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3 years ago

From the illustration of the potassium atom, fill in the periodic table selection.

Physics

1 answer:

stich3 [128]3 years ago

3 0

Answer:

2 in shell 1; 8 in shell 2; 8 in shell 3 and 1 in shell 4 ; 19 total protons and electrons

Explanation: Potassium is in Group 1 Period 4 and will have 1 valence electron that is 4s1 electron configuration.

Atomic mass =

39.10 = 19 protons + 20 neutrons

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In midair an M = 145 kg bomb explodes into two pieces of m1 = 115 kg and another, respectively. Before the explosion, the bomb w

Daniel [21]

Answer:

$v_2=-133.17m/s$ , the minus meaning west.

Explanation:

We know that linear momentum must be conserved, so it will be the same before ( $p_i$ ) and after ( $p_f$ ) the explosion. We will take the east direction as positive.

Before the explosion we have $p_i=m_iv_i=Mv_i$ .

After the explosion we have pieces 1 and 2, so $p_f=m_1v_1+m_2v_2$ .

These equations must be vectorial but since we look at the instants before and after the explosions and the bomb fragments in only 2 pieces the problem can be simplified in one dimension with direction east-west.

Since we know momentum must be conserved we have:

$Mv_i=m_1v_1+m_2v_2$

Which means (since we want $v_2$ and $M=m_1+m_2$ ):

$v_2=\frac{Mv_i-m_1v_1}{m_2}=\frac{Mv_i-m_1v_1}{M-m_1}$

So for our values we have:

$v_2=\frac{(145kg)(24m/s)-(115kg)(65m/s)}{(145kg-115kg)}=-133.17m/s$

5 0

3 years ago

What is the amount of work done when a force of 10N moves a 20kg mass of 8 meters?

Nataly_w [17]

Yo sup??

we should know that

work done=force*displacement

W=F.s

F=10 N

s=8 m

therefore

W=10*8

=80 N

Hope this helps.

3 0

3 years ago

Which is greater, the gravitational force between earth and the moon, or the force between earth and the sun?

Karo-lina-s [1.5K]

Earth and the Sun
GF = 3.647x10^22 N

8 0

3 years ago

The total units by an objects as it changes position is called _____ ?

Alika [10]

Change in position of object = Displacment

7 0

3 years ago

Two sound waves have equal displacement amplitudes, but wave 1 has two-thirds the frequency of wave 2. What is the ratio of the

zlopas [31]

Answer:

$\dfrac{I_1}{I_2}=\dfrac{4}{9}$

Explanation:

c = Speed of wave

$\rho$ = Density of medium

A = Area

$\nu$ = Frequency

$\nu_1=\dfrac{2}{3}\nu_2$

Intensity of sound is given by

$I=\dfrac{1}{2}\rho c(A\omega)^2\\\Rightarrow I=\dfrac{1}{2}\rho c(A2\pi \nu)^2$

So,

$I\propto \nu^2$

We get

$\dfrac{I_1}{I_2}=\dfrac{\nu_1^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{\dfrac{2}{3}^2\nu_2^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{4}{9}$

The ratio is $\dfrac{I_1}{I_2}=\dfrac{4}{9}$

8 0

3 years ago