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Allushta [10]
3 years ago
14

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
2 years ago
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