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3 years ago

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What do the atoms of elements in the same group have in common? A. They have the same atomic numbers. B. They have the same aver

age atomic masses. C. They have the same number of electron shells. D. They have the same number of electrons in their outermost shells.

2 answers:

ad-work [718]3 years ago

5 0

I think it's D . They have the same number of electrons in their outermost shells

garik1379 [7]3 years ago

5 0

Answer : The correct option is, (D) They have the same number of electrons in their outermost shells.

Explanation :

Periods : A row in the periodic table is known as periods.

The common features about the elements of a period are :

Each periods has elements with the same number of electron shells or energy levels or shell.

They transition from metal to noble gas that means they move from metal to non-metal and then non-metal to noble gas.

Groups : A column in the periodic table is known as groups.

The common features about the atoms of elements in a same group are :

Each groups has elements with the same number of valance electrons or have same of electrons in their outermost shells.

The elements in the same group have similar chemical properties.

The elements in the same group have similar physical properties.

Hence, the correct option is, (D) They have the same number of electrons in their outermost shells.

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What bo you nmean by ABC rule ?

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A squared plus b squared equals c squared

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4 years ago

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When an object of mass m1 is hung on a vertical spring and set into vertical simple harmonic motion, it oscillates at a frequenc

Arlecino [84]

Answer:

$m_2/m_1=9.745$

The ratio m_2/m_1 of the masses is 9.745

Explanation:

Formula for frequency when mass m_1 is hung on the spring:

$f_1=\frac{1}{2\pi}\sqrt{\frac{k}{m_1}}$

where:

k is the spring constant

Formula for frequency when mass m_2 is hung on the spring along with m_1:

$f_2=\frac{1}{2\pi}\sqrt{\frac{k}{m_1+m_2}}$

where:

k is the spring constant.

In order to find ratio m_2/m_1, Divide the above equations:

$\frac{f_1}{f_2} =\frac{ \frac{1}{2\pi}\sqrt{\frac{k}{m_1}}}{\frac{1}{2\pi}\sqrt{\frac{k}{m_1+m_2}}}$

On Solving the above equation:

$\frac{f_1}{f_2} =\frac{\sqrt{\frac{k}{m_1}}}{\sqrt{\frac{k}{m_1+m_2}}}\\(\frac{f_1}{f_2})^{2} =\frac{m_1+m_2}{m_1} \\(\frac{11.8}{3.60})^2= \frac{m_1+m_2}{m_1} \\10.745=\frac{m_1+m_2}{m_1}\\10.745m_1=m_1+m_2\\m_2=10.745m_1-1m_1\\m_2/m_1=9.745$

The ratio m_2/m_1 of the masses is 9.745

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4 years ago

Two blocks with masses M1 and M2 hang one under the other.

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Answer:

(a)T= M2 × g, (b)T= (M1 + M2)g, (c)T= M2 (a + g) and (d)T=(M1 + M2) (a + g)

Explanation:

M1 is hanged upper and M2 is lower at Rest.

(a) For M2

T2 = Weight of the Body M2= M2 × g

(b) T1 = Weight of the Body M2 + Weight of the Body M2

T1 = M1 g + M2 g = (M1 + M2)g

M1 is hanged upper and M2 is lower at accelerated upwards ( F = T - W)

(c) For M2

⇒T = M2a + M2g = M2 (a + g)

(d) For M1

T = (M1 + M2) a + (M1 + M2) g

⇒ T = (M1 + M2) (a + g)

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3 years ago

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Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its

ruslelena [56]

Answer:

a

The orbital speed is $v= 2.6*10^{3} m/s$

b

The escape velocity of the rocket is $v_e= 3.72 *10^3 m/s$

Explanation:

Generally angular velocity is mathematically represented as

$w = \frac{2 \pi}{T}$

Where T is the period which is given as 1.6 days = $1.6 *24 *60*60 = 138240 sec$

Substituting the value

$w = \frac{2 \pi}{138240}$

$= 4.54*10^ {-5} rad /sec$

At the point when the rocket is on a circular orbit

The gravitational force = centripetal force and this can be mathematically represented as

$\frac{GMm}{r^2} = mr w^2$

Where G is the universal gravitational constant with a value $G = 6.67*10^{-11}$

M is the mass of the earth with a constant value of $M = 5.98*10^{24}kg$

r is the distance between earth and circular orbit where the rocke is found

Making r the subject

$r = \sqrt[3]{\frac{GM}{w^2} }$

$= \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }$

$= 5.78 *10^7 m$

The orbital speed is represented mathematically as

$v=wr$

Substituting value

$v= (5.78*10^7)(4.54*10^{-5})$

$v= 2.6*10^{3} m/s$

The escape velocity is mathematically represented as

$v_e = \sqrt{\frac{2GM}{r} }$

Substituting values

$= \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }$

$v_e= 3.72 *10^3 m/s$

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4 years ago

If the outside air temperature increases during a flight at constant power and at a constant indicated altitude, the true airspe

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The true airspeed will increase and true altitude will increase.

<h3>What is true air speed?</h3>

True airspeed is the airspeed of an aircraft relative to undisturbed air.

It's the aircraft speed relative to the airmass in which it's flying.

<h3>How does outside air temperature affect air speed?</h3>

If the outside air temperature increases during a flight at constant power and at a constant indicated altitude, the true airspeed will increase and true altitude will increase.

Thus, the true airspeed will increase and true altitude will increase.

Learn more about true airspeed here: brainly.com/question/13257916

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2 years ago