Lifting hands and the down by one student at a time best describe the presentation of the transverse wave by students. Option D is correct.
<h3>
What is a Transverse wave?</h3>
- The wave in which the oscillation of particles is is perpendicular to the direction of energy transfer.
- The students can make a transverse wave by raising their hands up and then down, one student at a time.
- The raised hand represents the oscillation of particles while the sequence of the raising hand represents the direction of energy transfer.
Therefore, lifting hands and the down by one student at a time best describe the presentation of the transverse wave by students.
Learn more about Transverse waves:
brainly.com/question/3813804
It’s either 0.05 or 20. Assuming that the coefficient friction is a damping factor, I feel like 0.05 would be correct m
Answer:
2.84 seconds
Explanation:
t = ?
distance = 125
Velocity origianal = 60 m/hr = 88 ft/s
AVERAGE velocity = 88/2 = 44 ft/s
44 t = 125
t = 125/44 = 2.84 s
Answer:
a) x = 0.200 m
b)E = 3.84*10^{-4} N/C
Explanation:


DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m
by relation for electric field we have following relation

according to question E = 0
FROM FIGURE
x is the distance from left point charge where electric field is zero

solving for x we get

x = 0.200 m
b)electric field at half way mean x =0.25

E = 3.84*10^{-4} N/C