A deductive argument is always a stronger argument than an inductive argument
O True
False
A Moving to another question will save this

Elements are arranged on the periodic table in terms of valence electrons based on their atomic number and electron configuration.

1. Elements are arranged on the periodic table in terms of valence electrons based on their atomic number and electron configuration. The valence electrons are the outermost electrons in an atom's electron shell, and they are crucial in determining the chemical properties and reactivity of elements.

2. Evidence from data tables can be shown by examining the electron configuration and the group and period numbers of various elements on the periodic table. Here is a simplified example:

Element   | Electron Configuration | Group | Period |

--------------------------------------------

Hydrogen  | 1s^1                              | 1     | 1      |

Lithium       | [He] 2s^1                     | 1     | 2      |

Carbon       | [He] 2s^2 2p^2          | 14    | 2      |

Oxygen      | [He] 2s^2 2p^4          | 16    | 2      |

Neon          | [He] 2s^2 2p^6          | 18    | 2      |

--------------------------------------------

3. The evidence from the data table supports the claim that the arrangement of elements on the periodic table is based on valence electrons.

- Group Trend: Elements within the same group (vertical columns) share the same number of valence electrons. In the example table, Hydrogen, Lithium, and Neon are all in Group 1, indicating they have 1 valence electron.

- Period Trend: Elements within the same period (horizontal rows) have the same number of electron shells. In the example table, Hydrogen and Lithium are in Period 1, indicating they have their valence electron in the first energy level. Carbon, Oxygen, and Neon are in Period 2, indicating they have their valence electrons in the second energy level.

By examining the electron configurations, group numbers, and period numbers, we can clearly see the trends and patterns in the number of valence electrons for both groups and periods on the periodic table. This evidence supports the claim that the arrangement of elements on the periodic table is based on their valence electrons, which play a crucial role in determining their chemical behavior and properties.

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Answer:

Models,Mathematics

am not sure pliz mark brainliest

Explanation:

Answer:

a = 2.22 [m/s^2]

Explanation:

First we have to convert from kilometers per hour to meters per second

[tex]40 [\frac{km}{h}]*[\frac{1h}{3600s}]*[\frac{1000m}{1km}] = 11.11 [m/s][/tex]

We have to use the following kinematics equation:

[tex]v_{f}= v_{i}+(a*t)[/tex]

where:

Vf = final velocity = 11.11 [m/s]

Vi = initial velocity = 0

a = acceleration [m/s^2]

t = time = 5 [s]

The initial speed is taken as zero, as the car starts from zero.

11.11 = 0 + (a*5)

a = 2.22 [m/s^2]

Refer to the attachment.

Answer:

un café y a seguir pensando porque no se la respuesta xd

Answer:

social impact, focused business strategy, economic viability

Explanation:

Just like the journalism industry, the factors of sustainability are ever-evolving. Community Wealth Partners recommends that nonprofit organizations consider how they are performing across five key drivers of sustainability which is  social impact, focused business strategy, economic viability

9514 1404 393

Answer:

  A and B started at the same time

Explanation:

Speed is the slope of the position vs. time curve. Here, both curves have positive slope (positive speed) at t=0, so both A and B started at the same time.

Answer:

9.8 seconds

Explanation:

Let's start by changing the speeds of both animals into meters per second:

107 km/h = 107000 m/3600s = 29.72 m/s

71.8 km/h = 71800 m/3600s = 19.94 m/s

The distance covered equation for the cheetah is:

Dc = 29.72 t

The distance covered equation for the gazelle considering as origin the cheetah's starting location, is:

Dg = 95.6 + 19.94 t

We make both equations equal at the time (t) the cheetah catches with the gazelle:

Dc =Dg

29.72 t = 95.6 + 19.94 t

solve for "t":

29.72 t - 19.94 t = 95.6

9.78 t = 95.6

t = 95.6 / 9.78

t = 9.775 seconds

We can approximate it to 9.8 seconds

Answer:

R_cm = 4.66 10⁶ m

Explanation:

The important concept of mass center defined by

         R_cm = 1 / M   ∑  x_i m_i

where M is the total mass, x_i and m_i are the position and masses of each body

Let's apply this expression to our case.

Let's set a reference frame where the axis points from the center of the Earth to the Moon,

       R_cm = 1 / M (m_earth 0 + m_moon d)

the total mass is

      M = m_earth + m_moon

the distance from the Earth is zero because all mass can be considered to be at its gravimetric center

let's calculate

      M = 5.98 10²⁴ + 7.35 10²²

      M = 6.0535 10₂⁴24 kg

we substitute

      R_cm = 1 / 6.0535 10²⁴ (0 + 7.35 10²² 3.84 )

      R_cm = 4.66 10⁶ m

Answer:

ΔE = 7.559 eV ,     λ = 1,645 10⁻⁷ m

Explanation:

For this exercise we can use the Bohr model for ionized atom with only one free electron,

         r_n = n² a₀ / Z

         E_n = -13,606 Z² / n²

Where a₀ is the Bohr radius of the hydrogen atom (a₀ = 0.0529 nm), Z is the atomic number of the atom under study and 13.606 eV is the energy of the ground state of Hydrogen.

In our case the Helium atom has two protons Z = 2

let's calculate the quantum number and the energy of each orbit

r_n = 0.30 nm

          n₁ = √ (r_n Z / a₀)

          n₁ = √ (0.30 2 / 0.0529)

Note that we do not have to reduce the radius since they are all in nanometers

          n₁ = 3.3

since n is an integer we approximate it to

         n₁ = 3

r_n = 0.20 nm

          n₂ = √ (0.2 2 / 0.0529)

          n₂ = 2.7

To approximate this value we must assume that there could be some error in the medicinal radio,

          n₂ = 2

having the quantum numbers of the two radius we can calculate their energy

        E₃ = - 13,606 2²/3²

        E₃ = - 6.047 eV

        E₂ = -13.606 2²/2²

         E₂ = -13.606 eV

the energy of the emitted photon is

          ΔE = E₃ - E₂

          ΔE = -6.047 + 13.606

          ΔE = 7.559 eV

You do not indicate in the exercise if you want the energy or the wavelength of the photon,

to find the wavelength We use the Planck relation

          E = h f

          c = λ f

          E = h c /λ

          λ = h c / E

we must reduce the energy to the SI system

          E = 7.559 ev (1.6 10⁻¹⁹ J / 1eV) = 12.09 10⁻¹⁹ J

          λ = 6.63 10⁻³⁴ 3 10⁸ / 12.09 10⁻¹⁹

          λ = 1,645 10⁻⁷ m

Answer: Well you didn't give any answers so my guess would be it depends how much force you put into it and where you throw it.

Explanation:

Answer:

I think its false...........

To determine the resultant electric field strength at point X, we need to calculate the individual electric field strengths created by each point charge and then add them as vectors.

The electric field strength created by a point charge is given by the equation:

E = k * |q| / r^2

where E is the electric field strength, k is the electrostatic constant (k ≈ 8.99 × 10^9 N m^2/C^2), q is the magnitude of the charge, and r is the distance from the charge.

For charge 1 (q1 = +2.0) at point X, the distance is 40 units:

E1 = (8.99 × 10^9 N m^2/C^2) * |2.0 C| / (40 m)^2

E1 = 11243750 N/C

For charge 2 (q2 = -3.0) at point X, the distance is also 40 units:

E2 = (8.99 × 10^9 N m^2/C^2) * |3.0 C| / (40 m)^2

E2 = 16865625 N/C

To determine the resultant electric field at point X, we need to add the two electric field vectors:

EResultant = E1 + E2

EResultant = 11243750 N/C + 16865625 N/C

EResultant = 28109375 N/C

Therefore, the resultant electric field strength at point X is 28109375 N/C.

To determine the electric force on a point charge (q3 = 0.50) placed at point X, we can use the equation:

F = q3 * EResultant

F = (0.50 C) * (28109375 N/C)

F = 14054687.5 N

Therefore, the electric force acting on the point charge at point X is 14054687.5 Newtons.

Answer:

1. The magnitude of the vector is 26.9 m

2. The magnitude of direction of the vector is 21.8°.

Explanation:

From the question given, we obtained the following data:

Displacement along x (Bₓ) = 25 m

Displacement along y (Bᵧ) = 10 m

Magnitude of displacement (B) =?

Magnitude of direction (θ) =.?

1. Determination of the magnitude of displacement of the vector (B)

Displacement along x (Bₓ) = 25 m

Displacement along y (Bᵧ) = 10 m

Magnitude of displacement (B) =?

B² = Bₓ² + Bᵧ²

B² = 25² + 10²

B² = 625 + 100

B² = 725

Take the square root of both side

B = √725

B = 26.9 m

Therefore, the magnitude of the vector is 26.9 m

2. Determination of the magnitude of the direction of the vector.

Displacement along x (Bₓ) = 25 m

Displacement along y (Bᵧ) = 10 m

Magnitude of direction (θ) =.?

The magnitude of direction of the vector can be obtained by using Tan ratio as illustrated below:

Adjacent = Bₓ = 25 m

Opposite = Bᵧ = 10 m

Magnitude of direction (θ) =.?

Tan θ = Opposite /Adjacent

Tan θ = Bᵧ / Bₓ

Tan θ = 10 /25

Tan θ = 0.4

Take the inverse of Tan

θ = Tan¯¹ (0.4)

θ = 21.8°

Therefore, the magnitude of direction of the vector is 21.8°.

Explanation:

This problem is solved by determining how long it takes the arrow to fall to the ground in exactly the same manner as if the arrow was dropped with no horizontal velocity. The time required for the arrow to fall to the ground will be the same time that the arrow flies horizontally at 60.0 m/s, so

d=1/2at^2

solved for t=√2d/a

t=√(2)(2.00m)/9.80m/s^2 =0.639s

The time required for the fall is multiplied by the horizontal velocity to get the horizontal distance.

d(horizontal)=v(horizontal)×time=60.0m×0.639s =38.3m

The arrow will fly 37.8 meters in the horizontal direction before it strikes the ground.

Projectile motion is a form of motion experienced by an object or particle that is projected near Earth's surface and moves along a curved path under the action of gravity only.

Given is an arrow fired perfectly along the horizontal direction with velocity of 60 m/s. The arrow was 2 m above the ground when it was released.

initial horizontal velocity = u[H] = 60 m/s

initial vertical velocity = u[V] = 0 m/s

Horizontal acceleration = a[H] = 0 m/s²

Vertical acceleration = a[V] = 9.8 m/s²

Vertical distance = S[V] = 2 m

Using second equation of motion for vertical direction to calculate time taken to reach the ground.

S[V] = u[V] × t + 1/2 a[V] × t²

S[V] =  1/2 × a[V] × t²

2 = 1/2 x 9.8 x  t²

4/9.8 =  t²

t² = 0.4081

t = 0.63 seconds

Horizontal distance covered = S[H] = u[H] x t

S[H] = 60 X 0.63 = 37.8 meters

Therefore, the arrow will fly 37.8 meters in the horizontal direction before it strikes the ground.

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Answer:velocity =-.65 speed=.65

Explanation:

i got the answer right on kahn academy

Answer:

Option (a)

Explanation:

Given that,

Mass of a car, m = 1200 kg

Force exerted by the engine, F = 600 N

Noe force,F = ma

a is the acceleration of the engine

[tex]a=\dfrac{F}{m}\\\\a=\dfrac{600\ N}{1200\ kg}\\\\a=0.5\ m/s^2[/tex]

So, the acceleration of the car is 0.5 m/s².

Answer:

0.5 m/s^2

Hope it works!

Explanation:

Answer:

a)    #_pastes = N / 8 ,  b) 5 tablespoons oil

Explanation:

a) For this exercise we can use the inverse proportions rule. So if each pasta has more volume in the same box size, tame pasta will fit.

Let's find the volume of the new pasta

         V = 2L 2W 2H

         V = 8 L W H

          V = 8 V₀

where V₀ is the volume of the usual pasta and we assume that there is N quantity in the box. therefore for the new pasta

          #_pastas = N V₀ / 8V₀

          #_pastes = N / 8

unfortunately you do not give the number of usual pastes to finish the calculation.

b) Let's analyze the situation, if the size of the box is the same, I initially need 5 tablespoons for charring and then I have the same box I will need the same amount of oil since the number of pastes is less

Answer:

4.7 N

Explanation:

Use ratios.  3.0 kg out of 4.5 kg total means 2/3 of the applied force = 4.7 N

Answer:

The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

Explanation:

Given that,

Mass of block = 1.0 kg

Dependent force = F(x)

Frictional force = F(f)

Suppose, the following information would students need to test the hypothesis,

(A) The function F(x) for 0 < x < 5 and the value of F(f).

(B) The function a(t) for the time interval of travel and the value of F(f).

(C) The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(D) The function a(t) for the time interval of travel, the time it takes the block to move 5 m, and the value of F(f).

(E) The block's initial velocity, the time it takes the block to move 5 m, and the value of F(f).

We know that,

The work done by a force is given by,

[tex]W=\int_{x_{0}}^{x_{f}}{F(x)\ dx}[/tex].....(I)

Where, [tex]F(x)[/tex] = net force

We know, the net force is the sum of forces.

So, [tex]\sum{F}=ma[/tex]

According to question,

We have two forces F(x) and F(f)

So, the sum of these forces are

[tex]F(x)+(-F(f))=ma[/tex]

Here, frictional force is negative because F(f) acts against the F(x)

Now put the value in equation (I)

[tex]W=\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}[/tex]

We need to find the value of [tex]\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}[/tex]

Using newton's second law

[tex]\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{ma\ dx}[/tex]...(II)

We know that,

Acceleration is rate of change of velocity.

[tex]a=\dfrac{dv}{dt}[/tex]

Put the value of a in equation (II)

[tex]\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{m\dfrac{dv}{dt}dx}[/tex]

[tex]\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{v_{0}}^{v_{f}}{mv\ dv}[/tex]

[tex]\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}[/tex]

Now, the work done by the net force on the block is,

[tex]W=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}[/tex]

The work done by the net force on the block is equal to the change in kinetic energy of the block.

Hence, The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

Answer:

Kepler's three laws of planetary motion can be described as follows: ... Kepler's first law is rather simple - all planets orbit the sun in a path that resembles ... Newton was able to combine the law of universal gravitation with circular motion ... The net centripetal force acting upon this orbiting planet is given by the relationship ...

People also ask

Explanation:

Answer:

The answer is #4: there is a period of time when the deep space network won't be gathering data.

Explanation:

I think this is the answer. I hope it helps.

Explanation:

i hope i have been useful buddy.

good luck ♥️♥️♥️.

Given parameters:

Distance 1 = 319 miles  

Time taken 1 = 6 hours

Distance 2 = 221 miles

Time taken 2 = 3 hours

Unknown:

Average speed  = ?

To find the average speed we need find the total distance traveled and divided by the total time taken for the travel.

Speed is a scalar quantity that expresses the distance covered per unit of time.

Mathematically;

   Average speed  = [tex]\frac{Total distance }{Total time taken }[/tex]

                                = [tex]\frac{distance 1 + distance 2}{time taken 1 + time taken 2}[/tex]

                               = [tex]\frac{319 + 221}{6 + 3}[/tex]

                               = 60miles/hr

The average speed of the bus is 60miles/hr

Answer:

You need to be patient with them. Also, a positive attitude is the single most important quality for anyone who works with children with special needs. I’ve seen highly trained specialists unable to interact with Louie because of their negative attitude and assumptions

Answer:

The vacuum of space

Heat management problems

The difficulty of re-entry

Orbital mechanics

Micrometeorites and space debris

Cosmic and solar radiation

The logistics of having restroom facilities in a weightless

Explanation: