Are the two atoms shown below in the image from the same element?

The amplitude of a wave is the height of a wave as measured from the highest point on the wave________ to the lowest point on th

Brrunno [24]

The correct option is:

crest trough

In fact, the sentence can be completed as follows:

"The amplitude of a wave is the height of a wave as measured from the highest point on the wave crest to the lowest point on the wave trough."

3 0

2 years ago

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An electron is trapped in a one-dimensional infinite well of width 340 pm and is in its ground state. What are the (a) longest,

Nesterboy [21]

Answer:

(a) 1.2703×10⁻⁷ m

(b) 4.7636×10⁻⁸ m

(c) 2.5406×10⁻⁸ m

Explanation:

Given:

Width of the infinite well, L = 340 pm = 340×10⁻¹² m.

The formula for energy of the electron in nth state is:

$E_n=\frac {n^2\times h^2}{8mL^2}$

The expression for the difference in energy between the levels having quantum numbers n(initial) to n(final) is:

$\Delta E_n=\frac {({n_f}^2-{n_i}^2)\times h^2}{8mL^2}$

According to Planks theory:

E = hv

where, v is the frequency

Also,

Frequency×Wavelength = Speed of light

So,

$E=\frac {hc}{\lambda}$

$\lambda=\frac {hc}{E}$

Also, using energy from above formula as:

$\lambda=\frac {hc}{\frac {({n_f}^2-{n_i}^2)\times h^2}{8mL^2}}$

$\lambda=\frac {c\times {8mL^2}} {({n_f}^2-{n_i}^2)\times h}}$

For longest wavelength ni = 1 and nf = 2

m= mass of the electron = 9.1 ×10⁻³¹kg

c = 3×10⁸m/s

h = 6.625×10⁻³⁴J.sec

$\lambda_{Longest}=\frac {3\times 10^8\times {8\times 9.1\times 10^{-31}(340\times 10^{-12})^2}} {({2}^2-{1}^2)\times 6.625\times 10^{-34}}}$

<u>Longest wavelength = 1.2703×10⁻⁷ m</u>

For second longest wavelength ni = 1 and nf = 3

$\lambda_{Second\ Longest}=\frac {3\times 10^8\times {8\times 9.1\times 10^{-31}(340\times 10^{-12})^2}} {({3}^2-{1}^2)\times 6.625\times 10^{-34}}}$

<u>Second longest wavelength = 4.7636×10⁻⁸ m</u>

For third longest wavelength ni = 1 and nf = 4

$\lambda_{Third\ Longest}=\frac {3\times 10^8\times {8\times 9.1\times 10^{-31}(340\times 10^{-12})^2}} {({4}^2-{1}^2)\times 6.625\times 10^{-34}}}$

<u>Third longest wavelength = 2.5406×10⁻⁸ m</u>

3 0

3 years ago

A red car has a head-on collision with an approaching blue car with the same magnitude of momentum. A green car driving with the

WARRIOR [948]

Answer:

The head on Collison because you have both cars going (for the sake of it 30mph) and they both collide the energy from that is 60mph because the speed is combined with the two cars.

5 0

3 years ago

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7. Mrs. Everhart developed the theory that all bulldogs are great pets. After all her bulldog Meatball is a

Advocard [28]

Answer: No.

Explanation:

A scientific theory is an explanation of some aspect of the natural world that can be tested several times, and follows the scientific method. And, when possible, theories are tested with experiments.

Now, a scientific theory must meet some criteria, and one of those is that it must be supported by several (and independent) strands of evidence.

In this case, we have only one, Meatball, and it is not enough, then it can not be a scientific theory.

4 0

3 years ago

A wire is formed into a circle having a diameter of 10.0cm and is placed in a uniform magnetic field of 3.00mT . The wire carrie

Paul [167]

The range of potential energies of the wire-field system for different orientations of the circle are -

θ U

0° 375 π x $10^{-7}$

90° 0

180° - 375 π x $10^{-7}$

We have current carrying wire in a form of a circle placed in a uniform magnetic field.

We have to the range of potential energies of the wire-field system for different orientations of the circle.

<h3>What is the formula to calculate the Magnetic Potential Energy?</h3>

The formula to calculate the magnetic potential energy is -

U = M.B = MB cos $$\theta$

where -

M is the Dipole Moment.

B is the Magnetic Field Intensity.

According to the question, we have -

U = M.B = MB cos $$\theta$

We can write M = IA (I is current and A is cross sectional Area)

U = IAB cos $$\theta$

U = Iπ $r^{2}$ B cos $$\theta$

For $$\theta$ = 0° →

U(Max) = MB cos(0) = MB = Iπ $r^{2}$ B = 5 × π × $( 0.05 ) ^{2}$ × 3 × $10^{-3}$ =

375 π x $10^{-7}$ .

For $$\theta$ = 90° →

U = MB cos (90) = 0

For $$\theta$ = 180° →

U(Min) = MB cos(0) = - MB = - Iπ $r^{2}$ B = - 5 × π × $( 0.05 ) ^{2}$ × 3 × $10^{-3}$ =

- 375 π x $10^{-7}$ .

Hence, the range of potential energies of the wire-field system for different orientations of the circle are -

θ U

0° 375 π x $10^{-7}$

90° 0

180° - 375 π x $10^{-7}$

To solve more questions on Magnetic potential energy, visit the link below-

brainly.com/question/13708277

#SPJ4

3 0

1 year ago