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icang [17]
3 years ago
8

Which element does not normally form chemical bonds

Physics
1 answer:
NNADVOKAT [17]3 years ago
7 0

The "noble gases" do not normally form chemical bonds.

Those are helium, neon, argon, krypton, xenon, and radon.

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A 10.0 cm3 sample of copper has a mass of 89.6
Romashka-Z-Leto [24]
Density is mass divides by volume, so
89.6g / 10cm^3 =8.96g /cm^3

*cm^3 is a standard unit of volume*
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4 years ago
What is the time period and frequency of a sound wave if it completes 500 vibrations in 50 seconds
Degger [83]

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10 vibrations per second

Explanation:

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3 years ago
what distance is a book from the floor if the book contains 196 joules f potential energy and has a mass of 5 kg?
AleksAgata [21]
E=mgh.   196=5kg*9.81m/s^2*h.  So h=196/(5*9.81)=4m
5 0
4 years ago
Aang and Appa are flying over Ba Sing Se. They covered 500 m in 30 seconds while heading east. What was their velocity?
barxatty [35]

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Explanation:

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8 0
4 years ago
To meet a U.S. Postal Service requirement, employees' footwear must have a coefficient of static friction of 0.5 or more on a sp
liq [111]

Answer:

Minimum time interval (t2)=0.90 SECONDS

Explanation:

  • coefficient of friction for employees footwear = 0.5
  • coefficient of friction for typical athletic shoe = 0.810
  • frictional force = coefficient of friction X acceleration due to gravity X mass of body
  • Acceleration due to gravity is a constant = 9.81 m/s
  • Let frictional force for employee footwear = FF1
  • Let frictional force for athletic footwear =FF2

                 FF1 = O.5 X 9.81 X mass of body

                         = 4.905 x mass of body

                  FF2 = 0.810 X 9.81 X mass of body

                          = 7.9461 x mass of body

The body started from rest there by making the initial velocity zero ( u = 0)

From d= ut + 1/2 a x t^{2}

  •      d = \frac{1}{2} x a x t^{2}  .....................................i  

            where d= distance and it is given as 3.25m

  •          F =ma  ...................................ii

making acceleration subject of the formula from equation ii

  •              a =\frac{F}{m}

         Making t subject of formula from equation (i)

  • t=\sqrt{\frac{2d}{(f/m} }

    where

  • \frac{FF1}{Mass of body} = 4.905
  • \frac{FF2}{Mass of body} =7.9461

  Let

  •            t1 = minimum time taken for frictional force for employee foot wear
  •                                 t1 = \sqrt{\frac{6.5}{4.905} } =1.15 seconds

  •                                  t2 = \sqrt{\frac{6.5}{7.9461} } = 0.90 seconds

 

THANK YOU

5 0
3 years ago
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