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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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A 1500 kg truck is acted upon by a force that decreases its speed from 25 m/s to 15 m/s in 8 s. What is the magnitude of the for
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Answer:

F = 1,875 N

Explanation:

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and ∆V ----> change in velocity

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Needddd helppppppp!!!
yulyashka [42]

Answer:

2/9 times as strong.

Explanation:

From the question given above, the following assumptions were made:

Initial mass of 1st planet (M₁ ) = M

Initial mass of 2nd planet (m₁ ) = m

Initial distance apart (r₁) = r

Initial Force of attraction (F₁) = F

Final mass of 1st planet (M₂) = 2M

Final mass of 1st planet (m₂) = constant = m

Final distance apart (r₂) = 3r

Final force of attraction (F₂) =?

Next, we shall obtain an expression to determine the new force. This can be obtained as follow:

F = GMm / r²

Cross multiply

Fr² = GMm

Divide both side by Mn

G = Fr² / Mm

Since G is constant, then we have

F₁r₁² / M₁m₁ = F₂r₂² / M₂m₂

Finally, we shall determine the new force as follow:

Initial mass of 1st planet (M₁ ) = M

Initial mass of 2nd planet (m₁ ) = m

Initial distance apart (r₁) = r

Initial Force of attraction (F₁) = F

Final mass of 1st planet (M₂) = 2M

Final mass of 1st planet (m₂) = constant = m

Final distance apart (r₂) = 3r

Final force of attraction (F₂) =?

F₁r₁² / M₁m₁ = F₂r₂² / M₂m₂

Fr² / Mm = F₂ × (3r)² / 2M × m

Fr² / Mm = F₂ × 9r² / 2Mm

Cross multiply

Fr² × 2Mm = F₂ × 9r² × Mm

Divide both side by 9r² × Mm

F₂ = Fr² × 2Mm / 9r² × Mm

F₂ = F × 2 / 9

F₂ = 2/9 F

Thus, the new force is 2/9 times the original force i.e 2/9 times as strong.

4 0
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To solve this problem, we are going to use the formula for work which is Fd where x and y are measured separately.

 

X direction: W = 13.5 x 230 = 3105 Joules

Y direction: W = -14.3 x -165 = 2360 Joules

So the total work is getting the sum of the two: 3105 + 2360 = 5465 Joules

7 0
3 years ago
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