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ExtremeBDS [4]
3 years ago
5

Suppose you discovered a new element with 120 protons and 2 electrons in its outer lever. In what group does this new element be

long
Physics
2 answers:
Mars2501 [29]3 years ago
7 0

Answer:

This new element belongs to the group 2.

Explanation:

In the periodic table, the vertical columns are called groups (or families). The elements in the same group share similar chemical and physical properties.

One of these properties is the same amount of outer valence electrons.

For example, in the group 1 the elements have 1 outer valence electron (If we assume that the element is in a neutral state which is without electrical charges).

If this new element has 2 electrons in its outer level it will belong to the group 2. The group 2 is well-known as alkaline earth metals.

For example, the magnesium (Mg) and the calcium (Ca) belong to the group 2.

Lyrx [107]3 years ago
5 0
Assuming the 2 valence electrons are of the neutral state of the new element, the element would belong in Group 2, the Alkaline Earth Metals.
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In a long, straight, vertical lightning stroke, electrons move downward and positive ions move upward and constitute a current o
uranmaximum [27]

The number of revolutions the electron completes in 60.0-μs of the strike is 134.

A magnetic field, a vector field that describes the magnetic influence on moving electric charges, electric currents, and magnetic materials. When a charge moves through a magnetic field, a force that is perpendicular to both its own velocity and the magnetic field operates on it.

Electrons go downward and positive ions move upward in a long, straight, vertical lightning stroke, creating a current of magnitude I = 20.0 kA.

A free electron travels through the air at a speed of v = 300 m/s at a place r = 50.0 m east of the stroke's center.

Let the magnetic field be B, and F be the magnetic force.

Counterclockwise horizontal arcs of field lines are produced by the upward lightning current.

We have, B = 8 × 10⁻⁵ T and;

The mass of an electron is, m = 9.11 × 10⁻³¹ kg

The time interval is Δt = 60 μs = 60 × 10⁻⁶

The angular frequency is given as:

ω = qB /m = 2πN / Δt

Where the number of revolutions is N.

So,

N = qBΔt /2πm

N = (l.60 × l0⁻¹⁹)(8 × l0⁻⁵)(60 × 10⁻⁶) / 2π(9.11 × 10⁻³¹ kg)

N = 134 revolutions

Learn more about current here:

brainly.com/question/1100341

#SPJ4

7 0
2 years ago
2. The components of vector A are given as follows:
Stella [2.4K]

Answer:

50 degree.

Explanation:

Given that the components of vector A are given as follows: Ax = 5.6 Ay = -4.7

The angle between vector A and B in the positive direction of x-axis will be achieved by using the formula:

Tan Ø = Ay/Ax

Substitute Ay and Ax into the formula above.

Tan Ø = -4.7 / 5.6

Tan Ø = -0.839

Ø = tan^-1(-0. 839)

Ø = - 40 degree

Therefore, the angle between vector A and B positive direction of x-axis will be

90 - 40 = 50 degree.

3 0
3 years ago
1. A kangaroo hops 84 m to the east in 7 seconds.
DENIUS [597]

Explanation:

Given parameters:

Distance hopped  = 84m

Displacement  = 84m due east

Time  = 7s

Unknown:

Speed of kangaroo  = ?

Velocity of kangaroo  = ?

Solution:

To solve this problem,

    Speed  = \frac{distance}{time }   = \frac{84}{7}  = 12m/s

  Velocity  = \frac{displacement}{time}   = \frac{84}{7}   = 12m/s due east

3 0
3 years ago
Starting from rest, a solid sphere rolls without slipping down an incline plane. At the bottom of the incline, what does the ang
Marrrta [24]

Answer:

2/R*sqrt (g*s*sin(θ)) = w

Explanation:

Assume:

- The cylinder with mass m

- The radius of cylinder R

- Distance traveled down the slope is s

- The angular speed at bottom of slope w

- The slope of the plane θ

- Frictionless surface.

Solution:

- Using energy principle at top and bottom of the slope. The exchange of gravitational potential energy at height h, and kinetic energy at the bottom of slope.

                                         ΔPE = ΔKE

- The change in gravitational potential energy is given as m*g*h.

- The kinetic energy of the cylinder at the bottom is given as rotational motion: 0.5*I*w^2

- Where I is the moment of inertia of the cylinder I = 0.5*m*R^2

We have:

                              m*g*s*sin(θ) = 0.25*m*R^2*w^2

                              2/R*sqrt (g*s*sin(θ)) = w

- The angular velocity depends on plane geometry θ , distance travelled down slope s, Radius of the cylinder R , and gravitational acceleration g

3 0
3 years ago
E<br> 3.6 What force is needed to give a mass of<br> 20 kg an acceleration of 5 m/s??
luda_lava [24]

Explanation:

  • Mass(m)= 20kg
  • Acceleration (a)= 5m/s²
  • Force(F)= ?

We know that,

  • F=ma
  • F=20×5
  • F=100N

Hence, the needed force is 100N.

6 0
2 years ago
Read 2 more answers
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