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3 years ago

Compare and contrast rutherford's "planetary model" of the atom with our current understanding of an atom's internal structure.

1 answer:

5 0

<span>Rutherford felt that the atom had an internal structure much like our solar system, with a nucleus surrounded by orbiting electrons. Today, we know that the electrons inhabit specific shells and sub-shells, and there is an order to which these electrons are located.</span>

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A student lifts a physical science book off the table and above their head. Is there work being done?

DiKsa [7]

Answer: A force must cause a displacement in order for work to be done. A book falls off a table and free falls to the ground. Yes. This is an example of work.

Explanation:

7 0

2 years ago

explain how the composition of a star that will form a billon years in the future will differ the composition of our sun.

Grace [21]

Answer:

Explanation:

Basically the star slowly burns its hydrogen into Helium. Depending on the mass, the star will have a turbulent core where the Helium will be fully mixed or a radiative core where the helium will settle at the centre (remember it's heavier than Hydrogen). The second case is what happens in the Sun.

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2 years ago

2. Light waves of the wavelength of 650 nm and 500 nm produce interference fringes on a screen at a distance of 1m from a double

satela [25.4K]

Answer:

least distance= 13mm

ratio of the lattice = 1 : 0.71 : 0.58

Explanation:

given λ₁ = 650nm = 650×10⁻⁹m, λ₂ = 500nm = 500×10⁻⁹m

5 0

3 years ago

) A steel guitar string with a diameter of 1.00 mm is stretched between supports 80.0 cm apart. The temperature is 0.0°C. (a) Fi

ladessa [460]

Answer: a. Mass per unit length =0.0245kg/m

b. Tension =2.45x10^-8N

C. Tension = 2.45 x10^-8N

Fundamental frequency =200Hz

Explanation:

7 0

3 years ago

A 5.0 A electric current passes through an aluminum wire of 4.0~\times~10^{-6}~m^2 cross-sectional area. Aluminum has one free e

Serhud [2]

Answer: The electron number density (the number of electrons per unit volume) in the wire is $6.0 \times 10^{28} m^{-3}$ .

Explanation:

Given: Current = 5.0 A

Area = $4.0 \times 10^{-6} m^{2}$

Density = 2.7 $g/cm^{3}$ , Molar mass = 27 g

The electron density is calculated as follows.

$n = \frac{density}{mass per atom}\\= \frac{\rho}{\frac{M}{N_{A}}}\\$

where,

$\rho$ = density

M = molar mass

$N_{A}$ = Avogadro's number

Substitute the values into above formula as follows.

$n = \frac{\rho \times N_{A}}{M}\\= \frac{2.7 g/cm^{3} \times 6.02 \times 10^{23}/mol}{27 g/mol}\\= \frac{16.254 \times 10^{23}}{27} cm^{3}\\= 0.602 \times 10^{23} \times \frac{10^{6} cm^{3}}{1 m^{3}}\\= 6.0 \times 10^{28} m^{-3}$

Thus, we can conclude that the electron number density (the number of electrons per unit volume) in the wire is $6.0 \times 10^{28} m^{-3}$ .

8 0

2 years ago