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kondaur [170]
3 years ago
5

Which element of variation would not be affected by adding the data value 15 to the data set {3, 5, 6, 8, 9, 10, 13, 14}?

Physics
1 answer:
Anna11 [10]3 years ago
4 0
The answer to this would be 15
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The half-life of a certain isotope is 15 minutes. How much of a 400 g sample will remain after 90 minutes?12.5 g6.25 g26.7 g66.7
Virty [35]

There is a total of 6 half lives that need to take place.


ONE HALF LIFE = 200

TWO HALF LIFES = 100

THREE HALF LIFES  = 50

FOUR HALF LIFES = 25

FIVE HALF LIFES = 12.5

SIX HALF LIFES = 6.25


The answer is 6.25g



4 0
3 years ago
Plzz answer this question correctly
VikaD [51]

Answer:

the acceleration of A is three times that of B

3 0
3 years ago
In general, a (an)<br> metal will be more reactive than an alkaline earth metal in the same period.
Dmitriy789 [7]

Alkali metal

Explanation:

In general, an alkali metal will be more reactive than an alkaline earth metal in the same period.

What determines reactivity of metals?

The electropositivity of metals determines how reactive they are.

  • Electropositivity or metallicity is a measure of the tendency of atoms of an element to lose electrons.
  • It is closely related to ionization energy and the electronegativity of an element.
  • The lower the ionization energy of an element, the more electropositive or metallic it is
  • Also, the more reactive it will be because, it can lose electrons more readily.
  • Across a period from left to right, electropositivity decreases and from top to down a group, it increases.
  • Since alkali metals are more electropositive than alkali earth metals, they are more reactive and readily lose their only electron.
  • This is why the most reactive metal is francium found in the lower left corner on the periodic table.
  • For non-metals, electronegativity is the most important factor.

Learn more:

Alkali metals brainly.com/question/6324347

Electronegativity brainly.com/question/11932624

#learnwithBrainly

5 0
3 years ago
A solid nonconducting sphere of radius R has a charge Q uniformly distributed throughout its volume. A Gaussian surface of radiu
valentina_108 [34]

Answer:

E(4 \pi r^{2})={\frac{ Q r^{3}}{R^{3}\epsilon _{0}}}

or

E(4 \pi)={\frac{ Q r}{R^{3}\epsilon _{0}}}

Explanation:

We know that Gauss's law states that the Flux enclosed by a Gaussian surface is given by

E.S=\frac{q}{\epsilon_{0}}

Here , E is electric field and S is surface are and q is charge enclosed by the surface and e is electrical permeability of the medium.

Here the Gaussian is of radius r<R so area of surface is

S=4 \pi r^{2}

Also, charge enclosed by the surface is

Charge =\frac{Total \: Charge }{Total \:Volume} \times Volume \: of \: Gaussian \: surface

therefore,

q=\frac{Q}{\frac{4}{3} \pi R^{3} }\frac{4}{3} \pi r^{3} =\frac{ Q r^{3}}{R^{3}}

Here Q is total charge,

Insert values in Gauss's law

E(4 \pi r^{2})=\frac{\frac{ Q r^{3}}{R^{3}}}{\epsilon _{0}}

Rearrange them

E(4 \pi r^{2})={\frac{ Q r^{3}}{R^{3}\epsilon _{0}}}

on further solving

E(4 \pi)={\frac{ Q r}{R^{3}\epsilon _{0}}}

This is the required form.

6 0
4 years ago
During a phase change the temperature of a substance
Nostrana [21]
The temperature<span> remains constant as long as the </span>phase change<span> remains incomplete.</span>
3 0
3 years ago
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