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viva [34]
3 years ago
8

How are elements arranged in the modern periodic table?

Physics
2 answers:
LekaFEV [45]3 years ago
4 0

Elements that are on the periodic table are arranged by by atomic number, electron shells, and valence electrons. These elements are also arranged by families and groups. Families and groups go vertical on periodic table and they have the same number of e,evtrons In the outer shell and same number of valence electrons.

photoshop1234 [79]3 years ago
3 0
It is currently organised by the increasing atomic # based on the actual nuclear charge of the elements
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40 POINTS EASY
11111nata11111 [884]
We know, Mechanical Energy = K.E. + P.E.
As ball is at ground, P.E. would be zero. But as it is in motion, it must have some K.E. and that is:

K.E. = 1/2 mv²
K.E. = 1/2 * 1 * 2²
K.E. = 4/2
K.E. = 2 J

In short, Your Answer would be Option B

Hope this helps!
7 0
3 years ago
What is one use for infrared waves? A radar B Medical imaging C Thermal imaging cameras
Vilka [71]

Answer:

Im answering for free points sry

Explanation:

...

7 0
2 years ago
A merry-go-round on a playground consists of a horizontal solid disk with a weight
bezimeni [28]

Answer:α = 1.00 rad/s², τ = 90.0 N•m, KEr = 1.12 kJ

Explanation:

m = 795/9.81 = 81.0 kg

ω₁ = 47.79 rev/min(2π rad/rev) / (60 s/min) = 5.00 rad/s

α = (ω₁ - ω₀)/τ = (5.00 - 0.00)/5.00 = 1.00 rad/s²

I = ½mR² = ½(81.0)(1.49²) = 90.0 kg•m²

τ = Iα = 90.0(1.00) = 90.0 N•m

KEr = ½Iω² = ½(90.0)5.00² = 1,124.477 ≈ 1.12 kJ

5 0
3 years ago
What function does the shape of the star-nosed mole’s nose serve?
horsena [70]
Hello!

The shape of the star-nosed mole's nose helps it to take away the excess heat.

Letter b)

Hugs!
6 0
3 years ago
"Two identical positive charges exert a repulsive force of 6.7 × 10−9 N when separated by a distance 3.5 × 10−10 m. Calculate th
Tamiku [17]

Answer:

The charge of each charge is 3.02\times10^{-19} C

Explanation:

When yo have two charged particles they interact exerting an electrostatic force in the other particles, the magnitude of the electrostatic force between two particles is:

F_{e}=k\frac{\mid q_{1}q_{2}\mid}{r^{2}} (1)

with q1 and q2 the charges, r the distance between them and k the Coulomb's constant (k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}})

Because the charges we're dealing are identical positive q1=q2, then (1) is:

F_{e}=k\frac{\mid q^2 \mid}{r^{2}}

Using the values the problem give us:

6.7\times10^{-9}=8.98755\times10^{9}\frac{\mid q^2 \mid}{(3.5\times10^{-10})^{2}}

solving for q:

q=\sqrt{\frac{(3.5\times10^{-10})^{2}6.7\times10^{-9}}{8.98755\times10^{9}}}

q= 3.02\times10^{-19} C

8 0
3 years ago
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