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Ulleksa [173]
3 years ago
13

A spaceship from a friendly, extragalactic planet flies toward Earth at 0.203 0.203 times the speed of light and shines a powerf

ul laser beam toward Earth to signal its approach. The emitted wavelength of the laser light is 691 nm . 691 nm. Find the light's observed wavelength on Earth.

Physics
2 answers:
Serjik [45]3 years ago
7 0

Answer:

567.321nm

Explanation:

See attached handwritten document for more details

Rudik [331]3 years ago
7 0

Answer:

The observed wavelenght on earth will be 5.51x10^-7 m

Explanation:

Speed of light c = 3x10^8 m/s

Ship speed u = 0.203 x 3x10^8 = 60900000 = 6.09x10^7 m/s

Wavelenght of laser w = 691x10^-9 m

Observed wavelenght w' = w(1 - u/c)

u/c = 0.203

w' = 691x10^-9(1 - 0.203)

w' = 5.51x10^-7 m

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Two blocks are on a horizontal frictionless surface. Block a is moving with an initial velocity
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The final velocity of the block A will be 2.5 m/sec. The principal of the momentum conversation is used in the given problem.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

In a given concern, mass m₁ is M, mass m₂ is 3M. Initial speed for the mass m₁ and  m₂ will be u₁=5 and u₂=0 m/s respectively,

According to the law of conservation of momentum

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V=51.25 m/s

The velocity of block A is found as;

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If Vx = 7.00 units and Vy = -7.60 units, determine the magnitude of V⃗ .
Juliette [100K]

|V| = 10.33 units and the direction θ = -47.35° or 312.65°.

Given the x and y components of a vector, we can calculate the magnitude and direction from these components.

Applying the Pythagorean theorem we have that the magnitude of the vector is:

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|V| = \sqrt{(7.00units)^{2}+(-7.60units)^{2}} = \sqrt{106units^{2}} = 10.33units

The expression for the direction of a vector comes from the definition of the tangent of an angle:

tan θ = \frac{Vy}{Vx} ------>  θ = arc tan \frac{Vy}{Vx}

θ = arc tan \frac{-7.60units}{7.00units}

θ = -47.35° or 312.65°

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