Your new toaster has two separate toasting units, each of which consumes 600 watts of power when it is in use. When you operate one unit, a current of 5 amperes flowsthrough the wiring in your home and the wires waste about 1 watt of power handling that current. If you operate both toasting units at once, your toaster consumes 1200 watts and the current flowing through the wiring in your home doubles to 10 amperes. How much power will the wires in your home waste now

Answer:

[tex]\tau = 1\ ms[/tex]

Explanation:

First we need to find the capacitance of the capacitor.

The capacitance is given by:

[tex]C = \epsilon_0 * area / distance[/tex]

Where [tex]\epsilon_0[/tex] is the air permittivity, which is approximately 8.85 * 10^(-12)

The radius is 12/2 = 6 mm = 0.006 m, so the area of the capacitor is:

[tex]Area = \pi * radius^{2}\\Area = \pi * 0.006^2\\Area = 113.1 * 10^{-6}\ m^2[/tex]

So the capacitance is:

[tex]C = \frac{8.85 * 10^{-12} * 113.1 * 10^{-6}}{0.001}[/tex]

[tex]C = 10^{-12}\ F = 1\ pF[/tex]

The time constant of a rc-circuit is given by:

[tex]\tau = RC[/tex]

So we have that:

[tex]\tau = 10^{9} * 10^{-12} = 10^{-3}\ s = 1\ ms[/tex]

Answer: 10m

Explanation:

The magnitude of the vector would be 10

[tex]\sqrt{6^{2}+8^{2} } =10[/tex]

Answer:

The highest electric field is experienced by a 2 C charge acted on by a 6 N electric force. Its magnitude is 3 N.

Explanation:

The formula for electric field is given as:

E = F/q

where,

E = Electric field

F = Electric Force

q = Charge Experiencing Force

Now, we apply this formula to all the cases given in question.

A) A 2C charge acted on by a 4 N electric force

F = 4 N

q = 2 C

Therefore,

E = 4 N/2 C = 2 N/C

B) A 3 C charge acted on by a 5 N electric force

F = 5 N

q = 3 C

Therefore,

E = 5 N/3 C = 1.67 N/C

C) A 4 C charge acted on by a 6 N electric force

F = 6 N

q = 4 C

Therefore,

E = 6 N/4 C = 1.5 N/C

D) A 2 C charge acted on by a 6 N electric force

F = 6 N

q = 2 C

Therefore,

E = 6 N/2 C = 3 N/C

E) A 3 C charge acted on by a 3 N electric force

F = 3 N

q = 3 C

Therefore,

E = 3 N/3 C = 1 N/C

F) A 4 C charge acted on by a 2 N electric force

F = 2 N

q = 4 C

Therefore,

E = 2 N/4 C = 0.5 N/C

The highest field is 3 N, which is found in part D.

A 2 C charge acted on by a 6 N electric force

Answer:

rod end A is strongly attracted towards the balls

rod end B is weakly repelled by the ball as it is at a greater distance

Explanation:

When the ball with a negative charge approaches the A end of the neutral bar, the charge of the same sign will repel and as they move they move to the left end, leaving the rod with a positive charge at the A end and a negative charge of equal value at end B.

Therefore rod end A is strongly attracted towards the balls and

rod end B is weakly repelled by the ball as it is at a greater distance

Answer:

the magnitude is 7 and sign of the point charge on the surface shell is -13

Explanation:

Answer:

Make a hypothesis, conduct an experiment, Analyze the experimental data..

Answer:

1)copper wire

Explanation:

it is the best electric conductor

Before she jumped from the cliff, her gravitational potential energy was

GPE = (her weight) x (height of the cliff) .

That's exactly the GPE that she lost on the way down to the water.  So we can write

24,500 J = (her weight) x (44.0 m)

Divide each side by 44.0 m:

Her weight = 24,500 J / 44 m

Her weight = 556.8 Newtons

(about 125 pounds)

Answer:

At thestratosphere: it 20- 25km

Answer:

a) a = 3.09 m/s²

b) aₓ = 2.60 m/s²

Explanation:

a) The magnitude of her acceleration can be calculated using the following equation:

[tex] V_{f}^{2} = V_{0}^{2} + 2ad [/tex]

Where:

[tex]V_{f}[/tex]: is the final speed = 8.89 m/s

[tex]V_{0}[/tex]: is the initial speed = 0 (since she starts from rest)

a: is the acceleration

d: is the distance = 12.8 m    

[tex] a = \frac{V_{f}^{2}}{2d} = \frac{(8.89 m/s)^{2}}{2*12.8 m} = 3.09 m/s^{2} [/tex]

Therefore, the magnitude of her acceleration is 3.09 m/s².              

b) The component of her acceleration that is parallel to the ground is given by:

[tex] a_{x} = a*cos(\theta) [/tex]

Where:

θ: is the angle respect to the ground = 32.6 °

[tex] a_{x} = 3.09 m/s^{2}*cos(32.6) = 2.60 m/s^{2} [/tex]

Hence, the component of her acceleration that is parallel to the ground is 2.60 m/s².

I hope it helps you!

A skateboarder, starting from rest, rolls down a 12.8-m ramp the magnitude of the skateboarder's acceleration is approximately 3.07 [tex]m/s^2[/tex], the component of her acceleration that is parallel to the ground is approximately 1.66 [tex]m/s^2[/tex].

(a) The following kinematic equation can be used to calculate the skateboarder's acceleration:

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

[tex](8.89)^2 = (0)^2 + 2a(12.8)[/tex]

78.72 = 25.6a

a = 78.72 / 25.6

a = 3.07 [tex]m/s^2[/tex]

(b) Trigonometry can be used to calculate the part of her acceleration that is parallel to the ground. We are aware that the ramp's angle with the ground is 32.6°.

[tex]a_{parallel }= a * sin(\theta)[/tex]

Plugging in the values:

[tex]a_{parallel[/tex] = 3.07  [tex]m/s^2[/tex]* sin(32.6°)

[tex]a_{parallel[/tex]≈ 1.66  [tex]m/s^2[/tex]

Therefore, the component of her acceleration that is parallel to the ground is approximately 1.66  [tex]m/s^2[/tex].

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Explanation:

A) Draw free body diagrams of both blocks.

Force P is pushing right on block A, which will cause it to move right along the incline.  Therefore, friction forces will oppose the motion and point to the left.

There are 5 forces acting on block A:

Applied force P pushing to the right,

Normal force N pushing up and left 10° from the vertical,

Friction force Nμ pushing down and left 10° from the horizontal,

Reaction force Fab pushing down,

and friction force Fab μ pushing left.

There are 2 forces acting on block B:

Reaction force Fab pushing up,

And elastic force kx pushing down.

(There are also horizontal forces on B, but I am ignoring them.)

Sum of forces on A in the x direction:

∑F = ma

P − N sin 10° − Nμ cos 10° − Fab μ = 0

Solve for N:

P − Fab μ = N sin 10° + Nμ cos 10°

P − Fab μ = N (sin 10° + μ cos 10°)

N = (P − Fab μ) / (sin 10° + μ cos 10°)

Sum of forces on A in the y direction:

N cos 10° − Nμ sin 10° − Fab = 0

Solve for N:

N cos 10° − Nμ sin 10° = Fab

N (cos 10° − μ sin 10°) = Fab

N = Fab / (cos 10° − μ sin 10°)

Set the expressions equal:

(P − Fab μ) / (sin 10° + μ cos 10°) = Fab / (cos 10° − μ sin 10°)

Cross multiply:

(P − Fab μ) (cos 10° − μ sin 10°) = Fab (sin 10° + μ cos 10°)

Distribute and solve for Fab:

P (cos 10° − μ sin 10°) − Fab (μ cos 10° − μ² sin 10°) = Fab (sin 10° + μ cos 10°)

P (cos 10° − μ sin 10°) = Fab (sin 10° + 2μ cos 10° − μ² sin 10°)

Fab = P (cos 10° − μ sin 10°) / (sin 10° + 2μ cos 10° − μ² sin 10°)

Sum of forces on B in the y direction:

∑F = ma

Fab − kx = 0

kx = Fab

x = Fab / k

x = P (cos 10° − μ sin 10°) / (k (sin 10° + 2μ cos 10° − μ² sin 10°))

Plug in values and solve.

x = 500 N (cos 10° − 0.4 sin 10°) / (12000 (sin 10° + 0.8 cos 10° − 0.16 sin 10°))

x = 0.0408 m

x = 4.08 cm

B) Draw free body diagrams of both blocks.

Force P is pushing block A to the right relative to the ground C, so friction force points to the left.

Block A moves right relative to block B, so friction force on A will point left.  Block B moves left relative to block A, so friction force on B will point right (opposite and equal).

Block B moves up relative to the wall D, so friction force on B will point down.

There are 5 forces acting on block A:

Applied force P pushing to the right,

Normal force Fc pushing up,

Friction force Fc μ₁ pushing left,

Reaction force Fab pushing down and left 15° from the vertical,

and friction force Fab μ₂ pushing up and left 15° from the horizontal.

There are 5 forces acting on block B:

Weight force 750 n pushing down,

Normal force Fd pushing left,

Friction force Fd μ₁ pushing down,

Reaction force Fab pushing up and right 15° from the vertical,

and friction force Fab μ₂ pushing down and right 15° from the horizontal.

Sum of forces on B in the x direction:

∑F = ma

Fab μ₂ cos 15° + Fab sin 10° − Fd = 0

Fd = Fab μ₂ cos 15° + Fab sin 15°

Sum of forces on B in the y direction:

∑F = ma

-Fab μ₂ sin 15° + Fab cos 10° − 750 − Fd μ₁ = 0

Fd μ₁ = -Fab μ₂ sin 15° + Fab cos 15° − 750

Substitute:

(Fab μ₂ cos 15° + Fab sin 15°) μ₁ = -Fab μ₂ sin 15° + Fab cos 15° − 750

Fab μ₁ μ₂ cos 15° + Fab μ₁ sin 15° = -Fab μ₂ sin 15° + Fab cos 15° − 750

Fab (μ₁ μ₂ cos 15° + μ₁ sin 15° + μ₂ sin 15° − cos 15°) = -750

Fab = -750 / (μ₁ μ₂ cos 15° + μ₁ sin 15° + μ₂ sin 15° − cos 15°)

Sum of forces on A in the y direction:

∑F = ma

Fc + Fab μ₂ sin 15° − Fab cos 15° = 0

Fc = Fab cos 15° − Fab μ₂ sin 15°

Sum of forces on A in the x direction:

∑F = ma

P − Fab sin 15° − Fab μ₂ cos 15° − Fc μ₁ = 0

P = Fab sin 15° + Fab μ₂ cos 15° + Fc μ₁

Substitute:

P = Fab sin 15° + Fab μ₂ cos 15° + (Fab cos 15° − Fab μ₂ sin 15°) μ₁

P = Fab sin 15° + Fab μ₂ cos 15° + Fab μ₁ cos 15° − Fab μ₁ μ₂ sin 15°

P = Fab (sin 15° + (μ₁ + μ₂) cos 15° − μ₁ μ₂ sin 15°)

First, find Fab using the given values.

Fab = -750 / (0.25 × 0.5 cos 15° + 0.25 sin 15° + 0.5 sin 15° − cos 15°)

Fab = 1151.9 N

Now, find P.

P = 1151.9 N (sin 15° + (0.25 + 0.5) cos 15° − 0.25 × 0.5 sin 15°)

P = 1095.4 N

Answer:

Explanation:

Ex(z,t) = Eocos(kz - ω t + φ)

k = 2π/λ  , ω = 2π f

φ = +30° , E₀ = 10³ V .

z/λ = 0.25 , ft = 0.125

Ex(z,t) = Eocos(2πz/λ - 2πf t + φ)

Putting the values given above

Ex(z,t) = 10³ cos ( 2π / 4 - 2π x .125 + 30⁰ )

= 1000cos (90⁰ - 45+30)

= 1000 cos 75

=258.8  V .

Answer:

The thermal conductivity of the insulated metal rod is [tex]202.92\,\frac{W}{m\cdot K}[/tex].

Explanation:

This is a situation of one-dimensional thermal conduction of a metal rod in a temperature gradient. The heat transfer rate through the metal rod is calculated by this expression:

[tex]\dot Q = \frac{k_{rod}\cdot A_{c, rod}}{L_{rod}}\cdot \Delta T[/tex]

Where:

[tex]\dot Q[/tex] - Heat transfer due to conduction, measured in watts.

[tex]L_{rod}[/tex] - Length of the metal rod, measured in meters.

[tex]A_{c,rod}[/tex] - Cross section area of the metal rod, measured in meters.

[tex]k_{rod}[/tex] - Thermal conductivity, measured in [tex]\frac{W}{m\cdot K}[/tex].

Let assume that heat conducted to melt some ice was transfered at constant rate, so that definition of power can be translated as:

[tex]\dot Q = \frac{Q}{\Delta t}[/tex]

Where Q is the latent heat required to melt the ice, whose formula is:

[tex]Q = m_{ice}\cdot L_{f}[/tex]

Where:

[tex]m_{ice}[/tex] - Mass of ice, measured in kilograms.

[tex]L_{f}[/tex] - Latent heat of fussion, measured in joules per gram.

The latent heat of fussion of water is equal to [tex]330000\,\frac{J}{g}[/tex]. Hence, the total heat received by the ice is:

[tex]Q = (6.15\,g)\cdot \left(330\,\frac{J}{g} \right)[/tex]

[tex]Q = 2029.5\,J[/tex]

Now, the heat transfer rate is:

[tex]\dot Q = \frac{2029.5\,J}{(10\,min)\cdot \left(60\,\frac{s}{min} \right)}[/tex]

[tex]\dot Q = 3.382\,W[/tex]

Turning to the thermal conduction equation, thermal conductivity is cleared and computed after replacing remaining variables: ([tex]L_{rod} = 0.75\,m[/tex], [tex]A_{c,rod} = 1.25\times 10^{-4}\,m^{2}[/tex], [tex]\Delta T = 100\,K[/tex], [tex]\dot Q = 3.382\,W[/tex])

[tex]\dot Q = \frac{k_{rod}\cdot A_{c, rod}}{L_{rod}}\cdot \Delta T[/tex]

[tex]k_{rod} = \frac{\dot Q \cdot L_{rod}}{A_{c,rod}\cdot \Delta T}[/tex]

[tex]k_{rod} = \frac{(3.382\,W)\cdot (0.75\,m)}{(1.25\times 10^{-4}\,m^{2})\cdot (100\,K)}[/tex]

[tex]k_{rod} = 202.92\,\frac{W}{m\cdot K}[/tex]

The thermal conductivity of the insulated metal rod is [tex]202.92\,\frac{W}{m\cdot K}[/tex].

Answer:

Coefficient of friction.

Explanation:

The amount of friction divided by the weight of an object is equal to the coefficient of friction. It is a dimensional less number. It can be given by :

[tex]F=\mu N[/tex]

N is normal force.

[tex]\mu[/tex] = coefficient of friction

[tex]\mu=\dfrac{F}{N}[/tex]

Answer:

F = ILB

Explanation:

To find the net force on the conducting bar you take into account the following expression:

[tex]\vec{F}=I( \vec{L}X \vec{B})[/tex]

I: current in the conducting bar

L: length of the bar

B: magnitude of the magnetic field

In this case the direction of the magnetic field and the motion of the bar are perpendicular between them. The direction of the bar is + i, and the magnetic field poits upward + k. The cross product of these vector give us the direction of the net force:

+i X +k = +j

The direction of the force is to the right and its magnitude is F = ILB

Complete Question

The complete question is shown on the first uploaded image

Answer:

The distance is [tex]r_2 = 4 \ AU[/tex]

Explanation:

From the question we are told that

    The size of Jupiter is  [tex]s_2 = 143,000 \ km[/tex]

    The  length of the International Space Station is [tex]r_1 = 380\ km[/tex]

    The  size of the International Space Station is  [tex]s_1 = 90 \ m =0.09 \ km[/tex]

The angular size where the same one night and this angular size is mathematically represented as

      [tex]\theta = \frac{s}{r}[/tex]

Since  [tex]\theta[/tex] is constant

        [tex]\frac{s_1}{r_1} = \frac{s_2}{r_2}[/tex]

substituting values

      [tex]\frac{0.09}{380} = \frac{143000}{r_2}[/tex]

=>   [tex]r_2 = 6.04 * 10^{9} \ km[/tex]

Now we are told to convert to AU and  1 AU  [tex]= 1.5 * 10^8 \ km[/tex]

  So

      [tex]r_2 = \frac{6.04 * 10^8}{1.5*10^{8}}[/tex]

      [tex]r_2 = 4 \ AU[/tex]

Answer:

The correct answer is option (c) a success because it did not detect a shift in the interference pattern.

Explanation:

In Michelson and Morley experiment  it was considered to be successful.

They both found out that the experiment that was carried out was not a failure  since it did not detect any shift in the interference pattern.

With this findings it was widely regarded as correct and precise.

Answer:

it takes the shape and size of the container that it is in

Explanation:

Answer:

it takes the shape and size of a container

1)  100 ° C

2) 323 K

hope it helps youuuuuu

Answer:

32mi

Explanation:

If 1lb contains 3,500 Cal

It means the number of hours required to burn 3500cal would be;

3500/331 = 10.57hours

But a brisk walk is 3.0 mi/h,

It means a distance of 3.0 × 10.57 mi would be covered = 31.71 miles

32miles{ approximated to the nearest whole}

Note Distance = speed × time

Answer:

a) 0.147 N

b) 9.408 N

c) 9.261 N

Explanation:

The tension on the cord is the only force keeping the block in circular motion, thus representing the entirety of its centripetal force [tex]\frac{mv^{2} }{r}[/tex]. Plugging in values for initial and final states and we get answers for a and b. The work done by the person causes the centripetal force to increase, and thus is the difference between the final tension and the initial tension.

An object's resistance to any change in motion is the Inertia of the object.

Answer:

a)Car E = Car D  > (Car F = Car B = Car C) > Car A

b)Car E = Car D  > (Car F = Car B = Car C) > Car A

Explanation:

Car A: mass = 500 kg; speed = 10 m/s

Car B: mass = 2000 kg;speed = 5 m/s

Car C:mass = 500 kg; speed = 20 m/s

Car D: mass = 1000 kg; speed = 20 m/s

Car E:mass = 4000 kg; speed = 5 m/s

Car F: mass = 1000 kg; speed = 10 m/s

Part a) Now we know that momentum of each car is product of mass and velocity , so we will have

CarA:

[tex]P_1 = m \times v\\P_1 = (500)(10)\\P_1 = 5 \times 10^3 kg m/s[/tex]

Car B:

[tex]P_2 = m v\\P_2 = (2000)(5)\\P_2 = 10^4 kg m/s[/tex]

Car C:

[tex]P_3 = m v\\P_3 = (500)(20)\\P_3 = 10^4 kg m/s[/tex]

Car D:

[tex]P_4 = m v\\P_4 = (1000)(20)\\P_4 = 2\times 10^4 kg m/s[/tex]

Car E:

[tex]P_5 = m v\\P_5 = (4000)(5)\\P_5 = 2\times 10^4 kg m/s[/tex]

Car F:

[tex]P_6 = m v\\P_6 = (1000)(10)\\P_6 = 10^4 kg m/s[/tex]

So the momentum is given as ,

Car E = Car D  > (Car F = Car B = Car C) > Car A

Part b)Impulse is given as change in momentum so here we can say that final momentum of all the cars will be zero as they all stops and hence the impulse is same as initial momentum of the car

so the order of impulse from largest to least is given as

Car E = Car D  > (Car F = Car B = Car C) > Car A

Answer:

4.09 MeV

Explanation:

Find the given attachment

Answer:

If, instead, that rocket was on a planet with the same mass as Earth but half the diameter, the escape velocity would be 15.8 km/s Any object that is smaller than its Schwarzschild radius is a black hole – in other words, anything with an escape velocity greater than the speed of light is a black hole.

Explanation:

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Answer:

The velocity is  [tex]v_2= 0.45 \ m/s[/tex]

Explanation:

From the question we are told that

      The initial speed of the hot water is  [tex]v_1 = 0.85 \ m/s[/tex]

     The pressure from the heater  [tex]P_1 = 450 \ KPa = 450 *10^{3} \ Pa[/tex]

      The height of the hot water before flowing is  [tex]h_1 = 0 \ m[/tex]

      The height of bathtub above the heater is [tex]h_2 = 3.70 \ m[/tex]

       The pressure in the pipe is [tex]P_2 = 414 KPa = 414 *10^{3} \ Pa[/tex]

       The density of water is [tex]\rho = 1000 \ kg/m^3[/tex]

Apply Bernoulli equation

      [tex]P_1 + \rho gh_1 +\frac{1}{2} \rho v_1^2 = \rho g h_2 + \frac{1}{2}\rho v_2 ^2[/tex]

Substituting values

     [tex](450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) = (1000 * *9.8*3.70) + (0.5*1000*v_2^2 )[/tex]

=>   [tex]v_2^2 = \frac{ (450 *10^{3}) + (1000 * 9.8 *0 ) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}[/tex]

=>   [tex]v_2= \sqrt{ \frac{ (450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}}[/tex]

=>    [tex]v_2= 0.45 \ m/s[/tex]

Answer:

☟

Explanation:

Fuels made from oil mixtures containing large hydrocarbon molecules are not efficient as they do not flow easily and are difficult to ignite. Crude oil often contains too many large hydrocarbon molecules and not enough small hydrocarbon molecules to meet demand. This is where cracking comes in.

Cracking allows large hydrocarbon molecules to be broken down into smaller, more useful hydrocarbon molecules. Fractions containing large hydrocarbon molecules are heated to vaporise them. They are then either:

heated to 600-700°C

passed over a catalyst of silica or alumina

These processes break covalent bonds in the molecules, causing thermal decompositionreactions. Cracking produces smaller alkanesand alkenes (hydrocarbons that contain carbon-carbon double bonds). For example:

hexane → butane + ethene

C6H14 → C4H10 + C2H4

Some of the smaller hydrocarbons formed by cracking are used as fuels, and the alkenes are used to make polymers in plastics manufacture. Sometimes, hydrogen is also produced during cracking.

Fractional distillation of crude oil

Fractional distillation separates a mixture into a number of different parts, called fractions.

A tall fractionating column is fitted above the mixture, with several condensers coming off at different heights. The column is hot at the bottom and cool at the top. Substances with high boiling points condense at the bottom and substances with lower boiling points condense on the way to the top.

Crude oil is a mixture of hydrocarbons. The crude oil is evaporated and its vapours condense at different temperatures in the fractionating column. Each fraction contains hydrocarbon molecules with a similar number of carbon atoms and a similar range of boiling points.

Oil fractions

The diagram below summarises the main fractions from crude oil and their uses, and the trends in properties. Note that the gases leave at the top of the column, the liquids condense in the middle and the solids stay at the bottom.

As you go up the fractionating column, the hydrocarbons have:

lower boiling points

lower viscosity (they flow more easily)

higher flammability (they ignite more easily).

Other fossil fuels

Crude oil is not the only fossil fuel.

Natural gas mainly consists of methane. It is used in domestic boilers, cookers and Bunsen burners, as well as in some power stations.

Coal was formed from the remains of ancient forests. It can be burned in power stations. Coal is mainly carbon but it may also contain sulfur compounds, which produce sulfur dioxide when the coal is burned. This gas is a cause of acid rain. Also, as all fossil fuels contain carbon, the burning of any fossil fuel will contribute to global warming due to the production of carbon dioxide.

In fractional distillation, the crude oil is added to the chamber and heated. The components with the highest boiling point will condense in the lower part of the column and the components with the lower boiling point will condense at the top of the column. Petrol with a low boiling point is collected from the top of the column.

Fractional distillation can be described as the separation of a mixture into its component fractions. The chemical compound is separated by heating them to a temperature at which fractions of the mixture will vaporize.

Generally, the components have boiling points that differ by less than 25 °C  from each other under one atmosphere. When the mixture is heated, the component with the lower boiling point boils and changes to vapours.

The more volatile component remains in a vapour state and repeated distillations are used in the process, and the mixture is separated into component parts.

Therefore, the petrol from the crude oil can easily be separated as it has a boiling point of about 25-60°C.

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Answer:

f = 1.96 revolutions per minute

Explanation:

The formula for the the frequency of revolution of a satellite, to develop an artificial gravity, with the help of centripetal acceleration is given as follows:

f = (1/2π)√(ac/r)

where,

f = frequency of rotation = ?

ac = centripetal acceleration= apparent gravity or artificial gravity = 2.2 m/s²

r = radius of station or satellite = diameter/2 = 104 m/2 = 52 m

Therefore,

f = (1/2π)√[(2.2 m/s²)/(52 m)]

f = (0.032 rev/s)(60 s/min)

f = 1.96 revolutions per minute