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3 years ago

Scientists create models of atoms because atoms _____.

Physics

2 answers:

Marianna [84]3 years ago

5 0

Answer:

Are very small.

Explanation:

Scientists use models to study atoms because atoms are extremely small and can't even be seen.

marin [14]3 years ago

5 0

Answer:

Are very small.

Explanation:

The main reason scientists use models to study atoms is that atoms are extremely small and abstract.

Hope this helps :)

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3. What distance will a car, traveling 65 km/hr, cover in 3.0 hrs? (195 km)

qwelly [4]

Answer:

$\boxed{\sf Distance \ travelled \ by \ car = 195 \ km}$

Given:

Speed = 65 km/hr

Time = 3.0 hrs

To Find:

Distance

Explanation:

Distance = Speed × Time

= 65 × 3

= 195 km

7 0

4 years ago

Calculate the Fermi energy and the conductivity at room temperature for germanium containing 5×10^16 arsenic atoms per cubic cen

Cloud [144]

Answer:

The Fermi energy is 0.568 eV and the conductivity at room temperature is 31.24 (Ωcm)⁻¹

Explanation:

Data given:

Nd = doping concentration = 5x10⁶/cm³

According the mass action law, the hole concentration is:

$P_{0} =\frac{n_{i}^{2} }{n_{0} } ,n_{i}=2x10^{13} /cm^{3}$

$E_{f} =E_{i} +KTln(\frac{n_{D} }{n_{i} } ),K=8.61x10^{-19} ,KT=0.026eV$

$E_{i} =\frac{E_{0} }{2} =\frac{0.63}{2} =0.315eV\\E_{f}=0.315+0.203=0.568eV$

The conductivity of n-type of semi-conductor is equal to:

The mobility of Germanium is 3900 cm²/Vs

σ = 1.6x10⁻¹⁹ * 5x10¹⁶ * 3900 = 31.24 (Ωcm)⁻¹

6 0

3 years ago

What is the average speed between the times t = 4s and t = 12 s?

ziro4ka [17]

You need distance and time to find average speed.

3 0

3 years ago

A circular rod has a radius of curvature R = 9.09 cm and a uniformly distributed positive charge Q = 6.49 pC and subtends an ang

Digiron [165]

Answer:

E = 1.19 N/C

Explanation:

Let's first determine the length of the arc which can be given as:

L= Rθ

where:

L = length of the arc

R = radius of curvature

θ = angle in radius

L = (9.09×10⁻²m)(2.59)

L = (0.0909)(2.59)

L = 0.235431 m

Then, the magnitude of electric field that Q produces at the center of curvature can be calculated by using the formula:

$E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}-sin(-\frac{\theta}{2})]$

$E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}+sin(\frac{\theta}{2})]$

$E= \frac{2\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}]$

Since $\lambda = \frac{Q}{L}$

where;

L = length

Q = charge

λ = density of the charge;

then substituting $\frac{Q}{L}$ for λ, we have :

$E= \frac{2(\frac{Q}{L})}{4 \pi E_oR}[sin\frac{\theta}{2}]$

$E= \frac{2Q[sin\frac{\theta}{2}]}{4 \pi E_oLR}$

substituting our given parameter; we have:

$E= \frac{2(6.26*10^{-12}C)[sin\frac{2.59rad}{2}]}{4 \pi (8.85*10^{-12}C^2/N.m^2)(0.235431)(0.0909)}$

E = 1.1889 N/C

E = 1.19 N/C

∴ the magnitude of the electric field that Q produces at the center of curvature = 1.19 N/C

4 0

3 years ago

As an admirer of Thomas Young, you perform a double-slit experiment in his honor. You set your slits 1.01 mm apart and position

Snezhnost [94]

Answer:

2.316e-3 and 3.47e-3

Explanation:

Now, at that angle, we look at the bright spots on the screen.

tan θ = x / L (x is the horizontal distance from the centre of the screen, L is distance to screen)

For small angles, we can approximate that tan θ = sin θ.

nλ / d = x / L so then x = n λ L / d

First bright fringe, n = 1

x = (1) (641*10^-9 m) (3.65 m) / (1.01*10^-3 m) = 2.316e-3

For destructive interference (dark fringes), equation becomes:

x = (n - 0.5) λ L / d

Second dark fringe, n = 1.5

x = 1.5 (641*10^-9 m) (3.65 m) / (1.01*10^-3 m) = 3.47e-3

4 0

3 years ago