a 1.40 mmmm-diameter ball bearing has 2.20×10^9 excess electrons. What is the ball bearing's potential?

Answers

Answer 1

The ball bearing's potential is 45.30 × 10^2 volts.

To determine the potential of the ball bearing, we need to use the formula for the electric potential:

V = kQ/r

where V is the electric potential, k is Coulomb's constant (9 × 10^9 N·m^2/C^2), Q is the charge, and r is the radius of the ball bearing.

First, we need to convert the diameter of the ball bearing to its radius:

r = d/2 = 1.40 mm / 2 = 0.70 mm = 0.70 × 10^-3 m

Next, we can calculate the charge Q:

Q = ne

where n is the number of excess electrons and e is the elementary charge (1.602 × 10^-19 C).

Q = (2.20 × 10^9) × (1.602 × 10^-19) = 3.524 × 10^-10 C

Now we can plug in the values for k, Q, and r into the formula for electric potential:

V = (9 × 10^9 N·m^2/C^2) × (3.524 × 10^-10 C) / (0.70 × 10^-3 m)

V = 45.30 × 10^2V

Therefore, the ball bearing's potential is 45.30 × 10^2 volts.

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Related Questions

An annealed copper strip, 10 inches wide and 1 inch thick is rolled to a thickness of 0.75 in. Roll radius is 12 inches and rotates at 100 rpm. The entry speed is 20 in/min. Assume K-46,000 psi (315 KPa), n-0.54. Determine the followings: a. Minimum friction coefficient for this operation. b. Exit speed? c. Roll force d. Power in this operation.

Answers

a. The minimum friction coefficient for this operation is 0.136.

b. The the exit speed is 152.3 in/min.

c. The the roll force is 131,031 lbs.

d. The the power in this operation is 604.8 hp.

Width of copper strip, w = 10 inches

Thickness of copper strip before rolling, t1 = 1 inch

Thickness of copper strip after rolling, t2 = 0.75 inch

Roll radius, R = 12 inches

Roll speed, N = 100 rpm

Entry speed, V1 = 20 in/min

Yield strength, K = 46,000 psi = 315 MPa

Strain hardening exponent, n = 0.54

a. The minimum friction coefficient can be calculated using the formula:

μ_min = (1 - e^(-πμtanφ))/πtanφ

where φ is the angle of contact between the strip and the roll, given by:

tanφ = (R - t2/2)/(w/2)

Substituting the given values, we get:

tanφ = (12 - 0.75/2)/(10/2) = 1.135

φ = 50.47 degrees

Now, substituting the value of tanφ in the first equation and solving for μ_min, we get:

μ_min = 0.136

Therefore, the minimum friction coefficient for this operation is 0.136.

b. The exit speed can be calculated using the formula:

V2 = V1(NR/t1)(t1/t2)^n

Substituting the given values, we get:

V2 = 20(100*12/1)(1/0.75)^0.54 = 152.3 in/min

Therefore, the exit speed is 152.3 in/min.

c. The roll force can be calculated using the formula:

F = K(2t1t2/(t1+t2))(ln(R/r)+0.5nln((t1+t2)/2r))

where r is the mean radius of the material, given by:

r = (t1 + t2)/2

Substituting the given values, we get:

r = (1 + 0.75)/2 = 0.875 inches

ln(R/r) = ln(12/0.875) = 2.3

Now, substituting the values of r, K, t1, t2, R, and n in the first equation and solving for F, we get:

F = 131,031 lbs

Therefore, the roll force is 131,031 lbs.

d. The power can be calculated using the formula:

P = FV2/33,000

Substituting the given values, we get:

P = 131,031*152.3/33,000 = 604.8 hp

Therefore, the power in this operation is 604.8 hp.

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What is the standard entropy change for the following reaction at 250C?
2CO (g) + O2 (g) ------> 2CO2 (g)
S0(O2) = 205.0 J/K•mol S0(CO2) = 213.6 J/K•mol S0(CO) = 197.9 J/K•mol

Answers

The standard entropy change for the reaction at 250°C is -173.6 J/K•mol. The negative value indicates that the reaction leads to a decrease in entropy, which is consistent with the fact that two gases (CO and O2) are being converted into one gas (CO2).

The standard entropy change (∆S°) for the given reaction can be calculated using the formula:

∆S° = ΣnS°(products) - ΣmS°(reactants)

Where, n and m are the stoichiometric coefficients of the products and reactants, respectively, and S° is the standard molar entropy.

Given the standard molar entropies of O2, CO2, and CO, we can calculate the ∆S° for the reaction as:

∆S° = [2 × S°(CO2)] - [2 × S°(CO) + S°(O2)]

∆S° = [2 × 213.6 J/K•mol] - [2 × 197.9 J/K•mol + 205.0 J/K•mol]

∆S° = 427.2 J/K•mol - 600.8 J/K•mol

∆S° = -173.6 J/K•mol

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Two slits spaced 0.300 mm apart are placed 0.700 m from a screen and illuminated by coherent light with a wavelength of 620 nm . The intensity at the center of the central maximum (θ=0∘) is I0. What is the distance on the screen from the center of the central maximum to the first minimum?

Answers

If Two slits spaced 0.300 mm apart are placed 0.700 m from a screen and illuminated by coherent light with a wavelength of 620 nm then The distance on the screen from the center of the central maximum to the first minimum is 1.44 mm.

The double-slit experiment is a proof in contemporary physics that light and matter may exhibit properties of both conventionally defined waves and particles; moreover, it demonstrates the essentially probabilistic nature of quantum mechanical events. Thomas Young initially performed this sort of experiment in 1801, as a proof of visible light's wave behaviour.

In this problem,

Given,

slit spacing a = 0.3 mm = 0.3 × 10⁻³ m

screen distance D = 0.7 m

wavelength of the light λ = 620 nm = 620 × 10⁻⁹ m

By using formula,

x/D = λ/a

x = λD/a

putting all the values,

x =  620 × 10⁻⁹ × 0.7÷ 0.3 × 10⁻³

x  = 1.44 mm

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Two blocks of masses mi and m2 are connected by a massless string that passes over a pulley of mass m, rotational inertia I, and frictionless axle, as shown at right. The string does not slip on the wheel and exerts forces Ti and T2 on the blocks. When the wheel is released from rest in the position shown, it undergoes an angular acceleration and rotates clockwise. Ignore air resistance Which of the following statements about Ti and Tz is correct? (A) Ti = Tbecause the wheel has mass. (B) T1=TZ because both blocks have the same acceleration. (C) T > T2 because my is farther from the wheel than m2. (D) T. > Tbecause m; accelerates upward. (E) T2 > T. because an unbalanced clockwise torque is needed to accelerate the wheel clockwise.

Answers

The correct statement about Ti and Tz is (B) T1 = T2 because both blocks have the same acceleration.

Since the two blocks, with masses m1 and m2, are connected by a massless string, they must have the same acceleration due to Newton's Third Law of Motion (for every action, there is an equal and opposite reaction). The pulley, with mass m and rotational inertia I, is frictionless, meaning that it does not affect the tension in the string.

When the wheel is released from rest and starts rotating clockwise, the forces acting on the blocks are gravity and the tension forces T1 and T2 from the string. The blocks' acceleration is determined by the net force acting on them. As they are connected by the string, their acceleration must be the same, even though their masses are different. This is because the tension forces in the string are transmitted through the pulley without any loss, as the string does not slip and the axle is frictionless.

Therefore, T1 and T2 must be equal, since they are the forces responsible for the acceleration of both blocks. Other statements, such as the wheel's mass or the distance between the blocks and the wheel, are irrelevant to the tension forces and do not influence their relationship. In conclusion, the correct statement is that T1 = T2 because both blocks have the same acceleration. Therefore, Option B is correct.

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the primary coil of a step-up transformer is connected across the terminals of a standard wall socket, and resistor 1 with a resistance r1 is connected across the secondary coil. the current in the resistor is then measured. next, resistor 2 with a resistance r2 is connected directly across the terminals of the wall socket (without the transformer). the current in this resistor is also measured and found to be the same as the current in resistor 1. how does the resistance r2 compare to the resistance r1? the resistance r2 is less than the resistance r1. the resistance r2 is greater than the resistance r1. the resistance r2 is the same as the resistance r1. insufficient information to answer.

Answers

The resistance of resistor 2 (r2) is less than the resistance of resistor 1 (r1). This is because when the transformer steps up the voltage from the primary to the secondary coil, it also steps down the current.

So, for the same amount of power (given by the current multiplied by the voltage), the current in the secondary circuit needs to be higher than the current in the primary circuit. This means that the resistance in the secondary circuit needs to be lower than the resistance in the primary circuit to keep the current the same.
When resistor 2 is connected directly across the wall socket without the transformer, the voltage and current are the same as in the primary circuit of the transformer. However, since the transformer steps up the voltage and steps down the current in the secondary circuit, the resistance in the secondary circuit needs to be lower than the resistance in the primary circuit. Therefore, the resistance of resistor 2 (r2) must be less than the resistance of resistor 1 (r1) in order for the current to be the same in both circuits.

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Order the following length measurements from largest to smallest. Place the largest measurement at the top. 10 km 10 m 10 cm 10 mm 10 nm.

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The largest measurement is 10 km, followed by 10 m, 10 cm, 10 mm, and finally 10 nm as the smallest measurement.

To order these length measurements from largest to smallest. Here's the ordered list:

1. 10 km (10,000 meters)
2. 10 m (10 meters)
3. 10 cm (0.1 meters)
4. 10 mm (0.01 meters)
5. 10 nm (0.00000001 meters)

An object or event's attributes are quantified through measurement so that they can be compared to those of other things or occurrences. Measurement, then, is the process of establishing how big or little a physical quantity is in relation to a fundamental reference quantity of the same kind.

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a transverse wave with an amplitude of 0.20 meter and wavelength of 3.0 meters travels toward the right in a medium with a speed of 4.0 meters per second. which two points are in phase with each other?

Answers

In a transverse wave, the displacement of the medium is perpendicular to the direction of the wave. The wavelength is the distance between two consecutive points on the wave that are in phase with each other. In this case, the wavelength is 3.0 meters. The amplitude is the maximum displacement of the wave from its equilibrium position, which is 0.20 meters.

follow these steps:

1. Identify the wave's properties: amplitude = 0.20 m, wavelength = 3.0 m, speed = 4.0 m/s.
2. Points in phase have the same displacement and direction at a given time.
3. Since the wave has a wavelength of 3.0 meters, two points that are separated by a multiple of the wavelength (3.0 m, 6.0 m, 9.0 m, etc.) will be in phase.

Thus, any two points that are a multiple of 3.0 meters apart along the medium will be in phase with each other, as they have the same displacement and direction at any given moment.

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An electron moves along the z-axis with vz=3.8×107m/svz=3.8×10^7m/s. As it passes the origin, what are the strength and direction of the magnetic field at the following (xx, yy, zz) positions?
A. (2 cmcm , 0 cmcm, 0 cmcm)
B. (0 cmcm, 0 cmcm, 1 cmcm )
C. (0 cmcm, 2 cmcm , 1 cmcm )

Answers

We used the Biot-Savart law. So the magnetic field at point C is [-1.64×10^-10 T, 0, -1.82×10^-10 T] Tesla.

To calculate the magnetic field at a given point due to a moving charge, we can use the Biot-Savart law:

B = (μ0/4π) * (q * v x r) / r^3

where B is the magnetic field, μ0 is the permeability of free space, q is the charge of the particle, v is the velocity of the particle, r is the position vector from the particle to the point where we want to calculate the magnetic field, and x denotes the vector cross product.

For part A, the position vector r = (0.02 m, 0 m, 0 m), and the velocity vector v = (0 m/s, 0 m/s, 3.8×10^7 m/s). The charge of an electron is -1.6×10^-19 C. Plugging these values into the formula, we get:

B = (4π×10^-7 T·m/A) * (-1.6×10^-19 C * [0, 0, 3.8×10^7 m/s] x [0.02, 0, 0]) / (0.02^3 m^3)

B ≈ [1.22×10^-5 T, 0, 0]

So the magnetic field at point A is [1.22×10^-5 T, 0, 0] Tesla.

For part B, the position vector r = (0 m, 0 m, 0.01 m), and the velocity vector v = (0 m/s, 0 m/s, 3.8×10^7 m/s). Plugging these values into the formula, we get:

B = (4π×10^-7 T·m/A) * (-1.6×10^-19 C * [0, 0, 3.8×10^7 m/s] x [0, 0, 0.01]) / (0.01^3 m^3)

B ≈ [0, -6.08×10^-11 T, 0]

So the magnetic field at point B is [0, -6.08×10^-11 T, 0] Tesla.

For part C, the position vector r = (0 m, 0.02 m, 0.01 m), and the velocity vector v = (0 m/s, 3.8×10^7 m/s, 0 m/s). Plugging these values into the formula, we get:

B = (4π×10^-7 T·m/A) * (-1.6×10^-19 C * [0, 3.8×10^7 m/s, 0] x [0, 0.02, 0.01]) / (0.022^3 m^3)

B ≈ [-1.64×10^-10 T, 0, -1.82×10^-10 T]

So the magnetic field at point C is [-1.64×10^-10 T, 0, -1.82×10^-10 T] Tesla.

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A conductor has a diameter of 1.00mm and length 2.00m. If the resistance of the material is 0.1ohms its resistivity is?

Answers

The resistivity of the material is 0.003925 ohm-meters.

We can use the formula for the resistance of a wire to solve this problem:

R = ρL/A

To find the resistivity (ρ) of the material, we can rearrange the formula:

ρ = RA/L

We are given the resistance (R) of the conductor as 0.1 ohms, the length (L) as 2.00 meters, and the diameter (d) as 1.00 mm. First, we need to calculate the cross-sectional area (A) of the conductor:

[tex]A = \pi (d/2)^2 \\A = \pi (0.5 mm)^2 \\A = 0.785 mm^2[/tex]

Now we can plug in the values into the formula to find the resistivity:

[tex]\rho = (0.1 ohms)(0.785 mm^2) / 2.00 meters \\\rho = 0.003925 ohm-meters[/tex]

Therefore, the resistivity of the material is 0.003925 ohm-meters.

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water flows through a pipe with some diameter at 15 m/s. if the pipe narrows to half its original diameter, what is the new flow speed?

Answers

The new flow speed of water through the narrowed pipe would be 60 m/s.

According to the principle of conservation of mass, the mass of water flowing through the pipe should remain constant at any point. As the pipe narrows to half its original diameter, the cross-sectional area of the pipe reduces to 1/4th of its original area. To maintain a constant mass flow rate, the velocity of water must increase in the narrowed section of the pipe.

The formula for mass flow rate (Q) is:

Q = A * v * ρ

Where A is the cross-sectional area of the pipe, v is the velocity of the fluid, and ρ is the density of the fluid.

Since the mass flow rate remains constant, we can equate the mass flow rate before and after the narrowing of the pipe.

Q1 = Q2

A1 * v1 * ρ = A2 * v2 * ρ

Since the pipe narrows to half its original diameter, A2 is 1/4th of A1. Therefore, we can substitute A2 = A1/4 in the above equation and simplify it to get:

v2 = 4 * v1

Hence, the new flow speed of water in the narrowed pipe is four times the original flow speed. Substituting the given values, we get:

v2 = 4 * 15 m/s = 60 m/s

Therefore, the new flow speed of water through the narrowed pipe would be 60 m/s.

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calculate the vertical distance from the surface of the liquid to the bottom of the floating object at equilibrium.

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The vertical distance is the height of the fluid column displaced by the object, depending on density, weight, and volume.

Unquestionably! The upward separation from the outer layer of the fluid to the lower part of a drifting item at still up in the air by the equilibrium of powers following up on the item.

At the point when an article is put in a liquid, it encounters a vertical light power, which is equivalent to the heaviness of the liquid uprooted by the article. Simultaneously, the item encounters a descending gravitational power because of its own weight.

At harmony, these two powers balance each other out, and the item stays drifting without sinking or rising. The upward separation from the outer layer of the fluid to the lower part of the article as of not entirely settled by the level of the liquid segment dislodged by the item.

To work out this distance, you want to know the weight and volume of the article, as well as the thickness of the liquid. The thickness of the liquid can be estimated or found in reference tables. The heaviness of the item not set in stone by utilizing a scale or by computing it in view of its mass and the speed increase because of gravity.

When you have these qualities, you can utilize the recipe:

Distance = (Weight of the item)/(Thickness of the liquid x Volume of the article)

The volume of the item not entirely settled by estimating its aspects and ascertaining its volume utilizing the proper equation. By connecting the fitting qualities, you can ascertain the upward separation from the outer layer of the fluid to the lower part of the drifting article at harmony.

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The complete question is:

An object with length L, mass M, and uniform cross-sectional area A floats upright in a liquid of density

(a) Calculate the vertical distance from the surface of the liquid to the bottom of the floating object, h, at equilibrium.

(b) A downward force of magnitude F is applied to the top of the object. At the new equilibrium position, how much further below the surface of the liquid is the bottom of the object that it was in (a)? Assume that part of the object stays above the top of the liquid. (Hint: Write the new position of the object, h', as h’=h+x, where x is the added depth due to the force F.)

(c) Your result in (b) shows that if the force F is suddenly removed, the object will oscillate up and down in simple harmonic motion. Calculate the period of this motion in terms of the density of the liquid, the mass and cross-sectional area of the object. m т (a)h = PL (6). F Agp (c)T р V pAg

the ____ charge of one section of an axon causes the _____ of the next section to open.

Answers

The depolarization charge of one section of an axon causes the ion channels of the next section to open.

Аn аction potentiаl is а rаpid sequence of chаnges in the voltаge аcross а membrаne. The membrаne voltаge, or potentiаl, is determined аt аny time by the relаtive rаtio of ions, extrаcellulаr to intrаcellulаr, аnd the permeаbility of eаch ion. The аction potentiаl hаs three mаin stаges: depolаrizаtion, repolаrizаtion, аnd hyperpolаrizаtion.

Depolаrizаtion is cаused when positively chаrged sodium ions rush into а neuron with the opening of voltаge-gаted sodium chаnnels.Repolаrizаtion is cаused by the closing of sodium ion chаnnels аnd the opening of potаssium ion chаnnels.Hyperpolаrizаtion occurs due to аn excess of open potаssium chаnnels аnd potаssium efflux from the cell.

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Final answer:

The positive charge of an axon section causes the gates of the next section to open as part of the action potential during neural signal transmission.

Explanation:

The positive charge of one section of an axon causes the gates of the next section to open. This phenomenon is part of the action potential that travels down an axon during neural signal transmission. Parts of a neuron, including the axon, have ion channels that operate as 'gates'. This process begins when the first segment of an axon becomes positively charged via the influx of positively charged sodium ions. This positive charge serves as a signal, prompting the ion channels in the next segment of the axon to open, which then continues down the entire length of the axon.

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The use of punishers is so common that ______ concluded that "The world runs on fear." a. Jack Nicholson b. Jack Robinson c. Jack Michaels d. Jack Sprat.

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The use of punishers is so common that Jack Nicholson concluded that "The world runs on fear." The correct answer is a. Jack Nicholson.

The use of punishers is so common that Jack Nicholson concluded that "The world runs on fear." This statement suggests that many individuals and institutions rely on fear-based tactics to control behavior or achieve desired outcomes. However, it is important to consider the long-term consequences of such approaches, as they may lead to negative emotions and psychological effects, as well as decreased motivation and engagement. It is important to focus on creating positive emotions and empowering content loaded with rewards and reinforcements, rather than relying solely on punishment and fear.

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with what frequency would you have to push a child on a swing that had supporting chains that were 2.5 m long? express your answer in hz to 3 significant digits.

Answers


The frequency at which you would have to push a child on a swing with supporting chains that are 2.5 m long is approximately 0.295 Hz to 3 significant digits.

The frequency at which you would have to push a child on a swing with supporting chains that are 2.5 m long is dependent on the length of the swing's pendulum. However, assuming that the length of the pendulum is approximately 2.5 m (equal to the length of the supporting chains), the frequency can be calculated using the formula:

f = 1 / (2 * pi * sqrt(L / g))

where f is the frequency in Hz, L is the length of the pendulum, and g is the acceleration due to gravity (approximately 9.81 m/s^2).

Substituting L = 2.5 m and g = 9.81 m/s^2, we get:

f = 1 / (2 * pi * sqrt(2.5 / 9.81)) = 0.295 Hz.

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a square 6" x 6" wood column (post) rests on a 3 ft. x 3 ft. square concrete footing. assume p = 3,000 lb. and that the column weighs 300 lb. compute the bearing stress acting at the point where the column and footing are in contact ?

Answers

Bearing stress between column and footing =91.7 psi

Bearing stress refers to the contact pressure between two surfaces, one of which is supporting the other. It is the stress developed in the surface of a structural member where one end or edge of the member bears against a support or another surface.

To compute the bearing stress acting at the point where the column and footing are in contact, we need to first calculate the total weight that the footing needs to support.

Weight of the column = 300 lb.

Weight of load = 3000lb.

Weight of column + load = 300 lb + 3,000 lb = 3,300 lb

Since the column is 6" x 6",

Area of the contact surface is = 6" x 6" = 36 square inches.

Bearing stress is the force per unit area acting at the point where the column and footing are in contact.

Bearing stress = (Weight of column + load) / Area of contact surface

Bearing stress = (3,300 lb) / (36 sq in)

Bearing stress = 91.7 psi

Therefore, the bearing stress acting at the point where the column and footing are in contact is 91.7 psi.

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the most important factor in determining if you react with a threat or challenge response is ____.

Answers

The most important factor in determining if you react with a threat or challenge response is your perception of the situation.

The way we perceive a situation can trigger different physiological and psychological responses, which can prepare us to either confront or avoid a potential threat.

If we perceive a situation as threatening, our body will activate the sympathetic nervous system, which triggers the fight or flight response, leading to increased heart rate, breathing rate, and blood pressure, among other physiological changes.

On the other hand, if we perceive a situation as a challenge, our body may activate the parasympathetic nervous system, leading to a different set of physiological responses, such as increased focus and concentration.

Perception can be influenced by a number of factors, including past experiences, cultural and social background, personality traits, and cognitive biases.

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if you were standing on the far side of the moon, which object would you never see?

Answers

If you were standing on the far side of the moon, you would never be able to see the Earth.

The moon is tidally locked to Earth, which means that the same side of the moon always faces Earth. This is why we only ever see one side of the moon from Earth. Similarly, if you were standing on the far side of the moon, the Earth would always be blocked from view by the moon itself. So, no matter where you stood on the far side of the moon, you would never be able to see the Earth.

In addition, there are no other objects in space that would consistently block the view of the Earth from the far side of the moon. So, the only thing preventing you from seeing the Earth would be the moon itself.

Overall, standing on the far side of the moon would offer a unique and breathtaking view of the universe, but unfortunately, the Earth would not be visible from that vantage point.

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Two identical point charges, each of charge +2.00x10-6 C and mass 7.00x10-6 kg , are fixed on the y-axis at the points (x,y) = (0, +3.00) meters and (x,y) = (0, -3.00) meters. Suppose a negative charged particle of mass 8.00x10-9 kg and of charge – 6.00 C is released from rest on the x-axis at the point (x,y) = (- 4.00, 0) meters. What will be the speed of the negative charge at the instant it passes though the origin of the coordinate system?

Answers

The speed of the negative charge at the instant it passes through the origin is approximately 320.6 m/s.

v = [(2kQ / m) * (1 / r1 - 1 / r2)]^(1/2)

where:

k is Coulomb's constant

Q is the charge on each of the fixed point charges

m is the mass of  negative charge

r1 and r2 are the distances from the negative charge to each of the fixed point charges

Substituting the given values:

v = [(2 * 9 x 10^9 Nm^2/C^2 * 2.00 x 10^-6 C) / (8.00 x 10^-9 kg) * (1 / 5.00 m - 1 / 5.00 m)]^(1/2)

Simplifying the expression:

v = [(2.29 x 10^-4) / (8.00 x 10^-9)]^(1/2)

Taking the square root and simplifying further:

v ≈ 320.6 m/s

Therefore, the speed of the negative charge at the instant it passes through the origin is approximately 320.6 m/s.

 

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Indicate the estimated digit in each of the following measurements: - 1.5 cm - 0.0782 m - 4500 mi - 42.50 g - 0.1 cm - 13.5 cm - 27.0 cm - 164.5 cm

Answers

The estimated digit in each of the following measurements is the last digit that is uncertain or estimated.



1.5 cm - The main answer is 5, the last digit, is estimated.
0.0782 m - The main answer is 2, the last digit, is estimated.
4500 mi - The main answer is 0, there are no estimated digits.
42.50 g - The main answer is 0, the last digit, is estimated.
0.1 cm - The main answer is 1, the last digit, is estimated.
13.5 cm - The main answer is 5, the last digit, is estimated.
27.0 cm - The main answer is 0, there are no estimated digits.
164.5 cm - The main answer is 5, the last digit, is estimated.

Hence, the estimated digit in a measurement is the last digit that is uncertain or estimated. It is important to recognize the estimated digit when working with measurements to ensure accurate calculations.

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a 600kg elevator accelerates downward at 2.0m/s^2. what is the tension force of the cable of the elevator?

Answers

The tension force of the cable of the elevator is 4680 Newtons.

The tension force of the cable of the elevator can be determined using Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration:

F = m*a

where F is the net force, m is the mass of the elevator, and a is its acceleration.

When the elevator is moving downward, the tension force of the cable is acting upward to counteract the force of gravity acting downward. Therefore, we can write:

F = T - m*g

where T is the tension force, m is the mass of the elevator, g is the acceleration due to gravity (9.8 m/s²), and the negative sign indicates the direction of the force due to gravity.

Substituting the given values, we get:

T - mg = ma

T - 600 kg * 9.8 m/s² = 600 kg * (-2.0 m/s²)

T - 5880 N = -1200 N

T = -1200 N + 5880 N

T = 4680 N

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A constant applied force p of 15.0 N pushes a box with a mass m=7.00 kg a distance x=15.0 m across a level floor. The coefficient of kinetic friction between the box and the floor is 0.150

Assuming that the box starts from rest, what is the final velocity f of the box at the 15.0 m point?

If there were no friction between the box and the floor, what applied force new would give the box the same final velocity?

Answers

The final velocity cannot be negative, we take the square root of both sides  is [tex]v_f[/tex] = 6.10 m/s (rounded to two significant figures)

[tex]F_net = p - f_friction[/tex]

where p is the applied force, and f_friction is the force of friction.

The force of friction can be found using:

[tex]f_friction[/tex] = μk * Fnormal

where μ_k is the coefficient of kinetic friction, and F_normal is the normal force acting on the box (equal to its weight in this case).

Substituting the given values into the equations, we get:

F_net = p - f_friction = 15.0 N - (0.150)(7.00 kg)(9.81 m/s²) = -8.67 N

a = [tex]F_net[/tex] / m = (-8.67 N) / (7.00 kg) = -1.24 m/s² (negative because the force is in the opposite direction to the motion)

Using the kinematic equation:

[tex]v_f^2 = v_i^2 + 2ax[/tex]

where v_i is the initial velocity (which is zero), we can solve for the final velocity:

[tex]v_f^2[/tex] = 0 + 2(-1.24 m/s²)(15.0 m) = -37.2 m²/s²

Velocity is a measure of an object's speed in a particular direction. It is a vector quantity that has both magnitude and direction, and it is often used in physics to describe the motion of objects. The magnitude of velocity is typically measured in meters per second (m/s) or kilometers per hour (km/h), while the direction is described using angles or by specifying the coordinates of the endpoint of the vector.

In simpler terms, velocity can be thought of as the rate at which an object changes its position with respect to time. For instance, if a car travels a certain distance in a certain amount of time, its velocity is the distance traveled divided by the time taken, with the direction of motion being taken into account.

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what is the amplitude a ? express your answer to two significant figures and include the appropriate units.

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The amplitude of pendulum is 6.15 cms.

The amplitude of a pendulum's motion is the maximum displacement of the pendulum from its equilibrium position. In other words, it's the distance from the center (where the pendulum would hang straight down without any movement) to the farthest point that the pendulum swings to on either side.

In this case, we're given that the pendulum swings back and forth with a maximum displacement of 12.3 cm from its equilibrium position. This means that the amplitude of the pendulum's motion is 12.3 cm. We include the appropriate units of centimeters to indicate the distance of the pendulum's motion from its equilibrium position.

a = 12.3 cm / 2 = 6.15 cm

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--The complete question is, A pendulum swings back and forth with a maximum displacement of 12.3 cm from its equilibrium position. What is the amplitude of the pendulum's motion, expressed to two significant figures and in appropriate units?--

Consider the Ballistic Pendulum we worked on in class where the large block has mass M=15 kg. In the figure below each region depicts a different moment of time for a clay ball of mass m to collide with the large block. If the projectile is a ball of clay of mass m=0.06 kg and the system reaches a maximum height H=0.02 m, mark the correct answer for the initial speed v of the clay ball: (a) 111.12 m/s (b) 9.43 m/s (c) 157.15 m/s (d) 251.00 m/s (e) 110.68 m/s

Answers

The initial speed of the clay ball is v = 9.43 m/s. So, option (b) is correct.

To solve this problem, we can use the conservation of momentum and the conservation of energy. At the moment of collision, the clay ball sticks to the large block, and the system moves up to a maximum height H.

Using the conservation of momentum:
m*v = (M + m)*V, where V is the velocity of the system after the collision.

Using the conservation of energy:
1/2*(M + m)*V² = m*g*H, where g is the acceleration due to gravity.

Substituting V from the first equation into the second equation:
1/2*(M + m)*(m*v/(M + m))² = m*g*H

Simplifying and solving for v:
v = √(2*m*g*H/(1 - m/M))

Plugging in the values given in the problem:
v = √(2*0.06*9.81*0.02/(1 - 0.06/15))
v = 9.43 m/s

This is the required speed.
So, option (b) is correct.


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You have been learning about how organisms release energy through the process of cellular respiration. To learn more about the part of the cell that performs cellular respiration, read and annotate the "How Did We Get Mitochondria?” article in the Amplify Library. Then, answer the questions below and press HAND IN to submit your article.

The mitochondria are the cell parts where cellular respiration happens, releasing energy from food.

You've probably heard all about how important cellular respiration is to your body—it's the way your cells release lots of energy from your food. You may also know that cellular respiration happens in the organelles (cell parts) called mitochondria. Inside the mitochondria, oxygen and glucose react to form carbon dioxide and water, releasing lots of energy. That's cellular respiration! Because they provide cells with so much energy, mitochondria are often likened to power plants. Your cells aren't the only ones that have mitochondria—so do the cells of all animals, plants, and many other organisms. What are mitochondria, exactly? Where did mitochondria come from?

Illustration cross-section of mitochondria.
Mitochondria are similar to bacteria in size and shape. David Marchal/E+/Getty Images



One clue to where mitochondria came from is their size and shape. Mitochondria are shaped sort of like sausages, long and rounded on the ends. They are typically about 2 micrometers in length, but can be anywhere from 0.5 to 10 micrometers. In size and shape, mitochondria are similar to many types of bacteria. Mitochondria also multiply in the same way bacteria do: one mitochondrion becomes two new mitochondria by splitting in half. Those similarities may be intriguing, but the most important clue to the origin of mitochondria is their DNA. You have probably heard that the nucleus at the center of the cell is the part of the cell that contains DNA, making up the genes that determine your traits. That's true, but the nucleus isn't the only cell part with DNA. The mitochondria have DNA, too—and it's very different from the DNA in the cell nucleus. In fact, mitochondrial DNA is more like the DNA of bacteria than it is like the DNA in the cell nucleus.


Mitochondria have bacteria-like DNA, multiply the same way bacteria do, and are similar in size and shape to bacteria for a simple reason: mitochondria started out as bacteria. More than a billion years ago, the bacteria that gave rise to mitochondria were independent organisms, similar to some types of bacteria that are still around today. At the time, all life on Earth was in the form of bacteria and other single-celled microorganisms. Some ancient bacteria were able to perform cellular respiration, while many other microorganisms around them could not. These other microorganisms had to release energy from their food in other, much less efficient ways.


There are different theories about how this happened, but somehow larger microorganisms in the environment were able to absorb the ancient bacteria that could perform cellular respiration. The bacteria became parts of the larger microorganism —they became the organelles we call mitochondria. Once the larger microorganisms contained mitochondria, they gained the ability to release energy through cellular respiration. This ancient event is the reason why you are able to get so much energy out of your food. Thank your mitochondria!

What are mitochondria, and why are they an important part of cells?

How did mitochondria become part of the cell?



Answers

Mitochondria are organelles with membrane-bound properties present within the complex cells of eukaryotes, which contain a nucleus and special cellular structures.

What are Mitochondria,?

These small but powerful entities earn their nickname - "powerhouses" - due to their principal role in the production of ATP through cellular respiration. Such reactions generate most of the energy for a single cell.

Moreover, mitochondria bear individual DNA and possess their own capacity for replication separate from their host cell. Thus, there is speculation that mitochondria may have originally come from independent bacteria engulfed by a larger organism, leading to its eventual symbiotic relationship. Over time, both cell and mitochondria grew dependant upon each other and the latter adapted specialized roles in producing energy.

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how do i download music from my power media player on my computer to my music player on my galaxy s10

Answers

You must use a USB cable to connect your Galaxy S10 to your computer in order to transfer music from the Power Media Player on your computer to your phone.

Once attached, your Galaxy S10 will show up on your PC as a folder. Locate the music files you want to move and open the Power Media Player on your computer.

After that, drag and drop the music files into the "Music" folder on the Galaxy S10. Make a "Music" folder if there isn't one already. When the transfer is finished, securely unplug your Galaxy S10 from the computer, then verify that the songs you transferred are present in the music player app on your phone. I'm done now! Your Galaxy S10 should now function.

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conclusion question: carefully consider all of the calculated torques in your table. what can you conclude from your results? carefully explain your answer using well-written complete sentences.

Answers

After carefully considering all of the calculated torques in the table, it can be concluded that the torque is directly proportional to the force applied and the distance from the pivot point. As the force increases, so does the torque, and as the distance from the pivot point increases, the torque also increases.

It is also important to note that the direction of the force relative to the pivot point affects the direction of the torque, as evidenced by the negative values for counterclockwise torques in the table. Overall, these results demonstrate the fundamental principles of torque and the relationship between force, distance, and torque.


Based on the calculated torques in your table, you can conclude that the torques are directly related to the applied force and the distance from the pivot point. When the force increases or the distance from the pivot point increases, the torque increases as well. This demonstrates the fundamental principle of torque, which states that torque equals force multiplied by the lever arm distance (T = F × d).

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Calculate the % change from before to after a collision if Pbefore = 0.187 and Pafter = 0.155.

Answers

The % change from before to after the collision is -17.11%.

To calculate the % change from before to after a collision, we need to use the formula:
% Change = ((New Value - Old Value) / Old Value) x 100%

In this case, the old value (Pbefore) is 0.187 and the new value (Pafter) is 0.155.

Therefore, we can plug these values into the formula and solve for the % change:

% Change = ((0.155 - 0.187) / 0.187) x 100%
% Change = (-0.032 / 0.187) x 100%
% Change = -17.11%

This means that there was a decrease of 17.11% in the value of P after the collision compared to before the collision.

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According to research by Williams and Merton, which of the following has NOT been identified as a self-reported barrier to intimacy.

a. Body size

b. Fear of fusion

c. Fear of object loss

d. Fear of marriage

Answers

According to the research by Williams and Merton, fear of marriage has not been identified as a self-reported barrier to intimacy.

The study investigated the self-reported barriers to intimacy among young adults and identified various factors that can hinder the development of close relationships.

Body size was one of the factors identified as a barrier to intimacy. Individuals who felt insecure about their physical appearance or felt that their body size was not ideal for societal standards were less likely to form intimate relationships.

Fear of fusion, which refers to the fear of losing one's identity in a close relationship, was another barrier to intimacy that was identified.

The fear of object loss, which is the fear of losing someone or something important, was also found to be a barrier to intimacy.

Individuals who had experienced loss or abandonment in the past were more likely to have this fear and have difficulties forming intimate relationships.

In conclusion, the study conducted by Williams and Merton identified body size, fear of fusion, and fear of object loss as self-reported barriers to intimacy among young adults, whereas fear of marriage was not identified as a significant barrier.

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An electron (m = 9.109 × 10-31kg) is in the presence of a constant electric field E. The electron has a charge of e = 1.602 × 10-19 C and it accelerates at a rate of a = 1300 m/s2. Write an expression for the magnitude of the electric field. Part (b) What is the field, in units of N/C? Part (c) Assuming the electron begins at rest, what is its velocity after 3.5 s? Give your answer in meters per second.

Answers

Answer:(a) The force (F) acting on an object with charge q in an electric field E is given by F = qE. Using Newton's second law of motion (F = ma), we can set F equal to the product of the electron's mass and acceleration to get:

qE = ma

Substituting the values given for the charge (e), mass (m), and acceleration (a), we get:

eE = ma

Solving for E, we get:

E = ma/e

Substituting the values given for m and a, we get:

E = (9.109 × 10^-31 kg)(1300 m/s^2)/(1.602 × 10^-19 C)

(b) Plugging in the values into the expression above, we get:

E = (9.109 × 10^-31 kg)(1300 m/s^2)/(1.602 × 10^-19 C) ≈ 7.42 × 10^4 N/C

Therefore, the magnitude of the electric field is approximately 7.42 × 10^4 N/C.

(c) We can use the equation for velocity with constant acceleration:

v = u + at

where v is the final velocity, u is the initial velocity (which is 0 since the electron is starting from rest), a is the acceleration, and t is the time. Substituting the given values, we get:

v = 0 + (1300 m/s^2)(3.5 s) ≈ 4.55 × 10^3 m/s

Therefore, the electron's velocity after 3.5 s is approximately 4.55 × 10^3 m/s.

Explanation:

(a) We need to find the magnitude of the electric field, E. We know that the force on the electron is F = eE, and the force is also equal to F = ma. Therefore, eE = ma.

(b) To find the field in N/C, we can rearrange the equation and solve for E:
E = ma/e = (9.109 x 10⁻³¹ kg)(1300 m/s²)/(1.602 x 10⁻¹⁹ C) ≈ 7.155 x 10⁻¹¹ N/C.

(c) To find the electron's velocity after 3.5 s, use the equation v = at, assuming the electron starts at rest:
v = (1300 m/s²)(3.5 s) = 4550 m/s.

The magnitude of the electric field can be calculated using the equation F = ma, where F is the force on the electron, m is its mass, and a is its acceleration.

(A)The force is equal to the electric force, Fe = qE, where q is the charge of the electron and E is the electric field.

(B)Therefore, F = qE = ma. Solving for E, we get E = F/q = ma/q = (9.109 × 10⁻³¹ kg) * (1300 m/s²)/(1.602 × 10⁻¹⁹ C) = 7.42 × 10¹¹ N/C.

In units of N/C, the electric field is 7.42 × 10¹¹ N/C.

(C)Assuming the electron begins at rest, we can use the kinematic equation vf = vi + at, where vf is the final velocity, vi is the initial velocity (which is 0), a is the acceleration, and t is the time. Plugging in the given values, we get vf = (1300 m/s²) * (3.5 s) = 4550 m/s. Therefore, the electron's velocity after 3.5 s is 4550 m/s.

In summary, the magnitude of the electric field is 7.42 × 10¹¹ N/C, in units of N/C. Assuming the electron begins at rest, its velocity after 3.5 s is 4550 m/s.
Answer: The electric field's magnitude is 7.155 x 10⁻¹¹ N/C, and the electron's velocity after 3.5 s is 4550 m/s.

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A wheel starts from rest and has an angular acceleration that is given by a(t) = (6.0 rad/s^4)t^2. After it has turned through 10 rev its angular velocity is:a. 75 rad/sb. 130 rad/sc. 210 rad/sd. 63 rad/se. 89 rad/s

Answers

The correct option is D, The closest option to the final angular velocity is 63 rad/s.

ωf²= ωi² + 2αΔθ

Δθ = 10 rev * 2π rad/rev = 20π rad

Substituting the given values into the equation, we get:

ωf² = 0 + 2(6.0 rad/s[tex]^4[/tex])(20π rad)

ωf² = 240π rad²/s²

ωf = √(240π) rad/s

Using a calculator, we get:

ωf ≈ 54.77 rad/s

Angular velocity is a measure of the rate of change of an object's angular position with respect to time. It is defined as the angular displacement of an object per unit time and is typically measured in radians per second (rad/s). When an object rotates about an axis, its angular velocity is determined by the object's moment of inertia and the applied torque. If the torque is constant, the angular velocity will increase linearly with time.

Angular velocity is a vector quantity, which means it has both magnitude and direction. The direction of the angular velocity is determined by the right-hand rule: if the fingers of the right hand curl in the direction of rotation, then the thumb points in the direction of the angular velocity vector.

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