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4 years ago

A red ladybug appears in white light, in red light, and in blue light.

Physics

2 answers:

Andrej [43]4 years ago

7 0

A red ladybug appears in White light,Red in red light, and Black in blue light.

kirza4 [7]4 years ago

3 0

A red ladybug appears red in white light,red in red light, and black in blue light.

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Which of these forces determines the chemical properties of an atom? friction force nuclear force gravitational force electrical

zysi [14]

The answers is an electrical force.
Under normal conditions, atoms interact with each other via electrons that are furthest away from the nucleus. These electrons from the what is called the outer shell of the atom, electrons from the outer shell that can participate in chemical reactions are called valence electrons.

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3 years ago

High mass stars are more massive produce more energy burn brighter and have shorter life cycle than low mass stars

Zarrin [17]

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3 years ago

The angular width of localizers varies between ______ and ______ degrees in order to provide a signal width of approximately 700

natali 33 [55]

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4 years ago

What happens when two minerals have different arrangements of Atoms

zaharov [31]

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3 years ago

An electron is projected with an initial speed of 5 3.2 10 / ⋅ m s directly toward a proton that is fixed in place. If the elect

Dvinal [7]

Answer:

The distance is $d =1.66*10^{-9}m$

Explanation:

From the question we are told that

The initial speed of the electron is $v_i = 3.2 *10^5 m/s$

The mass of electron is $m = 9.11*10^{-31}kg$

Let $d$ be the distance between the electron and the proton when the speed of the electron instantaneously equal to twice the initial value

Let $KE_i$ be the initial kinetic energy of the electron \

Let $KE_d$ be the kinetic energy of the electron at the distance $d$ from the proton

Considering that energy is conserved,

The energy at the initial position of the electron = The energy at the final position of the electron

i.e

$KE_i +U_1 = KE_d + U_2$

$U_1 \ and \ U_2$ are the potential energy at the initial position of the electron and at distance d of the electron to the proton

Here $U_1 = 0$

So the equation becomes

$\frac{1}{2} mv_i^2 = \frac{1}{2} mv_d^2 + \frac{kq_1 q_2}{d}$

Here $q_1 \ and \ q_2$ are the charge on the electron and the proton and their are the same since a charge on an electron is equal to charge on a proton