Answer:
Part a) When collision is perfectly inelastic
Part b) When collision is perfectly elastic
Explanation:
Part a)
As we know that collision is perfectly inelastic
so here we will have
so we have
now we know that in order to complete the circle we will have
now we have
Part b)
Now we know that collision is perfectly elastic
so we will have
now we have
Answer: They are in the same group because they have similar chemical properties, but they are in different periods because they have very different atomic numbers.
Explanation: On Edgenuity!!
<span>A.) If a sideways force of 300 N is applied to the motor, how far will it move sideways?</span>
Complete Question:
A beam of white light is incident on the surface of a diamond at an angle 
, since the index of refraction depends on the light's wavelength, the different colors that comprise white light will spread out as they pass through the diamond. For example, the indices of refraction in diamond are 
for red light and 
for blue light. Thus, blue light and red light are refracted at different angles inside the diamond. The surrounding air has 
.
Now consider θc, the angle at which the blue refracted ray hits the bottom surface of the diamond. If θc is larger than the critical angle θcrit, the light will not be refracted out into the air, but instead it will be totally internally reflected back into the diamond. Find θcrit. Express your answer in degrees to four significant figures.
Answer:
Explanation:
Only the blue refracted ray is related to the critical angle in this question
The relationship between the critical angle(
), 
and 
can be given as 
Charge on can A is positive.
Charge on can C is negative.
Punctuation and capitalization are very useful things to pay attention to and this question would be a lot easier to understand if you had actually used both capitalization and punctuation. If I'm understanding the question, you have 3 metal can that are insulated from the environment and initially touching each other in a straight line. Then a negatively charged balloon is brought near, but not touching one of the cans in that line of cans. While the balloon is near, the middle can is removed. Then you want to know the charge on the can that was nearest the balloon and the charge on the can that was furthermost from the balloon.
As the balloon is brought near to can a, the negative charge on the balloon repels some of the electrons from can a (like charges repel). Some of those electrons will flow to can b and in turn flow to can c. Basically you'll have a charge gradient that's most positive on that part of the can that's closest to the balloon, and most negative on the part of the cans that's furthest from the balloon. You then remove can B which causes cans A and C to be electrically isolated from each other and prevents the flow of elections to equalize the charges on cans A and C when the balloon is removed. So you're left with a deficiency of electrons on can A, so can A will have a positive overall charge, and an excess of electrons on can C, so can C will have a negative overall charge.
