By what percentage would an earth-orbiting satellite average orbital radius need to be decreased in order to reduce its orbital period by a factor of one-half?

Answer:

Those substances which can conduct electricity are called conductors, while those substances which don't conduct electricity are called insulators.

Resistance is the obstruction provided by the material through which the current passes,so since conductors conduct electricity and insulators don't,so the obstruction i.e resistance provided by the conductor must be less,while insulators being unable to conduct electricity,has very high resistance.

Example of conductor is copper

Example of insulator is plastic

it seems that the rest of the question is missing. Could you please provide the full question and any relevant context so that I can better understand and provide an accurate answer. I can provide the definition of net force and how to calculate it.

The net force is the total force acting on an object, taking into account the magnitude and direction of all the individual forces acting on it. It is also referred to as the resultant force.

The net force on an object can be calculated using Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, this can be expressed as:

Net force = mass x acceleration

or

Fnet = ma

where Fnet is the net force, m is the mass of the object, and a is its acceleration.

To calculate the net force on an object, you first need to identify all the individual forces acting on it, including their magnitudes and directions. Then, you can use vector addition to find the net force, taking into account the direction of each force.

Hence, If the forces are acting in the same direction, you can simply add their magnitudes to find the net force. If they are acting in opposite directions, you subtract the smaller force from the larger force, and the direction of the net force is in the direction of the larger force. If the forces are acting at right angles to each other, you can use the Pythagorean theorem to find the magnitude of the net force, and trigonometry to determine its direction.

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Answer:

(2x + 4)(x - 4)=2x^2-4x-16

To calculate the electric force between two charges, we can use Coulomb's law:

F = k * |q1 * q2| / r^2

where F is the electric force, k is the electrostatic constant (k ≈ 8.99 × 10^9 N m^2/C^2), q1 and q2 are the magnitudes of the charges, and r is the distance between the charges.

In this case, q1 = 0.0042 C, q2 = -0.0050 C, and r = 0.0030 m.

Substituting these values into the equation:

F = (8.99 × 10^9 N m^2/C^2) * |0.0042 C * (-0.0050 C)| / (0.0030 m)^2

F = (8.99 × 10^9 N m^2/C^2) * 0.000021 C^2 / 0.000009 m^2

F ≈ 1977.78 N

Therefore, the electric force between the two charges is approximately 1977.78 Newtons.

Answer: Matter is how much space or opacity an object takes up. In short anything that take up space. Things like balls, trees, and even people are all made of matter. Mass is how much matter a object has. Air also has mass also, even though we can't see it. Things like cars, buildings, even planets have mass.

Explanation: Paragraphs are sometimes 4-6 sentences.

Answer:

31.84

Explanation:

To find the magnitude of the sum of the three vectors, first, we need to add the vectors component-wise. Then, we can find the magnitude of the resulting vector.

Add the vectors component-wise:

Vector A + Vector B + Vector C = [13+13+0, 8+0+1, 0+8+8] = [26, 9, 16]

Find the magnitude of the resulting vector:

The magnitude of a vector [x, y, z] is given by the formula:

magnitude = sqrt(x^2 + y^2 + z^2)

Plugging in the values from the resulting vector:

magnitude = sqrt(26^2 + 9^2 + 16^2)

magnitude = sqrt(676 + 81 + 256)

magnitude = sqrt(1013)

magnitude ≈ 31.84

The magnitude of the sum of the three vectors is approximately 31.84.

Hope this helps!

Answer:

- They need oxygen to burn fuel

- Aerodynamics

- Extreme temperatures

- Radiation

- Pressure issues

Explanation:

A airplane is a heavier-than-air aircraft kept aloft by the upward thrust exerted by the passing air on its fixed wings and driven by propellers, jet propulsion, etc.

Aeroplanes cannot travel in space for several reasons:

They need oxygen to burn fuel - Aeroplane engines rely on the oxygen in the atmosphere to burn fuel and generate thrust. In space, there is no atmosphere so there is no oxygen for the engines to work.

Aerodynamics - Aeroplane wings generate lift by interacting with the air. In space, there is no air so wings would be unable to generate any lift. Aeroplanes rely on aerodynamics to fly which does not work in space.

Extreme temperatures - In space, temperatures can range from -150 degrees Celsius to 150 degrees Celsius. Aeroplanes are designed to operate within a much narrower temperature range. The extreme cold and heat of space could damage aeroplane components.

Radiation - In space, there are high levels of radiation from the Sun and cosmic rays. Aeroplane bodies are not designed to shield against this type of radiation and it could damage electronics and affect aeroplane systems.

Pressure issues - Aeroplanes are designed to withstand air pressures at altitudes up to around 12 kilometers. In low-Earth orbit and beyond, the air pressure is essentially zero. This extreme change in pressure could cause structural damage to the aeroplane.

In summary, while aeroplanes are designed to fly through the Earth's atmosphere, they lack the key features needed to operate in the extreme environment of outer space like spaceships. Aeroplanes require things like oxygen, aerodynamics and being able to withstand changes in pressure - all of which do not exist or work the same way in space.

Explanation:

The wing is pushed up by the air under it. Large planes can only fly as high as about 7.5 miles. The air is too thin above that height. It would not hold the plane up.

Answer:

Tolerance: \(\pm 10\%\)

Explanation:

Resistor Color Codes

Resistor Color Coding uses colored bands to quickly identify the resistive value or resistors and its percentage of tolerance.

Since the question does not provide a specific color table, we'll use the table attached below.

The colors of the resistor shown in the question are:

First band: orange

Second band: blue

Third band: brown

Fourth band: silver

The colors relate to the following numbers respectively:

3, 6, 10Ω, \(\pm 1\%\)

The first two colors form the number 36

The third color is the multiplier: 36*10Ω = 360Ω

And the fourth color is the tolerance or the possible variation of the resistance \(\pm 1\%\)

Resistance: 360Ω

Tolerance: \(\pm 10\%\)

(a) Position vector expression ⇒ r = 16.4t i + ( 3.8t - 4.9t² ) j m

(b) Velocity vector expression ⇒ v = 16.4 i + ( 3.8 - 9.8t ) j m/s

(c) Acceleration vector expression ⇒ a = - 9.8 j m/s²

x-coordinate ⇒ x = 16.4t

y-coordinate ⇒ y = 3.8t - 4.9t²

(a) Position vector expression ⇒ r = 16.4t i + ( 3.8t - 4.9t² ) j m

(b) Velocity is the rate of change of position with respect to time. So differentiating the position vector expression,

v = dr / dt = d / dt [ 16.4t i + ( 3.8t - 4.9t² ) j ]

v = 16.4 i + ( 3.8 - 2 * 4.9 * t ) j

v = 16.4 i + ( 3.8 - 9.8t ) j m/s

(c) Acceleration is the rate of change of velocity with respect to time. So differentiating the velocity vector expression,

a = dv / dt = d / dt [ 16.4 i + ( 3.8 - 9.8t ) j ]

a= 0 i + ( 0 - 9.8 ) j

a = - 9.8 j m / s²

(d) t = 2.88 s

r ( 2.88 s ) =  16.4 ( 2.88 ) i + ( 3.8 * 2.88 - 4.9 ( 2.88 )² ) j

r ( 2.88 s ) = 47.23 i - 29.7 j m

v ( 2.88 s ) = 16.4 i + ( 3.8 - 9.8 * 2.88 ) j

v ( 2.88 s ) = 16.4 i - 24.42 j m / s

a ( 2.88 s ) =  - 9.8 j m / s²

Therefore, the required vectors are as follows

(a) Position vector expression ⇒ r = 16.4t i + ( 3.8t - 4.9t² ) j m

(b) Velocity vector expression ⇒ v = 16.4 i + ( 3.8 - 9.8t ) j m/s

(c) Acceleration vector expression ⇒ a = - 9.8 j m/s²

(d) r ( 2.88 s ) = 47.23 i - 29.7 j m

    v ( 2.88 s ) = 16.4 i - 24.42 j m / s

    a ( 2.88 s ) =  - 9.8 j m / s²

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Answer:

(i) To find the velocity of the crate just after losing contact with the roof, we can use the equation:

v^2 = u^2 + 2as

where v is the final velocity, u is the initial velocity (which is zero), a is the acceleration, and s is the displacement (which is 3 m).

Substituting the values, we get:

v^2 = 0 + 2 × 4.9 × 3

v^2 = 29.4

v = √29.4

v ≈ 5.42 m/s

Therefore, the velocity of the crate just after losing contact with the roof is approximately 5.42 m/s.

(ii) To find the velocity just before the crate hits the ground, we can use the conservation of energy principle. The initial potential energy of the crate at the top of the roof is converted into kinetic energy just before it hits the ground.

Initial potential energy = mgh

where m is the mass of the crate, g is the acceleration due to gravity, and h is the height of the roof.

Final kinetic energy = (1/2)mv^2

where v is the velocity just before the crate hits the ground.

Since there is no loss of energy, we can equate these two values:

mgh = (1/2)mv^2

Canceling out m on both sides and substituting the values, we get:

(1/2) × 1.17 × 9.81 × 3 = (1/2) × 1.17 × v^2

v^2 = 34.5

v = √34.5

v ≈ 5.87 m/s

Therefore, the velocity just before the crate hits the ground is approximately 5.87 m/s, and its direction is along the horizontal.

(iii) To find the time taken by the crate to hit the ground after leaving the roof, we can use the equation:

s = ut + (1/2)at^2

where s is the displacement (which is the height of the roof, 3 m), u is the initial velocity (which is zero), a is the acceleration, and t is the time taken.

Substituting the values, we get:

3 = 0 + (1/2) × 9.81 × t^2

t^2 = 0.612

t ≈ √0.612

t ≈ 0.78 s

Therefore, the time taken by the crate to hit the ground after leaving the roof is approximately 0.78 s.

(iv) Finally, to find the horizontal distance between the point directly below the roof and the landing point, we can use the equation:

s = ut + (1/2)at^2

where s is the horizontal distance, u is the initial velocity (which is zero), a is the acceleration (which is zero, since there is no force acting along the horizontal direction), and t is the time taken (which is 0.78 s).

Substituting the values, we get:

s = 0 + (1/2) × 0 × (0.78)^2

s = 0

Therefore, the horizontal distance between the point directly below the roof and the landing point is zero, which means the crate lands directly below the roof.

Explanation:

This is quite simple, we can solve this problem using kinematic equations of motion.

(i) The velocity of the crate just after losing contact with the roof:

We can use the kinematic equation, v^2 = u^2 + 2as, where u = 0 m/s (initial velocity), a = 4.9 m/s^2 (acceleration), and s = 3 m (distance).

v^2 = 0^2 + 2(4.9)(3) = 29.4

v = √29.4 = 5.42 m/s (velocity just after losing contact with the roof)

(ii) The velocity (magnitude and direction) just before it hits the ground:

We can use the kinematic equations of motion for motion in a straight line with constant acceleration.

First, let's find the time taken by the crate to reach the ground after leaving the roof. We can use the kinematic equation, s = ut + 1/2 at^2, where s = 3 m (vertical distance), u = 5.42 m/s (initial velocity), a = 9.8 m/s^2 (acceleration due to gravity), and t is the time taken to reach the ground.

3 = 5.42t - 1/2 (9.8)t^2

Simplifying this quadratic equation, we get t = 0.88 s (time taken to reach the ground after leaving the roof).

Now, let's find the velocity just before it hits the ground. We can use the kinematic equation, v = u + at, where u = 5.42 m/s (initial velocity), a = 9.8 m/s^2 (acceleration due to gravity), and t = 0.88 s (time taken to reach the ground after leaving the roof).

v = 5.42 + (9.8)(0.88) = 14.74 m/s (magnitude of the velocity just before hitting the ground)

The direction of the velocity just before hitting the ground is downwards, at an angle of 30 degrees below the horizontal (due to the slope of the roof).

(iii) The time the crate takes to hit the ground after leaving the roof:

We have already calculated this in part (ii), which is t = 0.88 s.

(iv) The horizontal distance between the point directly below the roof and the landing point:

We can use the formula, distance = velocity × time.

The horizontal velocity of the crate is constant and equal to 5.42 m/s, which is the same as the velocity just after losing contact with the roof (since there is no friction).

Therefore, the horizontal distance traveled by the crate before hitting the ground is:

distance = 5.42 m/s × 0.88 s = 4.77 m.

Answer:

D. 1.17 J

Explanation:

The potential energy stored in a spring is given by the following:

U = (K•x²)/2

U -> elastic Potential energy

K -> spring constant

x -> compression

So:

U = (53•0.21²)/2

U = 1.17

Answer1.17:

Explanation:

The magnitude of the displacement of the vector A is determined as 6.2 units.

The magnitude of the displacement of the vector is calculated as follows;

| A | = √[ (x₂ - x₁ )²  +  (y₂ - y₁ )² +  (z₂ - z₁ )²]

where;

The magnitude of vector A is calculated as;

| A | = √[ (2 - 0 )²  +  (3 - 0 )² +  (5 - 0 )²]

| A | = √ (4 + 9 + 25)

| A | = √ ( 38)

| A | = 6.2 units

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The complete question is below;

If vector A represents displacement from point O(x1, y1,z1) and P(x2, Yz, Z2). find the magnitude of the displacement of vector A.

The acceleration due to gravity at the surface of a celestial body can be calculated using the formula:

a = G * M / R^2

where G is the gravitational constant, M is the mass of the celestial body, and R is its radius.

Substituting the lunar and Martian fractional factors for mass over radius squared and multiplying the answers by 9.8, we have:

Acceleration due to gravity at the surface of the Moon:

a = G * Mmoon / Rmoon^2 = G * (0.0123ME) / (0.272RE)^2 = 1.62 m/s^2

Multiplying by 9.8, we have:

a = 1.62 * 9.8 = 15.88 m/s^2

Therefore, the acceleration due to gravity at the surface of the Moon is approximately 1.62 m/s^2, or 15.88 m/s^2 when multiplied by 9.8.

Acceleration due to gravity at the surface of Mars:

a = G * MMars / RMars^2 = G * (0.107ME) / (0.530RE)^2 = 3.71 m/s^2

Multiplying by 9.8, we have:

a = 3.71 * 9.8 = 36.41 m/s^2

Therefore, the acceleration due to gravity at the surface of Mars is approximately 3.71 m/s^2, or 36.41 m/s^2 when multiplied by 9.8.

Acceleration due to gravity is the acceleration gained by an object due to gravitational force. Its SI unit is m/s2. It has both magnitude and direction, hence, it’s a vector quantity. Acceleration due to gravity is represented by g. The standard value of g on the surface of the earth at sea level is 9.8 m/s2.

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Answer:

Distance from sun to Neptune = 4.495E9 km

Time for light to travel = 4.495E9 / 3E5 sec = 14,980 sec

That is from sun to Neptune time fof light = 250 min

Time for light to travel from sun to earth is about 8 min

So the time from Neptune would be 242 to 258 min depending on position of Neptune - Note that Neptune is about 30X as far from the sun as earth and

250 min / 8 min is roughly 30

The uniform motion of kinematics allows us to find the time it takes for light to arrive from Neptune to Earth, which varies between:

          t₁ = 1.45 10⁴ s and t₂₂= 1.55 10⁴ s

depending on the relative distance of the two planets

given parameters

to find

They  velocit of an electromagnetic wave is constant, so we can use the uniform motion relationships

             v = d / t

             t = d / v

where v is the speed of light, d the distance and y time, in this case the speed of the wave is the speed of light (v = c)

We look in the tables for the distances and the rotation periods around the sun

                           distance ( m)         period (s)

Sun Neptunium     4.50 10¹²             5.2 10⁹

Sun - Earth             1.5    10¹¹              3.2 10⁷

With the data of the period it is observed that the rotation of Neptune is much greater than that of Eart rotation around the sun, for which we will assume that Neptunium is fixed in space and the Earth may be in its aphelion or perihelion, maximum approach o away distance from the sun, consequently we calculate the time for the two cases:

Maximum approach

positions relative distance from the dos Plantetas is

         Δd = \(x_{Neptuno - Sum} - x_{Earth - Sum}\)d  

         Δd = 4.50 10¹²  - 1.5 10¹¹

         Δd = 43.5 10¹¹ m

the time it takes for Neptune's light to reach Earth is

        Δt = \(\frac{ 43.5 \ 10^{11} }{3 \ 10^8}\)  

        Δt = 14.5 10³ s

        Δt = 1.45 10⁴ s

We reduce to hours

        Δt = 1.45 10⁴ s (1 h / 3600 s) = 4.03 h

Maximum away

         Δd = \(x_{Neptune - Sum} + x_{Neptune-Sum}\)  

         Δd = 4.50 10¹² + 1.5 10¹¹

         Δd = 46.5 10¹¹

The time is

         Δt = \(\frac{46.5 \ 10^{11}}{ 3 \ 10^8}\)  

         Δt = 15.5 10³

         Δt = 1.55 10⁴ s

We reduce to hours

         Δt = 1.55 10⁴ s (1 h / 3600 s) = 4.31 h

In conclusion, the time it takes for light to arrive from Neptune to Earth varies between:

          t₁ = 1.45 10⁴ s and t₂ = 1.55 10⁴ s

depending on the relative distance of the two plants

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Answer:

The greater the rotation angle in a given amount of time, the greater the angular velocity. Angular velocity ω is analogous to linear velocity v. We can write the relationship between linear velocity and angular velocity in two different ways:

The total resistance of the circuit is 49 Ω. The current strength in the circuit, when connected to a voltage source of 56 V, is approximately 1.14 A.

To calculate the total resistance of the circuit, we need to determine the equivalent resistance of the resistors connected in a series.

Given:

R1 = 4 Ω

R2 = 30 Ω

R3 = 10 Ω

R4 = 5 Ω

Calculate the equivalent resistance (RT) of R1 and R2, as they are connected in series:

RT1-2 = R1 + R2

RT1-2 = 4 Ω + 30 Ω

RT1-2 = 34 Ω

Calculate the equivalent resistance (RTotal) of RT1-2 and R3, as they are connected in parallel:

1/RTotal = 1/RT1-2 + 1/R3

1/RTotal = 1/34 Ω + 1/10 Ω

1/RTotal = (10 + 34) / (34 * 10) Ω

1/RTotal = 44 / 340 Ω

1/RTotal ≈ 0.1294 Ω

RTotal ≈ 1 / 0.1294 Ω

RTotal ≈ 7.74 Ω

Calculate the equivalent resistance (RTotalCircuit) of RTotal and R4, as they are connected in series:

RTotalCircuit = RTotal + R4

RTotalCircuit = 7.74 Ω + 5 Ω

RTotalCircuit ≈ 12.74 Ω

Therefore, the total resistance of the circuit is approximately 12.74 Ω.

To determine the current strength (I) when connected to a voltage source of 56 V, we can use Ohm's Law:

I = V / RTotalCircuit

I = 56 V / 12.74 Ω

I ≈ 4.39 A

Therefore, the current strength in the circuit, when connected to a voltage source of 56 V, is approximately 4.39 A (or 1.14 A, considering significant figures).

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Answer:

20.4 meters

Explanation:

KE = 1/2 m v^2 = mgh = PE (convert all kinetic energy to potential energy, because at the highest point the coaster can climb, it will be stopped!)

=> h = (1/2 m v^2) / (mg) = v^2/(2*g) = (20 m/s)^2 / (2*9.8 m/s^2) = 20.4 m

The tallest height that the roller coaster can climb is 20 m.

A roller coaster is a device in which energy conversions from potential to kinetic energy takes place. Hence the roller coaster can easily be used to demonstrate the conversion of mechanical energy from one form to another.

Given that the velocity of the roller coaster at the bottom of the hill is  20.0 m/s, potential energy is converted into kinetic energy.

Hence;

mgh = 1/2mv^2

gh =1/2v^2

h = 1/2v^2/g

h = 0.5 × (20)^2/10

h = 20 m

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Explanation:

Track and Field is a sport, which is includes disciplines of running, jumping, and throwing events. The sport traces back to Ancient Greece. The first recorded examples of this sport were at the Ancient Greek Olympics. In Ancient Greece, only one event was contested, the stadion footrace. Later on, the game expanded to more events.Events of track and field are divided into three: track events, field events, and combined events. Track events consist of Sprints, middle-distance, long distance, hurdles and relays; Field events consist of jumps and throws; while combined events consist of pentathlon, heptathlon, and decathlon. Track and field is usually played outdoors in stadiums. The usual features of a track and field stadium are the outer running track, and the field within the track

Explanation:

u = 0m/s

v = 30m/s

a = g = 10m/s²

s = H = ?

using the formula,

v² = u² + 2as

30² = 0² + 2(10)(H)

900 = 20H

H = 900/20

H = 45m

Answer:

the answer is

45m - height

The statement" The direction of the force on the charge placed in the electric field is upward" is false because the direction of the force on a negative charge (-6 C) placed in an upward-directed uniform electric field of magnitude 24 N/C would be downward.

The direction of the force on a charged particle placed in an electric field is determined by the charge of the particle and the direction of the electric field. In this case, a charge of -6 C is placed in an electric field directed upward with a magnitude of 24 N/C.

The force on a charged particle in an electric field can be calculated using the formula:

F = q * E

Where F is the force, q is the charge of the particle, and E is the electric field.

Since the charge q in this case is negative (-6 C) and the electric field E is directed upward, we can substitute the values into the formula:

F = (-6 C) * (24 N/C)

F = -144 N

The negative sign in the force value indicates that the force is in the opposite direction to the electric field. Therefore, the force on the charge placed in the electric field is downward, not upward.

The force on a negative charge is always opposite to the direction of the electric field. This is because negative charges experience an attractive force towards positive charges, and electric fields are directed from positive charges to negative charges.

Therefore, the statement "The direction of the force on the charge placed in the electric field is upward." is false.

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9ycy8c8t 7f fixfuozofuxt8lsrupsurpaurae6pUeoUe6eoUeFipzuroz6d0, 7d0z6e0z7e0zurpz6e0z

Explanation:

force newton N - m·kg·s-2

pressure, stress pascal Pa N/m2 m-1·kg·s-2

energy, work, quantity of heat joule J N·m m2·kg·s-2

power, radiant flux watt W J/s m2·kg·s-3

volume cubic meter m3

Answer: Hot and room temperature surface.

Explanation: from the picture the hot temperature surface has easy and loose intermolecular activity, there are more spaces between the molecules to escape. So, it is for the water molecules to evaporate faster which is directly propositional to the temperature. Simply when a water is boiled at 100 degrees that when the heat helps the molecules to evaporate fast through the steam. But when it is in the room temperature the molecules are slightly tightly packed to each other, that is the reason why molecules can't evaporate as fast as compared to the hot temperature. It mostly depends on the intermolecular bond and kinetic energy that helps the molecules to escape into the atmosphere.

Answer:82. Since you have a distance and a force, then the easiest principle to use is energy, i.e. work.

The work done by friction is F * d. This work cancels out the kinetic energy of the stone (1/2)mv^2

Fd = (1/2)mv^2

F = (1/2)mv^2/d.

Plug in m = 20 kg, v = 3 m/sec, d = 40 m.

83. With more mass, the kinetic energy is higher now. The work needed is higher. W = F * d and F is the same.

Explanation:Hope I helped :)

Answer:

i think its the top one

Explanation:

pls tell me if im wrong  

Explanation:

the gas will be hit by high energy particles as the radioactive decay happens and the decaying nuclei are breaking apart. these particles energize the gas and cause electrons to move around the orbitals. the flow of electrons is detected by the sensor.

Answer:

The same as its translational KE.

The easy way to do this is to make up numbers and use them.

So, I'll say m=2 and r=3. I will also say v=3 .

Rot. Inertia of a hoop is mr^2. So the rot KE is: 1/2 (mr^2)(w^2)

note: (1/2*I*w^2)

Translational kinetic energy is basically normal KE, so 1/2(m)(v^2)

Now, lets plug our made up values in:

Rot Ke : 1/2 (9*2)(3/3) *note w = v/r

Tran Ke: 1/2(2)(9)

Rot Ke: 9

Tran Ke: 9

9=9, same.