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igomit [66]
3 years ago
8

A spaceship from a friendly, extragalactic planet flies toward Earth at 0.201 times the speed of light and shines a powerful las

er beam toward Earth to signal its approach. The emitted wavelength of the laser light is 697 nm . Find the light's observed wavelength on Earth.
Physics
1 answer:
bagirrra123 [75]3 years ago
8 0

Answer:

The wavelength of observed light on earth is 568.5 nm

Explanation:

Given that,

Velocity of spaceship v= 0.201c

Wavelength of laser \lambda= 697\ nm

We need to calculate the wavelength of observed light on earth

Using formula of wavelength

\lambda_{0}=\lambda_{e}\times\sqrt{\dfrac{1-\dfrac{v}{c}}{1+\dfrac{v}{c}}}

\lambda_{0}=697\times10^{-9}\times\sqrt{\dfrac{1-\dfrac{0.201 c}{c}}{1+\dfrac{0.201c}{c}}}

\lambda_{0}=697\times10^{-9}\times\sqrt{\dfrac{1-0.201}{1+0.201}}

\lambda=5.685\times10^{-7}\ m

\lambda=568.5\times10^{-9}\ m

\lambda=568.5\ nm

Hence, The wavelength of observed light on earth is 568.5 nm

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Ilya [14]

Answer with Explanation:

We are given that

Area of loop=(20\times 20) cm^2=400\times 10^{-4} m^2

1 cm^2=10^{-4} m^2

Resistance, R=0.1\Omega

B=4t-2t^2

We know that magnetic flux

\phi=BA

Emf ,E=\mid \frac{d\phi}{dt}\mid =\mid\frac{d(BA}{dt}\mid =\mid A\frac{dB}{dt}=400\times 10^{-4}\times \frac{4t-2t^2}{dt}\mid =\mid400\times 10^{-4}\times(4-4t)\mid

Current, I=\frac{E}{R}

Current, I=\frac{\mid 400\times 10^{-4}(4-4t)\mid }{0.1}=1.6\mid (1-t)\mid

Substitute t=0 s

Then, I=1.6\mid (1-0)\mid=1.6 A

Substitute t=1 s

Then, I=1.6\mid (1-1)\mid=0

Substitute

t=2 s

Current, I=1.6\mid(1-2)\mid=1.6 A

8 0
2 years ago
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Gravitational potential energy is due to....
mote1985 [20]

Answer:

A

Explanation:

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5 0
3 years ago
Hi I’m having some difficulty with this question could really use some help!
Serhud [2]
\begin{gathered} v=\lambda f \\ v=4m\cdot2s^{-1} \\ v=8m/s \end{gathered}

3 0
1 year ago
From the edge of a cliff, a 0.41 kg projectile is launched with an initial kinetic energy of 1430 J. The projectile's maximum up
NemiM [27]

Answer:

v₀ₓ = 63.5 m/s

v₀y = 54.2 m/s

Explanation:

First we find the net launch velocity of projectile. For that purpose, we use the formula of kinetic energy:

K.E = (0.5)(mv₀²)

where,

K.E = initial kinetic energy of projectile = 1430 J

m = mass of projectile = 0.41 kg

v₀ = launch velocity of projectile = ?

Therefore,

1430 J = (0.5)(0.41)v₀²

v₀ = √(6975.6 m²/s²)

v₀ = 83.5 m/s

Now, we find the launching angle, by using formula for maximum height of projectile:

h = v₀² Sin²θ/2g

where,

h = height of projectile = 150 m

g = 9.8 m/s²

θ = launch angle

Therefore,

150 m = (83.5 m/s)²Sin²θ/(2)(9.8 m/s²)

Sin θ = √(0.4216)

θ = Sin⁻¹ (0.6493)

θ = 40.5°

Now, we find the components of launch velocity:

x- component = v₀ₓ = v₀Cosθ  = (83.5 m/s) Cos(40.5°)

<u>v₀ₓ = 63.5 m/s</u>

y- component = v₀y = v₀Sinθ  = (83.5 m/s) Sin(40.5°)

<u>v₀y = 54.2 m/s</u>

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zavuch27 [327]

Answer:

D

Explanation:

a neutron does not have a positive nor negative charge it remains neutral

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