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Pani-rosa [81]
3 years ago
6

The Earth orbits the Sun at a speed of 30 km/s. At that speed it completes one path around the Sun every year. Of course, as tha

t happens we are heading toward a different place in space constantly, always accelerating toward the Sun to stay on the circular path. Suppose at a paritcular instant we are headed in the direction of a distant star and we measure the speed of the light from that star as it arrives on Earth. In the vacuum of space we would measurea. 299,792,458 m/sb. 299,792,428 m/sc. 299,792,488 m/sd. 0 m/s
Physics
1 answer:
romanna [79]3 years ago
3 0

Answer:

a. 299,792,458 m/s

Explanation:

Since the speed of light in a vacuum is invariant and has the value of 299,792,458 m/s, we would measure this value of 299,792,458 m/s for the speed of light from the star as it arrives on Earth.

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The vitreous humor, a transparent, gelatinous fluid that fills most of the eyeball, has an index of refraction of 1.34. Visible
Leto [7]

Answer:

a)   298.5 nm ,  522.4 nm  and b)  radiation frequency does not change

Explanation:

When electromagnetic radiation reaches a medium with a different index of refraction, the medium vibrates the molecules, as if it were a resonance process, whereby the medium vibrates at the same frequency as the incident light.

On the other hand, when the light reaches another medium its average speed within the medium changes, it is now less than the speed of light in a vacuum (c) for this to happen as we saw that the frequency is constant there must be a change in the wavelength of the radiation that is characterized by the ratio

    λₙ = λ₀ / n

    λₙ = 400 nm    in the void

    λₙ = 400 / 1.34

    λₙ= 298.5 nm

   λ₀ = 700 nm

   λₙ = 700 / 1.34

   λₙ = 522.4 nm

The radiation frequency does not change

4 0
3 years ago
How much force is needed to accelerate a 68 kilogram-skier at a rate of 1.2 m/sec^2?
DiKsa [7]
<span>Answer: Force = 81.6 N

Explanation:
According to Newton's Second law:
F = ma --- (1)

Where F = Force = ?
m = Mass = 68 kg
a = Acceleration = 1.2 m/s^2

Plug in the values in (1):
(1) => F = 68 * 1.2
F = 81.6 N (The force needed to accelerate the skier at a rate of 1.2 m/s^2)</span>
4 0
3 years ago
Read 2 more answers
At the same moment from the top of a building 3.0 × 10 2 m tall, one rock is dropped and one is thrown downward with an initial
Bess [88]
The equation that relates distance, velocities, acceleration, and time is,
                   d = V₀t + 0.5gt²
where d is distance,
V₀ is the initial velocity,
t is time, and 
g is the acceleration due to gravity (equal to 9.8 m/s²)

(1) Dropped rock,
                  (3 x 10² m ) = 0(t) + 0.5(9.8 m/s²)(t²)
The value of t from this equation is 24.73 s

(2) Thrown rock with V₀ = 26 m/s
                (3 x 10² m) = (26)(t) + 0.5(9.8 m/s²)(t²)
The value of t from the equation is 5.61 s

The difference between the tim,
        difference = 24.73 s - 5.61 s
          difference = 19.12 s

<em>ANSWER: 19.12 s</em>
5 0
3 years ago
Read 2 more answers
6. When adding vectors graphically, the direction and length of each vector must:
Tanzania [10]

Answer:

Not be changed

Option: D

<u>Explanation:</u>

The physical quantity which has both ‘magnitude and direction’ is called vector. These vectors are represented by a line and an arrow, <em>the line represent the magnitude and arrow represent the direction of the physical quantity</em>. The vectors are added and subtracted according to the direction of the vectors.

According to the vector law addition while adding vectors direction and length of the vector is not be changed.<em> If the length of the vector changed the magnitude is also changed while so, while adding vectors length must not be changed. </em>

3 0
3 years ago
A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th
NISA [10]

Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2

v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)

v_{pf}=-31.4\ m/s

v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}-\dfrac{(0.11-70)\times 0}{m_1+m_2}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}

v_{gf} = 0.0988\ m/s

4 0
3 years ago
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