A particle with charge 5 . 0- µ C is placed at the corner of a cube. (Physics Help)? A particle with charge 5.0-µC is placed at the corner of a cube. The total electric flux in N * m^2 C through all sides of the cube is: B) 7.1 x 10^4 D) 1.4 x 10^5

An inductor is just a coil of wire with magnetic properties. It is a device that stores electromotive force or energy in the form of magnetic field.  A current yields a magnetic field around it through a conductor. The pattern of flux for this magnetic field would be the number of concentric circle perpendicular to the detection of current. The capacity of the inductor is affected by the number of coils which would give more inductance, the material that is used for the coil, the cross sectional area of the coil and the length of the coil. For instance you have an 8 meters diameter coil with seven loops of wire. You place the coil on an area where there is a parked car. The inductance of the coil will be much larger compared to the coil without a parked car nearby because of the presence of the steel from the car. It changes the magnetic field of the coil because it acts as a core of the inductor.

Let us say that triangle XYZ has sides of XY, YZ, and ZX.

Each of the corners X, Y, and Z are located at their own (x, y) points.

If all of the triangle are transformed through translation by a movement of 2 units left and 1 unit down on the coordinate plane then we generate a triangle X’Y’Z’.

Then each of the corners X’, Y’ and Z’ are now located at (x – 2, y – 1) coordinates.

Since all of the corners were moved then we can also say that:

XY is congruent to X’Y’

YZ is congruent to Y’Z’

ZX is congruent to Z’X’

Since all sides are congruent, therefore all angles are also congruent.

Therefore the measure of angle x’ is equal to the measure of angle x.

Answer:

40 degrees

One complete orbit of the earth around the sun is 365 and ¼ days. Because of this, the earth has to completely orbit around the sun in respect to the stars too no the sun only and so the earth spins 366.26 times every rotation. The ratio of the earth's orbital period about the sun to the earth's period of rotation about its own axis will then be 1 is to 366.26.

Final answer:

The ratio of the Earth's orbital period about the sun to its period of rotation about its axis is 365.26:1.

Explanation:

The ratio of the earth's orbital period about the sun to the earth's period of rotation about its own axis is 365.26 days to 1 day.

This means it takes approximately 365.26 days for the Earth to orbit the Sun once, while it takes 1 day for the Earth to complete one full rotation about its axis.

Therefore, the ratio can be simplified to 365.26:1.

Part A: Make a free-body diagram of this ball just before it hits the wall.

The only force acting on the ball is the pull of the Earth, this is its weight; so the diagram is a vertical vector downwards.

Part B: Make a free-body diagram of this ball just after it has bounced free of the wall.

Again, the only force acting on the ball is the pull of the Earth, its weight, and the free-body diagram is identical to that of the part A.

Part C: Make a free-body diagram of this ball while it is in contact with the wall.

Buoyant force due to the surrounding air on a man is 0.84 Newton

[tex]\texttt{ }[/tex]

The basic formula of pressure that needs to be recalled is:

Pressure = Force / Cross-sectional Area

or symbolized:

[tex]\large {\boxed {P = F \div A} }[/tex]

P = Pressure (Pa)

F = Force (N)

A = Cross-sectional Area (m²)

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

Density of Air = ρ_air = 1.20 kg/m³

Weight of the man = w = 700 N

Density of the man = ρ = 1000 kg/m³

Asked:

Buoyant Force = F = ?

Solution:

We will use Archimedes' principle to solve the problem as follows:

[tex]F = \rho_{air} g V[/tex]

[tex]F = \rho_{air} g \frac{m}{\rho}[/tex]

[tex]F = \rho_{air} g \frac{w}{g\rho}[/tex]

[tex]F = \rho_{air} \frac{w}{\rho}[/tex]

[tex]F = 1.20 \times \frac{700}{1000}[/tex]

[tex]F = 0.84 \texttt{ Newton}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Pressure

[tex]\texttt{ }[/tex]

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant , Liquid , Pressure

The buoyant force due to the surrounding air on a man weighing 700 N is approximately 0.84 N. The ratio of the buoyant force to his weight is approximately 0.00120, which shows the buoyancy effect in air is minimal compared to his weight.

To calculate the buoyant force due to the surrounding air on a man weighing 700 N (equivalent to a mass of approximately 71.4 kg assuming g = 9.81 m/s2), we can use Archimedes' principle. This principle states that the buoyant force on an object immersed in a fluid is equal to the weight of the fluid displaced by the object. Since the man's average density is the same as that of water (about 1000 kg/m3), his volume V can be calculated using the formula:

V = mass / density = 71.4 kg / 1000 kg/m3 = 0.0714 m3.

The buoyant force (Fb) in air can then be calculated with the density of air (1.20 kg/m3):

Fb = density of air × volume × g = 1.20 kg/m3 × 0.0714 m3 × 9.81 m/s2 ≈ 0.841 N.

Therefore, the buoyant force is approximately 0.84 N.

To find the ratio of the buoyant force to the man's weight, we divide the buoyant force by the weight:

Ratio = Fb / weight = 0.841 N / 700 N ≈ 0.00120.

So, the ratio of the buoyant force to the man's weight is approximately 0.00120, which implies the effect of buoyancy in air is quite small compared to the weight of the man.

Answer:

Mass, m = 2000

Explanation:

Given that,

Force acting on the truck, F = 2000 N

Acceleration of the truck, [tex]a=1\ m/s^2[/tex]

To find,

The mass of the car

Solution,

The second law of motion gives the relationship between the mass force and the acceleration. It is given by :

[tex]F=ma[/tex]

[tex]m=\dfrac{F}{a}[/tex]

[tex]m=\dfrac{2000\ N}{1\ m/s^2}[/tex]

m = 2000 kg

So, the mass of the car is 2000 kg.

The gravitational force the asteroids exert on each other will be one fourth of their initial force.

Any two bodies will be attracted to one another by the force of gravity, also known as gravity. There is an attraction between every thing in the cosmos, but most of the time it is too faint to be noticed due to the extreme distance between the objects. Furthermore, although the influence of gravity is weaker as objects are moved away, its range is infinite.

We know that, gravitational force acting between two bodies,

F=[tex]\frac{ G m_1 m_2}{r^2}[/tex]

Where, G = universal gravitational constant

m₁ and m₂ are masses of the two bodies and r is distance between them.

Let, the masses of the two asteroids are M₁ and M₂ and initial distance between them is R.

Hence, gravitational force they exert on each other, F₁ =[tex]\frac{ G M_1 M_2}{R^2}[/tex]

Now, when the distance between two asteroids is doubled, that is 2R, the gravitational force they exert on each other will, F₂ = [tex]\frac{ G M_1 M_2}{(2R)^2}[/tex] = [tex]\frac{1}{4}[/tex] [tex]\frac{ G M_1 M_2}{R^2}[/tex] = [tex]\frac{1}{4}[/tex] F₁

Hence, the final gravitational force is one-fourth of the initial one.

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Answer:

(a) Emerging adulthood.

Explanation:

Emerging adulthood is a period between teenagers' dependence on guardians and grown-ups' long haul duties in adoration and work, and during these years, rising grown-ups center around themselves as they build up the information, abilities, and self-understanding they will requirement for grown-up life.

So the correct option is (a)

Answer:

C. 1,314,718 J

Explanation:

The heat energy needed to raise the temperature of the coal is given by:

[tex]Q=m C_s \Delta T[/tex]

where:

m = 5 kg is the mass of the coal

[tex]C_s = 1314 J/kg ^{\circ}C[/tex] is the specific heat of coal

[tex]\Delta T= 220^{\circ}C-20^{\circ}C=200^{\circ}C[/tex] is the increase in temperature

Substituting into the formula, we find

[tex]Q=(5 kg)(1314 J/kg ^{\circ}C )(200^{\circ}C)=1,314,000 J[/tex]

So, the closest option is

C. 1,314,718 J

The minimum length runway this aircraft can use is about 4.1 km

Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem !

Let's look at the table in the attachment

Given:

u = 0 m/s

v = 79 m/s

a = 23 / 10 = 2.3 m/s²

Unknown:

d = ?

Solution:

Let's calculate the takeoff distance.

[tex]v^2 = u^2 + 2as[/tex]

[tex]79^2 = 0^2 + 2(2.3)s[/tex]

[tex]6241 = 4.6s[/tex]

[tex]s = 6241 \div 4.6[/tex]

[tex]s \approx 1400 ~ m[/tex]

The length of the runway must be three times the takeoff distance.

[tex]d = 3s[/tex]

[tex]d = 3(6241 \div 4.6)[/tex]

[tex]d \approx 4100 ~ m[/tex]

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate

To determine the minimum runway length for an aircraft, multiply the takeoff distance by three. If the takeoff distance is known, this simple multiplication will provide the minimum runway requirement.

The question asks for the minimum runway length required for an aircraft, given that the runway must be three times the takeoff distance for safety reasons in the event of an aborted takeoff. To determine this, if we know the takeoff distance required for the aircraft, we simply multiply that distance by three to find the minimum runway length the aircraft can use. For example, if an aircraft requires 1 kilometer (1,000 meters) to take off, the minimum runway length must be 3 kilometers (3,000 meters) for safety reasons.

Compared to energy-flow, which enters ecosystems as sunlight and leaves as heat, the flow of matter is continually recycled and conserved, obeying the law of conservation of mass.

Compared to energy-flow in ecosystems, the flow of matter is conserved and recycled. While energy enters an ecosystem, typically in the form of sunlight, and is eventually dissipated as heat, matter circulates within the ecosystem through various biotic and abiotic processes. The law of conservation of mass supports the notion that matter is neither created nor destroyed, but rather continuously reused and transformed. Substances like water, carbon, and nitrogen undergo recycling through ecosystems; essential for life, these elements are integral components of the food web, influencing the distribution and abundance of organisms.

Answer:

Force of Earth will be dominating and the object will fall towards Earth

Explanation:

As we know that

Mass of Earth

[tex]M_e = 5.98 \times 10^{24} kg[/tex]

Mass of Moon

[tex]M_m = 7.35 \times 10^{22} kg[/tex]

since we know that gravitational force depends on mass and the distance between two objects

so here if an object is placed midway between Moon and Earth then as we can see that mass of Earth is approx 100 times more than the mass of Moon

So here we can say that Force of Earth will be dominating and the object will fall towards Earth

A driver sets out on a journey. for the first half of the distance she drives at the leisurely pace of 30 mi/h; during the second half she drives 60 mi/h, the driver's average speed on this trip is 40 mi/h.

To find the average speed for the entire trip, we can use the formula for average speed:

Average Speed = Total Distance ÷ Total Time.

In this case, the driver covers the first half of the distance at 30 mi/h and the second half at 60 mi/h.

Let's assume the total distance of the trip is [tex]\(D\)[/tex] miles. The first half of the distance is [tex]\(D/2\)[/tex] miles, and the second half of the distance is also [tex]\(D/2\)[/tex] miles.

Let's calculate the time taken for each half of the distance:

Time taken for the first half = Distance / Speed

= [tex]\((D/2) \, \text{miles} / (30 \, \text{mi/h})\)[/tex].

Time taken for the second half = Distance / Speed

= [tex]\((D/2) \, \text{miles} / (60 \, \text{mi/h})\).[/tex]

Total time for the trip = Time for the first half + Time for the second half.

Total time = [tex]\((D/2) / (30) + (D/2) / (60)\)[/tex].

Total time = [tex]\(\dfrac{D}{60} + \dfrac{D}{120}\)[/tex].

Total time = [tex]\(\dfrac{3D}{120} = \dfrac{D}{40}\)[/tex].

Now we can calculate the average speed:

Average Speed = Total Distance / Total Time.

Average Speed = [tex]\(D / \dfrac{D}{40}\)[/tex].

Average Speed = [tex]\(40\) mi/h[/tex].

Thus, the driver's average speed on this trip is 40 mi/h.

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The driver's average speed on this trip is 30 mi/h.

To find the average speed on this trip, we can use the formula: average speed = total distance / total time. Let's assume the total distance is D miles. According to the given information, the driver spends the first half of the distance at a speed of 30 mi/h and the second half at a speed of 60 mi/h. Therefore, the total time taken for the trip is D / (30 mi/h) + D / (60 mi/h) = D / (1/30 h) + D / (1/60 h) = 2D / (1/60 h) = 120D h. We can substitute this value into the average speed formula and simplify: average speed = D / (120D h) = 1/120 h. So, the driver's average speed on this trip is 1/120 h = 0.00833 h = 0.00833 x 60 min = 0.5 min/h = 0.5 x 60 = 30 mi/h. Therefore, the driver's average speed on this trip is 30 mi/h.

Final answer:

A vehicle traveling at 55 miles per hour covers approximately 80.67 feet in one second, calculated by converting miles per hour to feet per second.

Explanation:

To find out how many feet a vehicle travels in one second at a speed of 55 miles per hour, we can perform a unit conversion from hours to seconds. First, we need to determine how many feet are in a mile and then convert miles per hour to feet per second.

There are 5,280 feet in a mile, therefore:

Since there are 3,600 seconds in an hour, we divide the total feet per hour by the number of seconds in an hour to find the distance in feet per second.

So, a vehicle traveling at 55 miles per hour travels approximately 80.67 feet in one second.

The best answers to complete this sentence would be the following:

“sidereal day”

“upper median”

It keeps the time, in the two successive upper meridian of the sun during the crossing. Also it is when the star crosses the celestial meridian.

Explanation:

“sidereal day”

“upper median”

It keeps the time, in the two successive upper meridian of the sun during the crossing. Also it is when the star crosses the celestial meridian.

Hope this helps!      

The force of impact is the same as driving your vehicle off a 10.0 story structure.

Given the following data:

Conversion:

In this exercise, you're required to compare the force of impact with an equivalent height. Thus, we would use the following formula to calculate the height:

[tex]H = \frac{V^2}{2g}[/tex][tex]H = \frac{V^2}{2g}[/tex]

Where:

Substituting the parameters into the formula, we have;

H = \frac{26.82^2}{2(9.8)}

H = 36.70 meters.

Assuming a distance of 3.6 meters:

Height = \frac{36.70}{3.6}

Height = 10.0 meters.

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The binding energy of electrons to the magnesium surface can be calculated using the equation KE = hf - BE, where KE is the kinetic energy of the ejected electrons, hf is the energy of the incident photons, and BE is the binding energy. Given that the maximum velocity of the ejected electrons is 3.45 × 10^5 m/s, we can calculate the kinetic energy using the equation KE = (1/2)mv^2. Using the given wavelength of the photons (310 nm), we can calculate the energy of the photons using the equation E = hc/λ, where h is Planck's constant (6.63 × 10^-34 J.s), c is the speed of light (3.00 × 10^8 m/s), and λ is the wavelength in meters. By rearranging the equation to solve for the binding energy, we find that the binding energy is equal to the energy of the incident photons minus the kinetic energy of the ejected electron.

The binding energy of electrons to the magnesium surface can be calculated using the equation KE = hf - BE, where KE is the kinetic energy of the ejected electrons, hf is the energy of the incident photons, and BE is the binding energy. Given that the maximum velocity of the ejected electrons is 3.45 × 10^5 m/s, we can calculate the kinetic energy using the equation KE = (1/2)mv^2.

Using the given wavelength of the photons (310 nm), we can calculate the energy of the photons using the equation E = hc/λ, where h is Planck's constant (6.63 × 10^-34 J.s), c is the speed of light (3.00 × 10^8 m/s), and λ is the wavelength in meters.

By rearranging the equation to solve for the binding energy, we find that the binding energy is equal to the energy of the incident photons minus the kinetic energy of the ejected electrons.

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The binding energy of electrons to the magnesium surface is calculated to be approximately 0.62 eV.

To determine the binding energy of electrons ejected from a magnesium plate by photons with a wavelength of 310 nm, we use the photoelectric effect equation:

Calculate the energy of the photon (E(photon)):

Given:

Convert the photon energy from joules to electron volts (eV):

Calculate the kinetic energy (Ke) of the ejected electrons:

Convert the kinetic energy into electron volts (eV):

Calculate the binding energy (Eb):

Therefore, the binding energy of electrons to the magnesium surface is approximately 0.62 eV.

The information, most important when passing near a lighthouse is the

Water's depth or depth of the water.

A lighthouse is simply a structure that emits a bright light that provides navigators with a constant or intermittent signal.

In conclusion, When passing near a lighthouse we consider looking at the water's depth around the area

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Miranda is the right answer

The helium atom exhibit more visible emission lines than hydrogen atom as hydrogen has one electron per atom, while helium atom has two electrons per atom.

Emission lines are generally used to determine the atoms and the molecules.

A emission line is formed when a electron falls behind the low-level state of energy by freeing a photon.

Difference between the helium visible emission lines and hydrogen visible emission lines-

Hence, the helium atom exhibit more visible emission lines than hydrogen atom as hydrogen has one electron per atom, while helium atom has two electrons per atom.

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Helium exhibits more visible emission lines than hydrogen due to its complex electron energy level structure. As helium has two electrons and more possible transitions, it results in more visible lines in its emission spectrum compared to hydrogen.

Helium exhibits more visible emission lines than hydrogen due to the complexity of its electron energy level structure. When energy is absorbed by an atom, electrons get excited and move to higher orbitals. When these electrons transition back to lower energy levels, they emit energy in the form of light at specific frequencies, which are visible as emission lines in an emission spectrum.

For Hydrogen, in its ground state, no electrons are in the higher-energy levels required to produce either emission or absorption lines in the visible part of the Balmer series, therefore, the spectral features of hydrogen in the visible range are limited primarily to the Balmer line that only excited hydrogen atoms produce.

Helium, on the other hand, has two electrons in different energy levels. As such, there are more possible transitions and hence more lines visible in its emission spectrum.

This concept is fundamental to our understanding of interstellar mediums and the chemical composition of celestial bodies. Recognizing the unique spectral features in visible light allows us to decode the mysteries of the universe.

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The form of mechanical weathering is exfoliation, which refers to the peeling off of the outer layers of a rock due to physical forces. Hydration, carbonation, and oxidation represent forms of chemical weathering.

The weathering process that exemplifies a form of mechanical weathering is option C, exfoliation. Mechanical weathering, also known as physical weathering, refers to the process where rock is broken down into smaller pieces by physical forces without any changes in its chemical composition.

Exfoliation is a form of mechanical weathering that occurs when the outer layers of rock peel off in layers due to differential heating and cooling, or freeze-thaw cycles. In contrast, options A (hydration), B (carbonation), and D (oxidation) all depict processes of chemical weathering, wherein the rock's mineral composition itself changes.

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