Sunspot cycles, solar flares, prominences, layers of the Sun, coronal mass<span> ejections, and </span>nuclear<span> reactions</span>
B should be the answer
Look at this example to help you
Momentum is conserved, so the total momentum before collision is equal to the total momentum after collision. Take the right direction to be positive. Then
(3.00 kg) (2.09 m/s) + (2.22 kg) (-3.92 m/s) = (3.00 kg) (-1.71 m/s) + (2.22 kg) <em>v</em>
where <em>v</em> is the velocity of the 2.22 kg block after collision. Solve for <em>v</em> :
6.27 kg•m/s - 8.70 kg•m/s = -5.13 kg•m/s + (2.22 kg) <em>v</em>
(2.22 kg) <em>v</em> = 2.70 kg•m/s
<em>v</em> = (2.70 kg•m/s) / (2.22 kg)
<em>v</em> ≈ 1.22 m/s
i.e. a velocity of about 1.22 m/s to the right.
Hi there!
a)
We can use the equation t = √2d/g to solve. (Let's let g = 10 m/s²)
**How to get this equation**
We have the equation:
Δd = vit + 1/2at²
For freefall, we know that vi = 0, so we are left with:
Δd = 1/2at²
We know that a = g. Rearrange in terms of t:
2Δd / a = t²
Square root both sides:
√(2d/a) = t
Plug in the height and gravity:
t = √2(400)/10 = √800/10 = √80 ≈ 8.94 sec
b)
Find the final speed using the following formula:
vf = √2gd
**How to derive**
We know the equation:
vf² = vi² + 2ad
vi = 0, so:
vf² = 2ad
Square root both sides:
vf = √2ad
vf = √2(400)(10) = √8000 ≈ 89.44 m/s
Answer:
Every object has a different density and therefore carries different properties. When rays of white light strike an object, each ray light strikes the object with different frequency and therefore is absorbed and reflected differently from the host object.
In case if all the frequencies are absorbed by the object, it turns out to be black in color. Whereas on the other hand, if it is a mix of absorption and reflection, it makes different colors based on its frequencies and other properties of the object.
