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3 years ago

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A spacecraft flies away from the earth with a speed of 4.80×106m/s relative to the earth and then returns at the same speed. The

spacecraft carries an atomic clock that has been carefully synchronized with an identical clock that remains at rest on earth. The spacecraft returns to its starting point 365 days (1 year) later, as measured by the clock that remained on earth. What is the difference in elapsed time on the two clocks,measured in hour?

1 answer:

marta [7]3 years ago

3 0

Answer:

1.1215 h

Explanation:

Δt=Δt_o/√(1-u^2/c^2)

(Δt° is the proper time in the rest frame,

Δt is the time intent measured in the second frame of reference u is the speed of the second frame with respect to the rest frame c is the speed of light )

a) the proper time is the time on spacecraft (rest frame);

Δt = 365 days = 8760 h

Δt=Δt_o/√(1-u^2/c^2)= 8761.1215 h

the difference = 8761.1215 h — 8760 h = 1.1215 h

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Which method of popcorn popping transfers heat into the kernels without any direct

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A single-slit diffraction pattern is formed by monochromatic electromagnetic radiation from a distant source passing through a s

tatiyna

Answer:

a. λ = 647.2 nm

b. I₀ 9.36 x 10⁻⁵

Explanation:

Given:

β = 56.0 rad , θ = 3.09 ° , γ = 0.170 mm = 0.170 x 10⁻³ m

a.

The wavelength of the radiation can be find using

β = 2 π / γ * sin θ

λ = [ 2π * γ * sin θ ] / β

λ = [ 2π * 0.107 x 10⁻³m * sin (3.09°) ] / 56.0 rad

λ = 647.14 x 10⁻⁹ m ⇒ λ = 647.2 nm

b.

The intensity of the central maximum I₀

I = I₀ (4 / β² ) * sin ( β / 2)²

I = I₀ (4 / 56.0²) * [ sin (56.0 /2) ]²

I = I₀ 9.36 x 10⁻⁵

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3 years ago

A ray of light traveling through air strikes a piece of diamond at an angle of incidence equal to 56 degrees. Calculate the angu

Montano1993 [528]

Answer:

The angle of separation is $\Delta \theta = 0.93 ^o$

Explanation:

From the question we are told that

The angle of incidence is $\theta _ i = 56^o$

The refractive index of violet light in diamond is $n_v = 2.46$

The refractive index of red light in diamond is $n_r = 2.41$

The wavelength of violet light is $\lambda _v = 400nm = 400*10^{-9}m$

The wavelength of red light is $\lambda _r = 700nm = 700*10^{-9}m$

Snell's Law can be represented mathematically as

$\frac{sin \theta_i}{sin \theta_r} = n$

Where $\theta_r$ is the angle of refraction

=> $sin \theta_r = \frac{sin \theta_i}{n}$

Now considering violet light

$sin \theta_r__{v}} = \frac{sin \theta_i}{n_v}$

substituting values

$sin \theta_r__{v}} = \frac{sin (56)}{2.46}$

$sin \theta_r__{v}} = 0.337$

$\theta_r__{v}} = sin ^{-1} (0.337)$

$\theta_r__{v}} = 19.69^o$

Now considering red light

$sin \theta_r__{R}} = \frac{sin \theta_i}{n_r}$

substituting values

$sin \theta_r__{R}} = \frac{sin (56)}{2.41}$

$sin \theta_r__{R}} = 0.344$

$\theta_r__{R}} = sin ^{-1} (0.344)$

$\theta_r__{R}} = 20.12^o$

The angle of separation between the red light and the violet light is mathematically evaluated as

$\Delta \theta = \theta_r__{R}} - \theta_r__{V}}$

substituting values

$\Delta \theta =20.12 - 19.69$

$\Delta \theta = 0.93 ^o$

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An ac series circuit contains a resistor of 15 ohms, an inductor of 11.5 mH and a variable capacitor. If the frequency of the ap

SpyIntel [72]

Answer:

C) 20.23 μF

Explanation:

R = resistance of the resistor = 15 ohm

L = inductance of the inductor = 11.5 mH = 0.0115 H

f = resonance frequency achieved = frequency of applied voltage = 330 Hz

C = Capacitance of the capacitor

For resonance to be possible

$2\pi fL = \frac{1}{2\pi fC}$

$2(3.14) (330) (0.0115) = \frac{1}{2 (3.14) (330) C}$

C = 20.23 x 10⁻⁶ F

C = 20.23 μF

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