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3 years ago

Which of the following models shows how unpredictable electrons are?

Physics

2 answers:

777dan777 [17]3 years ago

8 0

Answer:

on edge

Explanation:

Fantom [35]3 years ago

7 0

The exact location of electrons in electron shells of atoms cannot be exactly ascertained. This is why VSPER atomic models represent the position of electrons (s, p, d, & f) using the probability of where they would most be expected to be found.

Explanation:

This is because merely observing electrons changes their behavior. Remember that to observe something one has to shine light on it so it bounces back to the eye. Due to the negligible mass of electrons, mere photons of light will change their direction of movement, spin or other behaviors/properties.

Learn More:

For more on electron clouds check out;

brainly.com/question/12199882

brainly.com/question/11686000

#LearnWithBrainly

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the system shown above is released from rest. if friction is negligible, the acceleration of the 4.0 kg block sliding on the tab

JulsSmile [24]

The acceleration of the first block (4 kg) is -9.8 m/s².

The given parameters:

Mass of the first block, m₁ = 4.0 kg
Mass of the second block, m₂ = 2.0 kg

The net force on the system of the two blocks is calculated as follows;

$m_2 g - T = m_1 a$

where;

T is the tension in the connecting string due weight of the first block

$m_2 g - m_1 g = m_1 a\\\\a = \frac{m_2 g - m_1g}{m_1} \\\\a = \frac{g(m_2 - m_1)}{m_1} \\\\a = \frac{9.8(2-4)}{2} \\\\a = -9.8 \ m/s^2$

Thus, the acceleration of the first block (4 kg) is -9.8 m/s².

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2 years ago

A current of 0.92 a flows in a wire. how many electrons are flowing past any point in the wire per second? the charge on one ele

Fantom [35]

The current is defined as the ratio between the charge Q flowing through a certain point of a wire and the time interval, $\Delta t$ :
$I= \frac{Q}{\Delta t}$
First we need to find the net charge flowing at a certain point of the wire in one second, $\Delta t=1.0 s$ . Using I=0.92 A and re-arranging the previous equation, we find
$Q=I \Delta t= (0.92 A)(1.0 s)=0.92 C$

Now we know that each electron carries a charge of $e=1.6 \cdot 10^{-19} C$ , so if we divide the charge Q flowing in the wire by the charge of one electron, we find the number of electron flowing in one second:
$N= \frac{Q}{q} = \frac{0.92 C}{1.6 \cdot 10^{-19} C}=5.75 \cdot 10^{18}$

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3 years ago

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Dovator [93]

The answer is B, because it will lose potential energy.

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2 years ago

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kvasek [131]

Answer:

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Explanation: