Types of media that can influence one's social development include all of the following except

The successive ionization energies for an unknown element are listed below. To which family in the periodic table does the unkno

bagirrra123 [75]

<span>The jump from 1966 to 16347 is the largest one or simply we can say it is hard to lose the 3rd electron.Whereas, it is relatively easy to lose the first two electrons.

So there will be only 2 electrons in the outer most shell.

According to the information mentioned above we can conclude the </span><span>unknown element likely belongs to the second group.
</span><span>I2 = 1752 kj/mol</span>

4 0

3 years ago

Read 2 more answers

What is Newtons laws

pogonyaev

<h2>Law 1:</h2><h3>An object already in motion stays in motion, unless acted upon by a force.</h3><h3 /><h2>Law 2:</h2><h3> $f=ma$ </h3><h3>f = forces on an object</h3><h3>m = mass of that object</h3><h3>a = acceleration of that object</h3><h3 /><h2>Law 3:</h2><h3>Everything has an equal and opposite reaction.</h3><h3 /><h3>Hope this helps!</h3>

5 0

3 years ago

Write a rule for the sequence. 3, -3, -9, -15. A. Start with 3 and add -6 repeatedly B. Start with -6 and add 3 repeatedly C. St

faust18 [17]

Substract two consecutive terms of the sequence to see if there is a common difference:

$\begin{gathered} (-3)-(3)=-3-3=-6 \\ (-9)-(-3)=-9+3=-6 \\ (-15)-(-9)=-15+9=-6 \end{gathered}$

As we can see, there is a common difference of -6.

Then, if a number of the sequence is given, the next one can be found by adding -6 (which is the same as subtracting 6).

Notice that the first term of the sequence is 3.

Then, the rule for the sequence is to start with 3 and add -6 repeatedly.

Therefore, the correct choice is option A) Start with 3 and add -6 repeatedly.

7 0

1 year ago

Initially, a particle is moving at 5.25 m/s at an angle of 35.5Â° above the horizontal. Three seconds later, its velocity is 6.0

ivolga24 [154]

Answer:

a =( -0.32 i ^ - 2,697 j ^) m/s²

Explanation:

This problem is an exercise of movement in two dimensions, the best way to solve it is to decompose the terms and work each axis independently.

Break down the speeds in two moments

initial

v₀ₓ = v₀ cos θ

v₀ₓ = 5.25 cos 35.5

v₀ₓ = 4.27 m / s

$v_{oy}$ = v₀ sin θ

$v_{oy}$ = 5.25 sin35.5

$v_{oy}$ = 3.05 m / s

Final

vₓ = 6.03 cos (-56.7)

vₓ = 3.31 m / s

$v_{y}$ = v₀ sin θ

$v_{y}$ = 6.03 sin (-56.7)

$v_{y}$ = -5.04 m / s

Having the speeds and the time, we can use the definition of average acceleration that is the change of speed in the time order

a = ( $v_{f}$ - v₀) /t

aₓ = (3.31 -4.27)/3

aₓ = -0.32 m/s²

$a_{y}$ = (-5.04-3.05)/3

$a_{y}$ = -2.697 m/s²

6 0

4 years ago

An electron moves in a circular path perpendicular to a uniform magnetic field with a magnitude of 2.14 mT. If the speed of the

JulsSmile [24]

Answer:

(a) 3.9cm

(b) 1.66 x 10⁻⁸s

Explanation:

Since the electron is moving in a circular path, the centripetal acceleration needed to keep it from slipping off is provided by the magnetic force. This force (F), according to Newton's second law of motion is given by,

F = m x a --------------(i)

Where;

m = mass of the particle

a = acceleration of the mass

The centripetal acceleration is given by;

a = v² / r [v = linear velocity of particle, r = radius of circular path]

<em>Therefore, equation (i) becomes;</em>

F = m v²/ r --------------------(ii)

The magnitude of the magnetic force on a moving charge in a magnetic field as stated by Lorentz's law is given by;

F = qvBsinθ -------------(iii)

Where;

q = charge of the particle

v = velocity of the particle

B = magnetic field

θ = angle between the velocity and the magnetic field

<em>Combine equations (ii) and (iii) as follows;</em>

m (v² / r) = qvBsinθ [divide both side by v]

m v / r = qBsinθ [make r subject of the formula]

r = (m v) / (qBsinθ) ---------(iv)

(a) From the question;

v = 1.48 x 10⁷m/s

B = 2.14mT = 2.14 x 10⁻³T

θ = 90° [since the direction of velocity is perpendicular to magnetic field]

m = mass of electron = 9.11 x 10⁻³¹kg

q = charge of electron = 1.6 x 10⁻¹⁹C

Substitute these values into equation (iv) as follows;

r = (9.11 x 10⁻³¹ x 1.48 x 10⁷) / (1.6 x 10⁻¹⁹ x 2.14 x 10⁻³ sin 90°)

r = 3.9 x 10⁻²m

r = 3.9cm

Therefore, the radius of the circular path is 3.9cm

(b) The time interval required to complete one revolution is the period (T) of the motion of the electron and it is given by

T = d / v --------------(*)

Where;

d = distance traveled in the circular path in one complete turn = 2πr

v = velocity of the motion = 1.48 x 10⁷m/s

d = 2 π (3.9 x 10⁻²) [Take π = 22/7 = 3.142]

d = 2(3.142)(3.9 x 10⁻²) = 0.245m

Substitute the values of d and v into equation (*) as follows;

T = 0.245 / 1.48 x 10⁷

T = 0.166 x 10⁻⁷s

T = 1.66 x 10⁻⁸s

Therefore, the time interval is 1.66 x 10⁻⁸s

6 0

3 years ago