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3 years ago

• Most of the galaxies in the universe are moving away from

Physics

1 answer:

WARRIOR [948]3 years ago

4 0

The frequency of the light observed from the Earth is $4.945\cdot 10^{14}Hz$

Explanation:

First of all, we start by noticing that the galaxy is receding from Earth (moving away): this means that according to the Doppler effect, the frequency of the light as seen from the Earth must be shorter than the real frequency of the light emitted by the galaxy.

Furthermore, we can quantify the change in frequency of the light using the following equation:

$\frac{\Delta f}{f}=\frac{v}{c}$

where

$\Delta f$ is the change in frequency

f is the real frequency

v is the velocity of recession of the galaxy (negative if the galaxy is moving away)

c is the speed of light

In this problem, we have:

$f=5.00\cdot 10^{14} Hz$

$v=-3325 km/s = -3.325\cdot 10^6 m/s$

$c=3\cdot 10^8 m/s$

Substituting and solving for $\Delta f$ , we find

$\Delta f = \frac{v}{c}f=\frac{-3.325\cdot 10^6}{3\cdot 10^8}(5.00\cdot 10^{14})=-5.542\cdot 10^{12} Hz$

And therefore, the frequency of the light observed from the Earth is

$f'=f+\Delta f = 5.00\cdot 10^{14} +(-5.52\cdot 10^{12})=4.945\cdot 10^{14}Hz$

Learn more about frequency and waves:

brainly.com/question/5354733

brainly.com/question/9077368

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Answer:

1) The total charge of the top plate is 0.008 C

b) The total charge of the bottom plate is -0.008 C

2) The electric field at the point exactly midway between the plates is 0

3) The electric field between plates is approximately 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N

Explanation:

The given parameters of the parallel plate capacitor are;

The dimensions of the plates = 4 × 2 cm

The distance between the plates = 10 cm

The surface charge density of the top plate, σ₁ = 10 C/m²

The surface charge density of the bottom plate, σ₂ = -10 C/m²

The surface area, A = 0.04 m × 0.02 m = 0.0008 m²

1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C

b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C

2) The electrical field at the point exactly midway between the plates is given as follows;

$V_{tot} = V_{q1} + V_{q2}$

$V_q = \dfrac{k \cdot q}{r}$

Therefore, we have;

The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m

$V_{tot} = \dfrac{k \cdot q}{0.05} + \dfrac{k \cdot (-q)}{0.05} = \dfrac{k \cdot q}{0.05} - \dfrac{k \cdot q}{0.05} = 0$

The electric field at the point exactly midway between the plates, $V_{tot}$ = 0

3) The electric field, 'E', between plates is given as follows;

$E =\dfrac{\sigma }{\epsilon_0 } = \dfrac{10 \ C/m^2}{8.854 \times 10^{-12} \ C^2/(N\cdot m^2)} \approx 1.1294 \times 10^{12}\ N/C$

E ≈ 1.1294 × 10¹² N/C

The electric field between plates, E ≈ 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates

The charge on an electron, e = -1.6 × 10⁻¹⁹ C

The force on an electron in the middle of the two plates, $F_e$ = E × e

∴ $F_e$ = 1.1294 × 10¹² N/C × -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N

The force on an electron in the middle of the two plates, $F_e$ ≈ 1.807 × 10⁻⁷ N

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