the particle, and v is its velocity.

To calculate the associated de Broglie wavelength of electrons in an electron beam accelerated through a potential difference (pd) of 4000 V, use the equation:

λ = h/√(2meV)

where λ is the de Broglie wavelength, h is Planck's constant, me is the mass of an electron (approximately 9.11 x 10^-31 kg), and V is the potential difference.

To calculate the energy of an alpha particle with a de Broglie wavelength of 5.7 x 10^-15 m, use the equation:

E = hc/λ

where E is the energy of the particle, h is Planck's constant, c is the speed of light, and λ is the de Broglie wavelength. The energy can then be converted to MeV (million electron volts) by dividing by 1.6 x 10^-13 J/MeV.

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Such a machine is not likely to be constructed because: the first law of thermodynamics states that energy can neither be created nor destroyed, but it can be converted from one form to another.

Therefore, it is impossible for a perpetual-motion machine to be constructed because this machine would require more energy than it produces, thus violating the first law of thermodynamics. In terms of energy efficiency, a perpetual-motion machine would need to produce more energy than it consumes in order to continuously operate.

This is impossible because no energy can be created and the energy that is produced needs to be balanced by an equal amount of energy consumed. Therefore, the perpetual-motion machine violates the first law of thermodynamics.

Moreover, even if a machine could be created that is capable of producing more energy than it consumes, the machine would eventually run out of energy because energy cannot be created. Therefore, a perpetual-motion machine is impossible because it violates the first law of thermodynamics.

In conclusion, a perpetual-motion machine is impossible to construct because it violates the first law of thermodynamics. The machine would need to produce more energy than it consumes, but no energy can be created and the energy that is produced needs to be balanced by an equal amount of energy consumed. Therefore, a perpetual-motion machine is impossible.

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The magnitude of the ball's acceleration when it leaves the gun, as a multiple of g, if it is shot straight up is -1g.

When the ball is shot straight up, its initial velocity is zero. Therefore, its final velocity is zero when it reaches the highest point in its trajectory. Using the kinematic equation,

vf^2 = vi^2 + 2ad

where, vf, vi are final and initial velocity, a is acceleration, and d is displacement

At the highest point in the ball's trajectory, its final velocity is zero, and its initial velocity is also zero. Therefore, the equation becomes:

0 = 0 + 2ad

which gives us the displacement of the ball as it moves upward from the gun.

Using this equation to find the time it takes for the ball to reach its highest point, we get,

t = √(2d/a)

At the highest point in the ball's trajectory, its velocity is zero, and its acceleration is equal to the acceleration due to gravity, which is -g. Therefore, using the kinematic equation to find the displacement of the ball:

d = vit + 1/2at^2

Substituting for vi, t and a,

d = -1/2g(√(2d/a))^2d = -d/2

Therefore, the displacement of the ball is negative, which means that the ball is below its initial position.

Therefore, the magnitude of the ball's acceleration is -1g.

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The frictional force required to keep the car on the circular track will increase by a factor of four when the speed of the car is doubled.

The centripetal force required to keep a car moving in a circular path is given by the formula F = mv²/r, where m is the mass of the car, v is its speed, and r is the radius of the circular path. In this case, the force of friction provides the necessary centripetal force to keep the car on the track.

When the speed of the car is doubled, the centripetal force required to keep it on the circular path increases by a factor of four, because the velocity appears squared in the formula for centripetal force. Therefore, the frictional force needed to keep the car on the road must also increase by a factor of four.

This means that if the original frictional force was F, then the new frictional force needed will be 4F.

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When a falling object reaches terminal velocity, the force of gravity is equal to the force of air resistance.

The highest speed that an object can travel at when falling through a fluid, such as air or water, is known as terminal velocity. This speed is determined by the conflicting forces of gravity and air resistance, or drag.

When something is dropped for the first time, it falls faster since gravity is pulling it down. Yet, as an item moves faster, the force of air resistance, commonly known as drag, grows until it precisely equalizes the force of gravity. The object will then stop accelerating and begin to fall at its terminal velocity, which is a constant speed.

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Answer: maybe read your book

Explanation:

A stone is dropped from the top of a building, as the stone falls, does its de broglie wavelength decreases.

De Broglie's wavelength λ is defined as λ = h/ p where h is Planck's constant and p is the instigation of the object.

The nonrelativistic instigation p is related to the speed v of the object as p = mv, where m is the mass of the object.

By fitting Equation 2 into Equation 1, we get λ = h/ mv.

Now, in this case, when the gravestone falls, the Earth exerts an attractional gravitational force on it and the gravestone accelerates. therefore, the speed v increases. By looking at Eq.( 3), we conclude that de Broglie's wavelength λ decreases. Also, note that we have used the nonrelativistic expression for the instigation of the gravestone p. This is a valid' approximation' since the gravestone moves at pets much lower than the speed of light in vacuum c. Also, the gravestone is a macroscopic object. therefore, incontinently, we can conclude that the wavelength associated with it's fairly small and the gravestone exhibits flyspeck like parcels.

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Answer:

13. a) In Einstein's photoelectric equation, hf represents the energy of a single photon of the electromagnetic radiation, and mvmax² represents the maximum kinetic energy of the emitted photoelectrons.

b) i. To calculate the threshold frequency, we can use the equation c = fλ, where c is the speed of light, f is the frequency of the electromagnetic radiation, and λ is the wavelength:

c = fλ

f = c/λ

f = (3.00 x 10^8 m/s)/(6.5 x 10^-7 m)

f = 4.62 x 10^14 Hz

Therefore, the threshold frequency is 4.62 x 10^14 Hz.

ii. We can use the equation hf = Φ + KEmax, where hf is the energy of a single photon, Φ is the work function energy of the metal, and KEmax is the maximum kinetic energy of the emitted photoelectrons. We know that the wavelength of the incident electromagnetic radiation is 6.5 x 10^-7 m, which corresponds to a frequency of f = 4.62 x 10^14 Hz (as calculated in part i). We also know that this wavelength is the maximum for which photoelectrons are released, which means that the energy of the photons is equal to the work function energy:

hf = Φ

Substituting the values for h and f, we get:

(6.63 x 10^-34 J s)(4.62 x 10^14 Hz) = Φ

Φ = 1.93 x 10^-19 J

Converting this to electronvolts (eV), we get:

Φ = (1.93 x 10^-19 J)/(1.60 x 10^-19 J/eV)

Φ = 1.21 eV

Therefore, the work function energy of the metal is 1.9 eV.

c) If the intensity of the incident light is doubled, the rate of release of photoelectrons will also double. This is because the intensity of light is directly proportional to the number of photons incident on the metal surface per unit time. Each photon can cause the emission of one photoelectron, so doubling the number of photons will double the number of emitted photoelectrons per unit time. However, the kinetic energy of the emitted photoelectrons will not change, since this is determined only by the frequency (and therefore the energy) of the incident photons, and not by their intensity.

(a) The total energy is 0.098 J.

(b)The kinetic and potential energies as a function of time is, K = 1/2 m (0.808 m/s)^2 sin^2 (8.08t) and U = 1/2 (19.6 N/m) (0.100 m)^2 cos^2 (8.08t) respectively.

(c) The velocity when the mass is 0. 050 m from equilibrium is (0.808 m/s) sin (8.08t).

(d) The kinetic and potential energies at half amplitude is, K = 0 and U = 0.098 J.

The total energy of the simple harmonic oscillator can be found using the formula E = 1/2 k A^2, where k is the spring constant and A is the amplitude, E = 1/2 (19.6 N/m) (0.100 m)^2 = 0.098 J

The kinetic energy (K) and potential energy (U) can be expressed in terms of the displacement (x) and velocity (v) as follows:

K = 1/2 m v^2 = 1/2 m (0.808 m/s)^2 sin^2 (8.08t)

U = 1/2 k x^2 = 1/2 (19.6 N/m) (0.100 m)^2 cos^2 (8.08t)

To find the velocity when the mass is 0.050 m from the equilibrium, we substitute x = 0.050 m into the expression for v,

v = (0.808 m/s) sin (8.08t) = (0.808 m/s) sin (8.08t) when x = 0.050 m

At half amplitude (x = A/2 = 0.050 m), the kinetic energy is zero and the potential energy is equal to the total energy,

K = 0

U = E = 0.098 J

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The galvanometer briefly swung to one side. The galvanometer is in the other way. There is no galvanometer deflection. Electromagnetic induction is the relevant phenomenon. The galvanometer is in the opposite direction from that in the previous case. There is no galvanometer deflection.

The observations that will occur for each choice when a coil of insulated copper wire is coupled to a galvanometer are as follows:

(I) Electromagnetic induction will cause the coil to create an electric current if a bar magnet is put into it.

(ii) If a bar magnet is removed from the inside, electromagnetic induction will again induce current in the copper wire, but this time the direction of the current in the galvanometer will be reversed.

(iii) If a bar magnet is kept stationary inside the coil, no current is produced, and as a result, the galvanometer does not deflect.

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The diameter of the coil that produces a 1.0 T magnetic field at the center of the n-turn loop, given a 1.0 A current, is 0.32 meters.

The formula for magnetic field strength in a coil is:

B = μ₀ * n * I / r

where, B is the magnetic field strength, μ₀ is the permeability of free space, n is the number of turns, I is the current, r is the radius of the coil.

To make this equation work, we need to solve for r. We can rearrange the equation and substitute values:

B = μ₀ * n * I / r

r = μ₀ * n * I / B

Substitute n = 1 meter, I = 1 A, and B = 1.0 T to obtain r.

r= (4π * 10⁻⁷ T m / A) * (1 turn) * (1.0 A) / (1.0 T)

r = 4π * 10⁻⁷ m

The diameter of the coil is then calculated using the formula:

Diameter = 2 * radius

Diameter = 2 * (4π * 10⁻⁷ m)

Diameter = 0.32 m

Therefore, the diameter of the coil is 0.32 meters.

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Option (D) is the correct answer i.e. Point Q has a larger tangential acceleration than point P, but both points have the same tangential velocity.

As the DVD is rotating with an ever increasing speed, we can infer that the angular velocity and angular acceleration of the DVD are increasing with time.

Now, let's consider points P and Q on the DVD, where point Q is farther away from the center of rotation than point P.

A) Points P and Q have the same tangential velocity and the same tangential acceleration: This cannot be true because tangential velocity is directly proportional to the distance from the center of rotation, and point Q is farther away from the center than point P. Therefore, point Q will have a greater tangential velocity than point P. Similarly, tangential acceleration is also directly proportional to the distance from the center of rotation, and since point Q is farther away from the center than point P, it will experience a greater tangential acceleration.

B) Points P and Q have the same angular velocity and the same angular acceleration: This cannot be true because we have already established that the angular velocity and angular acceleration of the DVD are increasing with time.

C) Points P and Q have the same angular velocity but point Q has a larger angular acceleration: This cannot be true because if point Q has a larger angular acceleration, then its angular velocity would be increasing at a greater rate than that of point P. Therefore, point Q would have a greater angular velocity than point P, contradicting the initial assumption.

D) Point Q has a larger tangential acceleration but points P and Q have the same tangential velocity: This is true because as we established earlier, point Q is farther away from the center of rotation and therefore experiences a greater tangential acceleration. However, both points have the same tangential velocity because they are both rotating at the same angular velocity.

E) Angular velocity, angular acceleration, tangential velocity, and tangential acceleration are larger for point Q than for point P: This cannot be true because we have already established that the angular velocity of the DVD is increasing with time. Therefore, the angular velocity, angular acceleration, tangential velocity, and tangential acceleration of point Q may be greater than those of point P, but they cannot be greater than their own respective values at an earlier time.

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In a uniform field, a beam has no magnetic poles and therefore experiences a force equal to the product of its length and the field intensity. whereas bar magnet is influenced by the interaction between its magnetic poles and the field intensity.

When placed in a uniform field, the bar magnet experiences a torque, which causes it to align itself with the field direction. In contrast, a beam placed in a uniform field experiences a uniform force, which acts parallel to the field direction and perpendicular to the beam length. This force tends to cause the beam to move in the direction of the field intensity. The magnitude of the force on the beam is proportional to the length and field intensity, as well as the product of its magnetic permeability and the field intensity.

The force on the beam also tends to cause it to rotate about its center, in the direction of the field intensity. This rotation will result in the beam becoming oriented along the field direction, but it will not cause the beam to align itself with the field direction as with a bar magnet. Instead, the beam will experience a net force, which tends to move it in the direction of the field intensity.

In conclusion, the behavior of a beam in a uniform magnetic field differs from the behavior of a bar magnet in several ways. A beam placed in a uniform field experiences a uniform force, which acts parallel to the field direction and perpendicular to the beam length. This force tends to cause the beam to move in the direction of the field intensity and rotate about its center, in the direction of the field intensity.

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The student came up with a model that shows a loop of wire being rotated by some external force between two strong permanent magnets. this causes the charges in the loop to flow. The student make a model of c. both a motor and a generator

The student made a model of a generator. A generator is a device that transforms mechanical energy into electrical energy. When a coil of wire rotates in a magnetic field, the changing magnetic field creates a current in the wire, which produces an electrical current. Because of the process of electromagnetic induction, this can happen. A motor, on the other hand, converts electrical energy into mechanical energy.

The electromagnetic force that is generated when a current is passed through a wire is used to create rotational motion in a motor. In the case of the student's model, a loop of wire is rotated by some external force between two strong permanent magnets. This causes the charges in the loop to flow. This process is a clear example of the principles of electromagnetic induction, which are essential to the operation of generators. Hence, the student has created a model of a generator. Therefore, option B is the correct answer.

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Answer:A

Explanation: because her family doesn’t have any disease

A doctor would likely recommend regular health screenings for Patrice, despite her lack of family disease history. Screenings are universally recommended for early detection and treatment of potential issues. Vaccinations would also not be avoided based on family history alone.

Even if Patrice does not have a family history of diseases, her doctor is most likely to tell her that she should get regular health screenings. Having no family history of specific diseases does not mean one is entirely free from the potential of developing health issues. Regular health screenings can help detect potential problems early, when they're easier to treat or manage. These screenings can involve routine physical exams, blood tests, or specific tests like mammograms or colonoscopies, depending upon age, gender, and other factors. Vaccines, similarly, are not decided based on family history, but rather on the possibility of exposure to specific diseases and age-specific recommendations.

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The speed required to push the box is approximately 11.27 m/s.

Given the angle of inclination of the slope, θ = 18°. The distance covered is, d = 15 meters.

The component of gravity parallel to the slope is, g|| = g.sinθ

The force required to push the box up the slope is, F = m.g, The acceleration due to gravity is, g = 9.81 m/s².

The weight of the box is, m.g. The component of gravity parallel to the slope is, g|| = g.sinθ.

The force required to push the box up the slope is, F = m.g||

The force required to push the box up the slope is,F = m.g.sinθ.

The frictionless surface has no frictional force acting upon the object moving upon it. Therefore, no work is done by friction on the object. Hence, the energy of the object in the form of kinetic energy is conserved. This is as given below:

K.E. initial + work done = K.E. finfinalitial kinetic energy is zero. This is because the box is initially at rest. The final kinetic energy is K. This is because we need to push the box up the slope with the required force. The work done is equal to the force exerted multiplied by the distance moved in the direction of the force.

Hence,W = F.dThe final equation becomes,0 + F.d = ½.m.K²The final speed required is,K = √(2.F.d/m)Substituting the values in the above equation, we get the required final speed,K = √(2.m.g.sinθ.d/m)K = √(2.g.sinθ.d)Putting the values of d, g, and θ in the equation, we get,K = √(2.9.81.sin18°.15)≈ 11.27 m/s.

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Explanation:

To find the net (total) force on the ball, we need to add the two forces together. Since the two players are kicking the ball from opposite sides, we need to subtract the force of the player kicking in the opposite direction.

Therefore, the net force on the ball is:

63 N (force of Blue #5) - 50 N (force of Red #3) = 13 N

The net force on the ball is 13 Newtons in the direction of Blue #5's kick.

Living a physically active lifestyle can improve academic performance by increasing energy and focus, improving mood and mental health, enhancing memory and cognitive function, improving sleep quality, and reducing absenteeism.

Living a physically active lifestyle can have a positive impact on your academic performance in several ways:

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The wavelength of a wave is the distance between two consecutive points on the wave that are in phase.

The wavelength is the distance between two consecutive nodes or two consecutive antinodes.

The first harmonious wavelength on a rope is given by the equation

λ = 2L/ n

where λ is the wavelength, L is the length of the rope, and n is the harmonious number. For the first harmonious, n = 1.

In this case, the length of the rope is L = 4m. Substituting into the equation over, we get

λ = 2( 4m)/ 1

λ = 8m

thus, the first harmonious wavelength on the rope is 8 measures.

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The ammeter and voltmeter should be read simultaneously to determine the resistance of a circuit component because the resistance depends on the current passing through the component and the voltage drop across it. By measuring both the current and voltage, Ohm's law (V=IR) can be applied to calculate the resistance of the component.

The resistance of a circuit component represents how much it impedes the flow of current through it. According to Ohm's law, the voltage drop across a component is directly proportional to the current flowing through it, and the constant of proportionality is the resistance of the component. Therefore, to determine the resistance of a component, we need to know both the current passing through it and the voltage drop across it.

An ammeter is used to measure the current flowing through a circuit, while a voltmeter is used to measure the voltage difference between two points in the circuit. By reading both instruments simultaneously, we can use Ohm's law to calculate the resistance of the component being measured.

It is important to note that the readings must be taken simultaneously because the current and voltage can change over time, and any delay between taking the readings can result in inaccurate measurements. Therefore, it is necessary to read the ammeter and voltmeter at the same time to obtain an accurate measurement of the resistance of the component.

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The anthracite coal price per MMBtu is $3.00/MMBtu. Therefore, the correct option is A.

The heat content of anthracite coal is 15,000 BTU/pound. The question requires the conversion of an anthracite coal price of $90/ton to $/MMBtu. Let us use the given data to solve the problem. 1 ton equals 2000 pounds (lb). Hence,

2000 lb of anthracite coal has a heat content of 2000 lb x 15,000 BTU/lb = 30,000,000 BTU

Thus, the price per MMBtu can be calculated by the given formula;

Price per MMBtu = Price per ton / (BTUs per ton / MMBTUs per ton)

Price per MMBtu = $90 / (30,000,000 / 1,000,000)

Price per MMBtu = $90 / 30

Price per MMBtu = $3.00/MMBtu

Thus, the price per MMBtu of anthracite coal is $3.00/MMBtu. Therefore, the correct answer is option A: $3.00/MMBtu.

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The radial acceleration of a point on the second hand of a large clock is 0.14 cm/s². Using the formula for radial acceleration, the distance of the point from the axis of rotation is approximately 11.3 cm.

The radial acceleration of a point on a rotating object is the acceleration that points towards the center of rotation. This acceleration arises due to the object's circular motion and is given by the formula a = v²/r, where a is the radial acceleration, v is the tangential velocity of the object, and r is the radius of the circular path. In this case, we know that the radial acceleration of the point on the second hand of the clock is 0.14 cm/s². Since the second hand completes one full revolution in 60 seconds, we can find its tangential velocity using the formula v = 2πr/T, where T is the time taken for one revolution. Substituting T = 60 seconds, we get v = 0.1047 cm/s. We can now use the formula for radial acceleration to find the radius of the circular path followed by the point on the second hand. Rearranging the formula, we get r = v²/a, which gives us a value of approximately 11.3 cm for the distance of the point from the axis of rotation.

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The turning point's y-value is indeed the largest. As a consequence, we must discover the greatest quadratic function indicating that perhaps the ball would achieve a height limit of 72.5 feet. Which indicates:

[tex]x = 110.90[/tex] or [tex]x = -0.90[/tex]

Given that we're working with time,

[tex]x = 110.90[/tex]

We have 110.9 feet to the nearest decimal place.

Why is it referred to as a quadratic function?

A quadratic issue is a type of challenge in mathematics that deals with something like a variable multiplication by itself, which is defined as squaring. This language stems from the fact that the area of a square is equal to its line segment multiplied by itself. The term "quadratic" is derived from quadratum, which is the Latin word meaning square.

Describe the three different kinds of quadratic functions?

Quadratics are typically utilized in three ways:

1)Standard Form: y = an x 2 + b x + c y=ax2+bx+c y=ax2+bx+c y=ax2+bx+c.

2)Factored Form: y = a (x r 1) (x r 2) y=a(x-r 1)(x-r 2) y=a(xr1)(xr2) y=a(xr1)(xr2).

3)Vertex Form: y = a (x h) 2 + k y=a(x-h)2+k y=a(xh)2+k.

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The amount of time the baseball was in air for at a momentum of 0.85kgm/s is 0.399s

Momentum is the product of its mass and velocity, or the vector sum of the products of its masses and velocities. It can be calculated as follows:

Momentum = mass (m) × velocity (v)

However, the momentum of an object can be estimated using the following;

ΔM = F × ΔT

Where;

F = mass × acceleration

F = 0.213kg × 10m/s² = 2.13N

0.85 = 2.13 × t

t = 0.399s

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The statement that  best describes the result of moving the charge to  the point marked X is: B. Its electric potential energy increases because the electric field increases.

Moving the charge to the point marked X will change its electric potential energy because the electric field at that point is different from the electric field at its initial position.

At the initial position of the charge, the electric potential energy is determined by the electric potential at that point and the charge of the object. When the charge is moved to point X, the electric potential at that point changes, which means that the electric potential energy of the charge also changes. This is because the electric potential at a point is directly proportional to the electric field at that point.

Therefore, since the electric field is different at point X than at the initial position, the electric potential energy of the charge increases as it moves to point X.

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Select a white dwarf or other star that is known to be relativity close and steady in its brightness as your reference point. When Mars is on separate sides of the sun, see the target star twice, six months apart.

The parallax angle can be used to calculate distance. It is occupied at the two sites on Earth that are separated by b by the far-off celestial object.

When measuring the distance between the earth and other planets and stars using the parallax method, we estimate the position of a first from one position, then from another. Calculating the distance from a star or planet involves measuring the amount of shift that occurs in its location.

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1. The W14x48 of A992 steel with continuous lateral support can carry the given load.

2. The W14x48 of A992 steel with an unbraced length of 18 ft can also carry the given load.

The given DL = 350 lb/ft, LL = 180 lb/ft.

Load per unit length w = DL + LL = 350 + 180= 530 lb/ft

Length of the beam = L = 18 ft

The properties of W14x48 are:

Width = 8.01 in

Depth = 14.00 in

Thickness = 0.335 in

Area = 13.80 in²

Using the relation: σ = Mc / I

where

σ = stress

c = distance from the neutral axis to the extreme fiber

I = Moment of Inertia

M = Maximum bending moment

Therefore, M = WL² / 8= (530) (18²) / 8= 21285 lb-ft = 255420 lb-inI = bd³ / 12= (8.01) (14.00)³ / 12= 1628.06 in⁴c = d / 2= 14.00 / 2= 7.00 inσ = Mc / I= (255420) (7.00) / 1628.06= 1093.14 psi

The allowable bending stress for A992 steel is Fb = 0.66 Fy

where Fy is the yield strength of the steel for A992,

Fy = 50 ksi = 50000 psi

Therefore, Fb = 0.66 (50000)= 33000 psi

The flexural strength of the W14x48 beam is more than the allowable bending stress. Hence, the beam can carry such a load.

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