What is the potential difference between the plates of a 3.7-f capacitor that stores sufficient energy to operate a 75.0-w light bulb for one minute?

The annual 10-k report costco wholesale corporation is

a. 4.28 billion dollars.

b. 1.128 billion.

c. $2,393 billion

d. $481 million less.

Costco Wholesale Corporation: What is it?

The fee is a multibillion-dollar large container retail organisation that operates a network of comrades for the large container shops. it's miles the fifth-biggest retailer inside the globe as of the year 20220. It offers wine and natural goods for sale. From these operations, price receives $4,285 in coins. The coins spent on the brand new assets become system for sales of $2,393 million USD, and the depreciation expenditure changed into $1,127 USD. The value increases its holdings to roughly 481 million USD.

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Now the islands are moving more north compared to before compare to the direction of pacific plate motion relative to the hawaiian hotspot over the past million years.

The Pacific Ocean's ocean floor is made up of multiple plates. In relation to the plate holding North America and to hot areas rising through the mantle from below the plates, the largest one, the Pacific Plate, is moving north-west (they generate islands like Hawaii). The plate is travelling around Hawaii at a rate of 7 cm per year.

A hot spot that provided magma produced deep under the mantle of the earth, which was used to create the long volcanic chain, did so over the course of about 70 million years. Over the course of around 70 million years, a hot spot supplied magma that was formed deep under the earth's interior (mantle), pushing its way through the earth's surface and ocean floor to construct volcanic islands.

The hot point continuously created new volcanoes on the Pacific Plate, creating the volcanic chain, while the Pacific Plate was shifted by tectonic forces inside the Earth.The Pacific Plate will move in the direction and at the rate determined by the distances between and the approximate ages of several of the Hawaiian volcanoes.

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The number of walking paces as you walk down main street (0.25 miles) is 720 walking paces

1 mile = 5280 feet

Therefore,

0.25 mile = (0.25 mile × 5280 feet) / 1 mile

0.25 mile = 1320 feet

1 feet = 12 inches

Therefore,

1320 feet = (1320 feet × 12 inches) / 1 feet

1320 feet = 15840 inches

22 inches = 1 walking pace

Therefore,

15840 inches = (15840 inches × 1 walking pace) / 22 inches

15840 inches = 720 walking paces

Thus, we can conclude that as you walk down main street (0.25 miles), there are 720 walking paces

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The charge on the penny is necessary to levitate the penny, the magnitude is 3*10^-4C and sign is negative.

Charged matter experiences a force when it is exposed to an electromagnetic field due to the physical property of electric charge. Positive or negative charges can exist in an electric field (commonly carried by protons and electrons respectively). A neutral object is one that carries no net charge. The early understanding of how charged particles interact is today known as classical electrodynamics, and it is still true for issues that do not demand for taking into account quantum phenomena. Electric charge is a conserved property; the net charge of an isolated system, which is the sum of the positive and negative charges, cannot change.

According to Newton's second law -

∑fy= ma

fe-mg=ma

Fe= ma + mg

qE = 3*10^-3*0.19 + 3*10^3*9.81

qE= 30 * 10^-3

q= 30*10^3/100 = 30*10^-5

q= 3*10^-4C

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Answer:88.3 m/s

Explanation:

The renting car that is reported to achieve 10.1 km/L its mileage in miles/gal is: 23.75 miles/gal

To solve this problem, we have to do the conversion of units with the given information.

Information about the problem:

Converting the units from km/L to miles/gal, we get:

10.1 Km/L * (1 miles / 1.60934 km) * (3.785 L  1 gal) = 23.75 miles/gal

10.1 Km/L is equivalent to 23.75 miles/gal

It is the transformation of a value expressed in one unit of measurement into an equivalent value expressed in another unit of measurement of the same nature.

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The number of stars present in the galaxy is 10 raised to the power of 14. So the exponent is equal to 14.

Given in the question

Total mass of the particular galaxy = 10 to the power of 47 grams

The average mass of the star in the galaxy = 10 to the power of 33 grams

We know that the total mass of the galaxy which is a collection of several thousand stars is equal to the sum of all the stars present in it. Also, the total mass of the galaxy can be calculated by multiplying the average mass of a star in the galaxy by the number of stars that are there in the galaxy.

So, Total mass = Average mass × Number of stars in the galaxy

Number of stars in the galaxy = Total mass of the galaxy/ Average mass of the star

Put in the values, we get

Number of stars in the galaxy = 10^47/10^33

Number of stars in the galaxy = 10^(47-33)

Number of stars in the galaxy = 10^14

So, the number of stars present in the galaxy is 10 raised to the power of 14. So the exponent is equal to 14.

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The change in energy is -10.2 eV.

We need to know about transition energy to solve this problem. Electrons in a hydrogen atom can move to another level of energy. It is also called a transition. In this process, electrons will absorb or release their energy to move. The transition energy can be determined as

ΔE = E2 - E1

where ΔE is transition energy, E2 is the final state of energy and E1 is the initial state of energy.

The state energy of a hydrogen atom can be calculated by

E = -(13.6) / n² eV

where n is the number of energy states.

From the question above, we know that

n1 = 2

n2 = 1

By substituting the following parameter, we get

ΔE = E2 - E1

ΔE = -(13.6) / n2² - (-(13.6) / n1²)

ΔE = -(13.6) / 1² - (-(13.6) / 2²)

ΔE = 3.4 - 13.6

ΔE = -10.2 eV

The transition energy is minus because electrons emit the energy to move.

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Therefore, the net force acting on particle 3 is 139.68 × 10—5 N, and the angle made it by with at the +x axis is 19.1°.

The charge of particle 1 is q1.

q1 = -6.6nC

The charge of particle 2 is q2.

q2 = -13.2 nC

Sides of the equilateral triangle = 3 cm = 3 × 10 —² m.

The force acting on particle 1 is,

[tex]F _{1} = \frac{kq_{1} q_{3}}{a ^{2} } [/tex]

[tex] = \frac{9 \times 10 ^{9} \times 6.6 \times 10 ^{ - 9} \times 8 \times 10 ^{ - 9} }{9 \times 10 ^{ - 4} } [/tex]

[tex]F _{1} = 52.8 \times 10 ^{ - 5} \: N[/tex]

[tex]F _{1} = F _{1} \: cos(60°)i + F _{1} \: sin(60°)j[/tex]

[tex]= ( - 26.4 \: i \: + 45.7 \: j \: ) ×10 ^{ - 5} [/tex]

The force acting on particle 2 is,

[tex] = \frac{9 \times 10 ^{9} \times 3.2 \times 10 ^{ - 9} \times 8 \times 10 ^{ - 9} }{9 \times 10 ^{ - 4} } [/tex]

[tex] = - 105.6 \times 10 ^{ - 5} \: j[/tex]

The net force acting on particle 3 is,

[tex]F_{Net} =F _{1} + F _{2}[/tex]

[tex] = ( - 132 \times 10 ^{ - 5} \: i + 45.7 \times 10 ^{ - 5} \: j)[/tex]

[tex]F_{Net} =139.68 \times \: 10 ^{ - 5} \: N[/tex]

Thus, the net force acting on particle 3 is 139.68 × 10—5 N.

Let the angle made by it with at the +x axis be θ.

[tex]cos \: θ = \frac{ - 132 \times 10 ^{ - 5} }{139.6 \times 10 ^{ - 5} } [/tex]

θ = 19.1°

Therefore, the net force acting on particle 3 is 139.68 ×

10—5 N and the angle made it by with at the +x axis is 19.1°.

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D. The height of a student in the class from the given data is 70.

A stem plot is a way to plot data where the data is split into stems (the largest digit) and leaves (the smallest digits).

5, 25, 56, 92, 47, 76, 70, 1

Thus, the height of a student in the class from the given data is 70.

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The total time it takes a project fired straight up at 10m/s and Return to its starting point is about time, t= 2s.

At top speed v=0,

V= urate

0=10+ -g×t

t=1 sec

Top to Bottom it will take one more sec

t= 1+1

t=2 sec/s.

Unless there has been a break in service of less than twelve (12) months, in which case the employee will get an adjusted seniority date upon restoration to a Unit 6 classification, Total Time is calculated as the amount of time since the hire date.

During a gap in employment, an employee is not permitted to gain seniority. When an employee returns to a Unit 6 classification after more than twelve (12) months of absence, their starting seniority date is reset entirely.

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The acceleration ax = 3.03 m/s^2 and the ay = 1.75 m/s^2.

It is given that

h = 1.75m

[tex]\theta[/tex]= 30°

Vo = 5.87 m/s

Now Vox= Vo cos[tex]\theta[/tex]

= 5.87 x cos30° = 5.08 m/s

Voy = Vo sin[tex]\theta[/tex] = 2.94 m/s

Tan[tex]\theta[/tex] = h/x

⇒ x =h/tan[tex]\theta[/tex]

⇒ x = 3.03m

y = h = 1.75m

Now, v for the ball is given by

V = Vo + [tex]\sqrt{10gh/7}[/tex]

= Vo + [tex]\sqrt{10*9.81*1.75/7}[/tex]

V = 5.87 + 4.95 = 10.82m/s

⇒ Vx = Vcos[tex]\theta[/tex] = 9.37 m/s

and Vy = 5.41 m/s

a = 5/7 gsin[tex]\theta[/tex]

So, ax = 5/7 gsin[tex]\theta[/tex] x cos[tex]\theta[/tex]

ax = 5/14 gsin(2[tex]\theta[/tex])

ax = 5/14 x 9.81 x sin60°

ax= 3.03 m/s^2

and ay = 5/7gsin[tex]\theta[/tex] x sin[tex]\theta[/tex] = 5/7 gsin^2[tex]\theta[/tex]

= 5/7 x 9.81 x sin^2 30°

⇒ ay = 1.75 m/s^2

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The change in the internal energy of the gas is ΔE int =0 J.

Q=250J heat is absorbed by the gas in this process.

What is  ​isothermal process?

A thermodynamic process known as an isothermal process keeps the system's temperature constant. Because heat is transferred into or out of the system so slowly, thermal equilibrium is preserved.

Givens:

Vi =0.5 L=0.5×10^ −3 m^3

Vf =4.0 L=4.0×10^−3 m^3

W= 250 J

T= 300 K

Because the gas temperature is constant, the process is ISOTHERMAL. As a result, the system's internal energy is constant.

ΔE int =0 J

We know that, ΔE int =Q−W

And we are aware that there has been no change in the system's inherent energy.

Therefore,0=Q−W

Q=W=250J

Q=250J

​

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Question:

When a dilute gas expands quasi-statically from 0.50 to 4.0 L, it does 250 J of work. Assuming that the gas temperature remains constant at 300 K, (a) what is the change in the internal energy of the gas? (b) How much heat is absorbed by the gas in this process?

The additional charge that flows to the positive plate from the 8 Volts battery is  12 x 10⁻³ Coulomb's.

We have an uncharged capacitor which is charged in two different ways using battery of two different voltages. Firstly, it is connected to the terminals of a 4.0 V battery, and 12 μc of charge flows to the positive plate. Then, the 4.0 V battery is disconnected and replaced with a 8.0 V battery

We have to determine the Additional charge that flows to the Positive plate from this 8 V battery.

Capacitance is the ability of an object to collect and store energy in the form of an electrical charge. Mathematically -

C = Q/V

According to the question, we have -

Electric potential of battery 1 = V[1] = 4 V

Charge flow from battery 1 to the positive plate = Q[1] = 12 x 10⁻³ C

Electric potential of battery 2 = V[2] = 8 V

Now, using the following relation, we will find Q[2] {Charge flow from battery 2 to the positive plate} →

V[1] / V[2] = Q[1] / Q[2]

Substituting the values -

Q[2] = (12 x 10⁻³ x 8)/4 = 24 x 10⁻³ C

Additional Charge that flowed to the positive plate →

Q[2] - Q[1] = 12 x 10⁻³ Coulomb's

Hence, the additional charge that flows to the positive plate is  [tex]12\times 10^{-3}[/tex] Coulomb's.

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[ The question given is incomplete, the complete question is -

"An uncharged capacitor is connected to the terminals of a 4.0 v battery,  and 12 μc flows to the positive plate. the 4.0 v battery is then disconnected and replaced with a 8.0 v battery, with the positive and negative terminals connected in the same manner as before. How much additional charge flows to the positive plate" ]

The frequency of these vibrations if the car moves at 43.0 m/s is 1954.54 Hz.

Given that,

The speed of the car = 43m/s.

or the speed of the car = 4300cm/s

A tire has a tread pattern with a crevice = 2.20cm

Since the crevice hits every 2.20cm,

This happens,

= 4300cm / 2.20

= 1954.54.

Therefore,

The frequency is 1954.54Hz

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Answer:

According to Newton's second law of motion, the acceleration of a body is directly proportional to the force acting on the body and inversely proportional to its mass. The formula for this law is

a

=

F

m

, from which we get the formula

F

=

m

a

. When mass is in kg and acceleration is in

m/s/s

or

m/s

2

, the unit of force is

kgm/s

2

, which is read as kiligram-meter per second squared. This unit is replaced with an N in honor of Isaac Newton. Your problem can be solved as follows:

Known/Unknown:

m

=

3000kg

a

=

2m/s

2

Equation:

F

=

m

a

Solution:

F

=

m

a

=

3000kg

x

2m/s

2

=

6000kgm/s

2

=

6000N

Explanation:

A rock is thrown vertically upward at an initial velocity of 70 m/s.

The approximate acceleration of the rock when it reaches its maximum heigh is- zero

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The longest wavelength of light capable to remove an electron from a sample of potassium metal is 1.136 x 10^-19 m.

The binding energy is the amount of energy needed to bind the electron to the element.

If energy equal to or more than the binding energy is provided, electrons will be removed from the metal.

The energy(E) of a light wave is calculated using the following formula,

E = hc/λ

In the above formula

h = Planck’s Constant = 6.626 x 10^-34 J s

c = speed of light

λ = wavelength of light

Putting E = 1.76 x 10^6 J in the formula

1.76 x 10^6 = 6.626 x 10^-34 x 3 x 10^8/lambda

λ = 1.136 x 10^-19 m.

Thus, the longest wavelength of light capable to remove an electron from a sample of potassium metal is 1.136 x 10^-19 m.

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The longest wavelength of light capable to remove an electron from a sample of potassium metal is 1.136 x 10^-19 m.

The binding energy is the amount of energy needed to bind the electron to the element.

If energy equal to or more than the binding energy is provided, electrons will be removed from the metal.

The energy(E) of a light wave is calculated using the following formula,

E = hc/λ

In the above formula

h = Planck’s Constant = 6.626 x 10^-34 J s

c = speed of light

λ = wavelength of light

Putting E = 1.76 x 10^6 J in the formula

1.76 x 10^6 = 6.626 x 10^-34 x 3 x 10^8/lambda

λ = 1.136 x 10^-19 m.

Thus, the longest wavelength of light capable to remove an electron from a sample of potassium metal is 1.136 x 10^-19 m.

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By ideal gas approximation, the final to initial density ratio is 1/2.

We need to know about the ideal gas theory to solve this problem. The ideal gas is assumed that there is no interaction between particles in a gas. It can be determined by the equation

P . V = n . R . T

where P is pressure, V is volume, n is the number of moles gas, R is the ideal gas constant (8.31 J/mol.K) and T is temperature.

From the question above, we know that

P2 = 0.5 P1

Pressure is inversely proportional to pressure, hence it can be written as

P ~ (1/V)

Assuming the temperature does not change, the ratio of pressure is

P1 / P2 = (1/V1) / (1/V2)

P1 / P2 = V2 / V1

P1 / 0.5P1 = V2 / V1

V2 = 2 V1

Find the density ratio

ρ2 / ρ1 = (m/V2) / (m/V1)

ρ2 / ρ1 = V1 / V2

ρ2 / ρ1 = V1 / 2V1

ρ2 / ρ1 = 1 / 2

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Atoms, in which the number of electrons does not equal the number of protons, are  referred to what we call ions.

This refers to an atom or molecule which has a net electric charge as a result of the gain or loss of electrons which is the part of the subatomic particle actively involved in a chemical reaction.

Due to the electrons involvement in a reaction, it doesn't equal the number of protons and results in the formation of ions in which the positive ions are the ones which donate electrons and negative ions accept electrons.

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The acceleration of the ball is zero.

The term acceleration refers to the rate of change of velocity with time thus we can write;

a = d(v)/dt. This simply means the change in velocity with respect to time in the calculus notation.

This implies that before we can talk of acceleration the velocity must be changing. We can not have acceleration when the velocity is constant. It violates the very definition of acceleration. As such, the acceleration of the ball is zero.

Thus whether over a period of three seconds or over an infinite period of time, the acceleration would remain zero because the velocity is constant and does not change over the time interval as given.

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Answer:

metric

Explanation:

Option B has four significant figures.

The number of significant single digits (0 through 9 inclusive) in the coefficient of an expression in scientific notation is referred to as significant figures. The amount of precision or assurance an engineer or scientist uses to state a quantity is indicated by the number of significant figures in a statement. Rounding off an expression once a calculation is completed yields significant figures. The number of significant figures in the result of any computation must be equal to or less than the number of significant figures in the expression or component with the least accuracy. The digits in a number that have significance in terms of accuracy or precision are known as significant figures. Any digit that is not zero is among them. As in 3003 or 45.60009, there are zeros in between non-zero digits.

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Answer: 8.10m/s

Explanation:

using second eq of motion s=ut+1/2at

2

as u=0, a=

t

2

2s

​

=

5.21

2

2∗110

​

=8.10m/s

2

Answer:

1,25m.s^-1

Explanation:

Speed=distance /time

=90000/72000

In the Lewis formula for hydrazinium ion, N 2H 5 +, the total number of lone electron pairs around the two nitrogen atoms is 1.

What is Lewis structure?

Lewis structures, often referred to as Lewis dot formulas, Lewis dot structures, electron dot structures, or Lewis electron dot structures (LEDS), are diagrams that depict the interactions of atoms inside molecules as well as any lone pairs of electrons that may be present. Any molecule with a covalent link, as well as coordination compounds, can have a Lewis structure. Gilbert N. Lewis, who described it in his 1916 article, The Atom and the Molecule, gave the Lewis structure its name. Lewis structures add lines between atoms to represent shared pairs in a chemical bond, extending the idea of the electron dot diagram.

The two nitrogen atoms are surrounded by just one lone electron pair in the hydrazinium ion's Lewis formula, n2h5+.

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The time period remains the same = 1 sec.

We have a mass–spring system.

We have to determine the period of oscillation if mass–spring system is displaced from equilibrium by 10cm and not from 5cm.

The time period is given by -

[tex]$T = 2\pi \sqrt{\frac{M}{K} }[/tex]

According to the question -

For x = 5 cm the time period is 1 seconds.

The time period is directly proportional to [tex]$\sqrt{\frac{M}{K} }[/tex] and it is not dependent on x.

So the time period remains the same = 1 sec only and only is the spring - mass system is the same.

Hence, the time period remains the same = 1 sec.

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A. The ball is moving 17.15 m/s when it was 10 m above the ground

B. The time required for the ball to reach the ground is 2.26 s

h = ½gt²

15 = ½ × 9.8 × t²

15 = 4.9 × t²

Divide both side by 4.9

t² = 15 / 4.9

Take the square root of both side

t = √(15 / 4.9)

t = 1.75 s

v = gt

v = 9.8 × 1.75

v = 17.15 m/s

h = ½gt²

25 = ½ × 9.8 × t²

25 = 4.9 × t²

Divide both side by 4.9

t² = 25 / 4.9

Take the square root of both side

t = √(25 / 4.9)

t = 2.26 s

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By Applying Newton's law of gravitation to this speeds, we can deduce the mass of the cluster.

What is a galaxy?

A galaxy is a collection of stellar remnants, stars, interstellar gas, dust, and dark matter that is gravitationally bound to each other. The name comes from the Greek word galaxias, which translates to mean "milky" and refers to the Milky Way galaxy, which houses the Solar System. Galaxies range in size from dwarfs with fewer than one hundred million (108) stars to the largest galaxies known as supergiants with one hundred trillion (1014) stars, each orbiting its galaxy's center of mass. Galaxies are thought to contain an average of 100 million stars.

Therefore, the speeds of individual galaxies can be used to determine the mass of the cluster of galaxies  by the method of measuring the speeds and positions of the galaxies in a cluster as they orbit. And then, applying Newton's law of gravitation to this speeds, we can deduce the mass of the cluster.

Determination of mass of cluster using speeds of galaxies orbiting the center:

a) determine a cluster's average speed and redshift measurements

b) To determine each galaxy's individual speed, deduct this average from the galaxy's overall speed.

c) calculate cluster mass using newtons universal law of gravitation and individual speed

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