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Lunna [17]
4 years ago
14

Why would elements found in the same group on the periodic table have similar reactivity’s

Physics
1 answer:
9966 [12]4 years ago
5 0

Answer:

D. They have the same number of valence electrons

Explanation:

In order to answer this question we need to understand the following.

  • Valence electrons refers to the outermost electron(s) in the outermost energy level of an atom.
  • Elements in the same group in the periodic table have the same number of valence electrons. For instance, group 2A elements have 2 valence electrons.
  • The reactivity of elements in the periodic table is based on the number of valence electrons and the ease of gaining or losing electrons.
  • Elements in the same group have similar valence electrons and therefore share similar chemical properties such as reactivity.
  • Hence, elements in the same group have same number of valence electron and will thus have similar reactivity.
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Let the mass of 2500 kg car be m_1 and it's velocity be v_1 and the mass of 1500 kg car be m_2 and it's velocity be v_2 .

After the bumping the mass be M and it's velocity be V.

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                    V = 14000/4000 = 7/2 = 3.5 m/s

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Pls help i give brainliest
san4es73 [151]

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a.

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cuz whenever the ball was traveling up the staircase it was building up energy to use (potential energy) unlike b.

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At a given instant, the force on an electron is in the +z-direction (out of the page), which the electron is moving in the +x-di
Hitman42 [59]

Answer:

The direction of the B-field is in the +y-direction.

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The corresponding formula is

F_B = qv\times B

This means, we should use right-hand rule.

Our index finger is pointed towards +x-direction (direction of velocity),

our middle finger should point towards the direction of the B-field,

and our thumb should point towards the +z-direction (direction of the force).

Since our middle finger in this situation points towards +y-direction, the B-field should be in +y-direction.

\^{x} \times \^{y} = \^{z}

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4 years ago
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Phantasy [73]

Answer:

The magnetic field will be \large{\dfrac{1.4 \times 10^{-4}}{d}} T, '2d' being the distance the wires.

Explanation:

From Biot-Savart's law, the magnetic field (\large{\overrightarrow{B}}) at a distance 'r' due to a current carrying conductor carrying current 'I' is given by

\large{\overrightarrow{B} = \dfrac{\mu_{0}I}{4 \pi}} \int \dfrac{\overrightarrow{dl} \times \hat{r}}{r^{2}}}

where '\overrightarrow{dl}' is an elemental length along the direction of the current flow through the conductor.

Using this law, the magnetic field due to straight current carrying conductor having current 'I', at a distance 'd' is given by

\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}

According to the figure if 'I_{t}' be the current carried by the top wire, 'I_{b}' be the current carried by the bottom wire and '2d' be the distance between them, then the direction of the magnetic field at 'P', which is midway between them, will be perpendicular towards the plane of the screen, shown by the \bigotimes symbol and that due to the bottom wire at 'P' will be perpendicular away from the plane of the screen, shown by \bigodot symbol.

Given \large{I_{t} = 19.5 A} and \large{I_{B} = 12.5 A}

Therefore, the magnetic field (\large{B_{t}}) at 'P' due to the top wire

B_{t} = \dfrac{\mu_{0}I_{t}}{2 \pi d}

and the magnetic field (\large{B_{b}}) at 'P' due to the bottom wire

B_{b} = \dfrac{\mu_{0}I_{b}}{2 \pi d}

Therefore taking the value of \mu_{0} = 4\pi \times 10^{-7} the net magnetic field (\large{B_{M}}) at the midway between the wires will be

\large{B_{M} = \dfrac{4 \pi \times 10^{-7}}{2 \pi d} (I_{t} - I_{b}) = \dfrac{2 \times 10^{-7}}{d} = \dfrac{41.4 \times 10 ^{-4}}{d}} T

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3 years ago
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