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Makovka662 [10]
3 years ago
13

When all group iia elements lose their valence electrons, the remaining electron configurations are the same as for what family

of elements? family 1a, the alkali metals family 2a, the alkaline metals family 7a, the halogens family 8, the noble gases
Physics
2 answers:
Orlov [11]3 years ago
8 0
Noble Gases

I hope I helped

ΩΩΩΩΩΩΩΩΩΩ

ArbitrLikvidat [17]3 years ago
7 0
Im going to say, the answer is the halogens family 8. not completely sure<span />
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Water is being boiled in an open kettle that has a 0.52-cm-thick circular aluminum bottom with a radius of 12.0 cm. If the water
tangare [24]

Answer:

T_b=107.3784\ ^{\circ}C

Explanation:

Given:

  • thickness of the base of the kettle, dx=0.52\ cm=5.2\times 10^{-3}\ m
  • radius of the base of the kettle, r=0.12\ m
  • temperature of the top surface of the kettle base, T_t=100^{\circ}C
  • rate of heat transfer through the kettle to boil water, \dot Q=0.409\ kg.min^{-1}
  • We have the latent heat vaporization of water, L=2260\times 10^3\ J.kg^{-1}
  • and thermal conductivity of aluminium, k=240\ W.m^{-1}.K^{-1}

<u>So, the heat rate:</u>

\dot Q=\frac{0.409\times 2260000}{60}

\dot Q=15405.67\ W

<u>From the Fourier's law of conduction we have:</u>

\dot Q=k.A.\frac{dT}{dx}

\dot Q=k\times \pi.r^2\times \frac{T_b-T_t}{5.2\times 10^{-3}}

where:

A= area of the surface through which conduction occurs

T_b= temperature of the bottom surface

15405.67=240\times \pi\times 0.12^2\times \frac{T_b-100}{5.2\times 10^{-3}}

T_b=107.3784\ ^{\circ}C is the temperature of the bottom of the base surface of the kettle.

6 0
3 years ago
Which sequence shows electromagnetic waves arranged in a decreasing order of their wavelengths?
Hunter-Best [27]

Answer: Gamma rays, x-rays, ultraviolet rays, visible light, and infrared rays.

8 0
3 years ago
How many meters are in 45 centimeters?
adell [148]

Explanation:

100 CM = 1 m

45 CM = 45 / 100 = 0.45 m

hope it helps:)

6 0
3 years ago
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A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end.
Svet_ta [14]

Answer: a) 0.315 (V/L)

Explanation:

From Conservation of angular momentum, we know that

L1 = L2 ,

Therefore MV L/2 = ( Irod + Ib) x W

M/4 x V x L/2 = (M (L/2)^2 + 1/3xMxL^2) x W

M/8 X VL = (ML^2/16 + ML^2 /3 )

After elimination we have,

V/8 = 19/48 x L x W

W = 48/8 x V/19L = 6/19 x V/L

Therefore W = (0.136)X V/L

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3 years ago
A current is established in a gas discharge tube when a sufficiently high potential difference is applied across the two electro
pishuonlain [190]

Explanation:

It is given that the number of electrons passing through the cross-sectional area in 1 s is 3.4 \times 10^{18}. Also, we know that charge on an electron is -1.60 \times 10^{-19} C, then negative charge crossing to the left per second is  as follows.

         I- = 3.4 \times 10^{18} electrons \times -1.6 x 10^{-19} C/electrons

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As it is given that the number of protons crossing per second is 1.4 \times 10^{18}, as the charge on the proton is +1.60 \times 10^{-19} C, then positive charge crossing to the right per second is calculated as follows.

          I+ = 1.4 \times 10^{18} electrons \times 1.6 \times 10^{-19} electrons/C

            I+ = 0.224 A

          I = l I+ l + l I- l

So,    I = 0.544 + 0.224

            = 0.768 A

Thus, we can conclude that the current in given hydrogen discharge tube is 0.768 A.

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3 years ago
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