The correct answer among all the other choices is volts. This unit expresses the difference in energy. It is not watts or amps. Thank you for posting your question. I hope this answer helped you. Let me know if you need more help.
<span>The distance between the sources of a two point surface
interference determines the number of lines that can be created for the nodal
and anti-nodal lines. When the sources are moved closer together, the number of
lines becomes minimal. Thus, the spaces between the lines become bigger. The
answer to this item is letter A. The spacing increases. </span>
Answer: A) <u>Either source or listener must be moving.</u>
Explanation:
Frequencies can shift if an observer is moving relative to the wave’s source. This type of shift is called the Doppler effect (often used to analyze sounds). Think about the sound you hear when a police siren passes you and drives away; as the car increases its distance from you, the pitch of its sound becomes lower. This is because each wave is emitted from a greater distance, causing the wavelength to spread out or increase relative to you, the listener.
The amplitude of wave-c is 1 meter.
The speed of all of the waves is (12meters/2sec)= 6 m/s.
The period of wave-a is 1/2 second.
Answer:
A.) 4.81 seconds
B.) 44.6 m/s
Explanation:
He begins his dive by jumping up with a velocity of 5 (m/s).
Let us first calculate the maximum height reached by using third equation of motion
V^2 = U^2 - 2gH
At maximum height, V = 0
0 = 5^5 - 2 × 9.8H
19.6H = 25
H = 25 /19.6
H = 1.28 m
The time taken for the diver to reach the water from the maximum height can be calculated by using second equation of motion.
Where height h = 1.28 + 100 = 101.28 m
h = Ut + 1/2gt^2
As the diver drop from maximum height, U = 0
101.28 = 1/2 × 9.8 × t^2
4.9t^2 = 101.28
t^2 = 101.28/4.9
t^2 = 20.669
t = sqrt ( 20.669)
t = 4.55s
As the diver jumped up, the time taken to reach the maximum height will be
Time = 1.28 / 5 = 0.256
The time taken for him to hit the water below will be 0.256 + 4.55 = 4.81 seconds
B.) Velocity right before he hits the water will be
V^2 = U^2 + 2gH
But U = 0
V^2 = 2 × 9.8 × 101.28
V^2 = 1985.09
V = 44.6 m/s
