They would hit the ground at the same time. No matter the weight laws of physics.
Answer:
V₁ = 5.6 m/s
V₂ = 7.2 m/s
V₃ = 8.8 m/s
Explanation:
Average velocity: Average velocity can be defined as the ratio of the total displacement to the total time taken. The S.I unit of Average velocity is m/s.
For the first 2 s,
V₁ = Δd₁/t
Where V₁ = Average velocity for the first 2 s
Where Δd₁= distance, t = time
Δd₁ = 25.6-14.4 = 11.2 m t = 2 s
V₁ = 11.2/2
V₁ = 5.6 m/s
For the second 2 s,
V₂ =Δd₂/t
Where V₂ = average velocity for the second 2 s.
Δd₂= 40-25.6 = 14.4 m, t= 2 s
V₂ = 14.4/2
V₂ = 7.2 m/s
For the last 2 seconds,
V₃ =Δd₃/t
Where V₃ = average velocity for the last 2 s
where Δd₃ = 57.6- 40 = 17.6 m, t = 2 s
V₃ = 17.6/2
V₃ = 8.8 m/s.
Answer:
The combination of cells in which the negative terminal of a first cell is connected with the positive terminal of second cell and the negative terminal of a second cell is connected to the positive terminal of a third cell and so on is known as series combination of cells.
Explanation:
hope this helps to u
Answer: person C.
Explanation:
1) The frequency of the sound wave is perceived or detected as the pitch of the sound.
And the higher the frequency the the higher the pitch.
2) When a sound source is in motion there is an apparent change in the pitch perceived. This is the Doppler effect, which is defined as the change in the perception of the pitch in virtue of the relative motion between the listener and the emitter of the sound.
If the source of the sound is moving toward the listener, the pitch will be higher that if the sound is at rest. If the source of the sound is moving away from the listener, the pitch will be lower.
That is because the the motion of the emitter increases (or decreases, depending on whether the emitter is moving toward or away the listener) the speed of the sound waves and they arrive more frequently (or less frequently).
Then, since the sound is emitted by the motorbike and it is moving toward the person C, he will hear the highest pitch.
Answer:
Since the astronaut drops the rock, the initial velocity of the rock is 0 m/s
<u>We are given:</u>
initial velocity (u) = 0 m/s
final velocity (v) = v m/s
acceleration (a) = 1.62 m/s/s
height (h) = 1.25 m
<u>Solving for v:</u>
From the third equation of motion:
v²-u² = 2ah
replacing the variables
v² - (0)² =2 (1.62)(1.25)
v² = 1.62 * 2.5
v² = 4 (approx)
v = √4
v = 2 m/s
The speed of the rock just before it lands is 2 m/s
