Check Google it might have the answer sorry I couldn't be much help
Answer:
It will emerge at its initial speed not a slower speed.
Explanation:
It will emerge at the initial speed because the medium at the point of emergence is the same as the medium before incidence.
Light moves at a constant speed in any particular medium. Hence, the speed of light in air is constant in air and the speed of light in glass is constant in glass.
Answer:
(c) 16 m/s²
Explanation:
The position is ![r(t) = [3.0 \text{ m} - (4.00 \text{ m/s})t]\hat{i} + [6.0 \text{m} - (8.00 \text{ m/s}^2 )t^2 ]\hat{j}](https://tex.z-dn.net/?f=r%28t%29%20%3D%20%5B3.0%20%5Ctext%7B%20m%7D%20-%20%284.00%20%5Ctext%7B%20m%2Fs%7D%29t%5D%5Chat%7Bi%7D%20%2B%20%5B6.0%20%5Ctext%7Bm%7D%20-%20%288.00%20%5Ctext%7B%20m%2Fs%7D%5E2%20%29t%5E2%20%5D%5Chat%7Bj%7D)
.
The velocity is the first time-derivative of <em>r(t).</em>
<em />
<em />
The acceleration is the first time-derivative of the velocity.
Since <em>a(t)</em> does not have the variable <em>t</em>, it is constant. Hence, at any time,
Its magnitude is 16 m/s².
I have a strange hunch that there's some more material or previous work
that goes along with this question, which you haven't included here.
I can't easily find the dates of Mercury's extremes, but here's some of the
other data you're looking for:
Distance at Aphelion (point in it's orbit that's farthest from the sun):
<span><span><span><span><span>69,816,900 km
0. 466 697 AU</span>
</span>
</span>
</span>
<span>
Distance at Perihelion
(</span></span><span>point in it's orbit that's closest to the sun):</span>
<span><span><span><span>46,001,200 km
0.307 499 AU</span> </span>
Perihelion and aphelion are always directly opposite each other in
the orbit, so the time between them is 1/2 of the orbital period.
</span><span>Mercury's Orbital period = <span><span>87.9691 Earth days</span></span></span></span>
1/2 (50%) of that is 43.9845 Earth days
The average of the aphelion and perihelion distances is
1/2 ( 69,816,900 + 46,001,200 ) = 57,909,050 km
or
1/2 ( 0.466697 + 0.307499) = 0.387 098 AU
This also happens to be 1/2 of the major axis of the elliptical orbit.
Answer:
depende de que fenómenos nos referimos de acuerdo al los cuerpos de formación puede aver movimiento contante
