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3 years ago

Bohr's model is sometimes called the ________ model because it resembles a mini-solar system.

Physics

2 answers:

elena-14-01-66 [18.8K]3 years ago

7 0

His model was also called the Planetary model

CaHeK987 [17]3 years ago

7 0

Answer:

Bohr's model is sometimes called the Rutherford–Bohr model because it resembles a mini solar system.

Explanation:

Early 1900s Rutherford's model shows that an atom is mostly empty space, with electrons orbiting a fixed, positively charged nucleus in set, predictable paths. This model of an atom was developed by Ernest Rutherford, a New Zealand native working at the University of Manchester in England.

The Bohr Model as presented by Niels Bohr is a planetary model in which the negatively charged electrons orbit a small, positively charged nucleus similar to the planets orbiting the sun (except that the orbits are not planar).

The highlighted similarity between Rutherford and Bohr theory made them similar and a collaboration was been executed by both parties to form Rutherford-Bohr model.

Thank you

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When you connect an unknown resistor across the terminals of a 1.50 V D-cell battery having negligible internal resistance, you

g100num [7]

Answer:

Explanation:

Using ohm's law

a) V = IR where V is voltage in Volt, I is current in Ampere and R is resistance in ohms

R = V / I = 1.50 V/ ( 2.05 /1000) A = 731.71 ohms

b) Power = IV = $\frac{V}{R}$ × v = $\frac{V^{2} }{R}$ = $\frac{9^{2} }{731.71}$ = 0.1107 W

c) E = IR + Ir = ( 731.71 × 0.0036) + ( 35 × 0.0036) = 2.76 V

d) Power use by the resistor = I²R = 0.0036² × 731.71 = 0.00948 W = 0.00948 W = 0.000009483 kw × ( 18 / 60 ) H = 2.84 × 10⁻⁶ KW-H

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3 years ago

In foot x-ray, what is the distance/FFD and Why?

MakcuM [25]

Answer and Explanation:

FFD is the distance between the film on which the image is obtained and the center of the anode tube. The magnification and resolution of the image depends on the FFd By varying the FFD we can change the magnification and resolution of the image. The standard FFD is about 100 centimeters.

New studies have found that by changing the FFD to 130 cm the radiation dosage reduces while the image quality remains practically the same.

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3 years ago

Calculate the potential at the center of curvature of the arc if the potential is assumed to be zero at infinity.

yan [13]

the potential at the center of curvature of the arc = v = Q ∕ (4πε∘a) or $\frac{kQ}{a}$

ATQ,

We have density of charge,

λ = $\frac{Q}{L}$

Where L is the rod's length, in this case the semicircle's length L = πr

Q is the charge on the rod

The potential created at the center by an differential element of charge is:

$k = \frac{dQ}{r}$

where k is the coulomb's constant

r is the distance from dQ to center of the circle

v = ∫ $\frac{K dQ}{a}$ , Where a = radius, k = 1 / 4πε∘

v = $\frac{kQ}{a}$ or Q ∕ (4πε∘a)

To learn more about potential from the given link

brainly.com/question/25923373

#SPJ4

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2 years ago

A diffraction grating with 140 slits per centimeter is used to measure the wavelengths emitted by hydrogen gas. At what angles i

Degger [83]

Answer:

0.003181 radians

0.003005 radians

Explanation:

Number of slits = 140 /cm

λ = Wavelength = 434 nm = 434×10⁻⁹ m

m = 3 Third order spectrum

Space between slits

$d=\frac{0.01}{140} =7.14\times 10^{-5}\ m$

Now,

$dsin\theta = m\lambda\\\Rightarrow \theta=sin^{-1}\left(\frac{m\lambda}{d}\right)\\\Rightarrow \theta=sin^{-1}\left(\frac{3\times 434\times 10^{-9}}{7.14\times 10^{-5}}\right)\\\Rightarrow \theta=sin^{-1}0.018228\\\Rightarrow \theta=0.01823^{\circ}=0.01823\times \frac{\pi}{180}=0.003181 radians$

0.003181 radians

When λ = 410 nm = 410×10⁻⁹ m

$dsin\theta = m\lambda\\\Rightarrow \theta=sin^{-1}\left(\frac{m\lambda}{d}\right)\\\Rightarrow \theta=sin^{-1}\left(\frac{3\times 410\times 10^{-9}}{7.14\times 10^{-5}}\right)\\\Rightarrow \theta=sin^{-1}0.01722\\\Rightarrow \theta=0.01722^{\circ}=0.01722\times \frac{\pi}{180}=0.003005 radians$