Answer:
The items here are describing either a condition in a later interacton or a protogalactic cloud. The results matching with spiral and elliptical galaxy are:
For spiral galaxy are options 6,3,2 and 5.
and for elliptical galaxy are options 4 and 1.
Explanation:
Here it is given that astrnomers suspect that types of galaxy can be affected both by the conditions which occurs due to protogalactic cloud and then from it forms the initial conditions and then by the later interactions with the other galaxies.
so, both types of galaxies are matched with their respective items given:
A. Spiral galaxy:
2. A galaxy collision results tostripping of gas.
3. The protogalactic cloud rotates in a very slow motion.
5. The density of protogalactic cloud is very high.
6. when the protogalactic cloud shrinks cloud forms very rapidly.
B. Elliptical galaxy:
1. The protogalactic cloud has high angular momentum.
4. Most of the protogalactic gases settles down into a disk.
They don't lose OR gain electrons as they've already achieved the octet rule and have 8 valence electronsn
A 3rd harmonic of a tube open at both ends will have displacement antinodes at both ends.
In a tube of length L with two open ends, the longest standing wave has displacement antinodes (pressure nodes) at both ends. The fundamental or first harmonic is what it is known as. The second harmonic is the longest standing wave in a tube of length L with two open ends.
Learn more about harmonics here brainly.com/question/17315536
#SPJ10
Answer:
- The emf of the generator is 6V
- The internal resistance of the generator is 1 Ω
Explanation:
Given;
terminal voltage, V = 5.7 V, when the current, I = 0.3 A
terminal voltage, V = 5.1 V, when the current, I = 0.9 A
The emf of the generator is calculated as;
E = V + Ir
where;
E is the emf of the generator
r is the internal resistance
First case:
E = 5.7 + 0.3r -------- (1)
Second case:
E = 5.1 + 0.9r -------- (2)
Since the emf E, is constant in both equations, we will have the following;
5.1 + 0.9r = 5.7 + 0.3r
collect similar terms together;
0.9r - 0.3r = 5.7 - 5.1
0.6r = 0.6
r = 0.6/0.6
r = 1 Ω
Now, determine the emf of the generator;
E = V + Ir
E = 5.1 + 0.9x1
E = 5.1 + 0.9
E = 6 V
