Answer:
The observer hears a loud sound
Explanation:
In order to know if the observer hears a loud or a quiet sound, you need to know if there is a constructive or destructive interference between the sound waves of the loudspeakers.
You first calculate the distance between the observer and the loudspeakers.
The distances are given by:
d1: distance to loudspeaker A = 2.10m
d2: distance to loudspeaker B
Next, you calculate the wavelength of the sound waves by using the following formula:
vs: speed of sound = 343 m/s
f: frequency of the waves = 400Hz
λ: wavelength
Next, you calculate the path difference between the distance from the observer to the loudspeakers:
You obtain a constructive interference (loud sound) if the quotient between the wavelength of the sound and the difference path is an integer:
Then, there will be a constructive interference, and the sound who the observer hears is loud.
(a) The distance will be more than 2.0 meters.
In fact, you starts your fall after your friend has already fallen 2.0 meters. This means that your friend has already accelerated for a while, therefore his velocity will be greater than yours. But this statement will be actually true for the entire fall, since you has some delay, therefore when your friend will hit the water, the separation between you and him will be greater than the initial separation of 2.0 meters.
b) First of all we need to calculate the height of the bridge with respect to the water. We know that you take 1.6 s to fall down, therefore we can use the following equation:
We know that your friend will take 1.6 s to falls down. Instead, you start your jump after he has already fallen 2.0 m, therefore after a time given by the equation:
Using S=2.0 m,
So we know that you start your fall 0.64 s after your friend. Therefore, now we can find how much did you fall between the moment you started your fall (0.64 s) and the moment your friend hits the water (1.6 s). Using
we find
So, when your friend hits the water, you just covered 4.52 m, while he already covered 12.56 m. Therefore, the separation between you and your friend is more than 2 meters.
Answer:
v₁f = 0.5714 m/s (→)
v₂f = 2.5714 m/s (→)
e = 1
It was a perfectly elastic collision.
Explanation:
m₁ = m
m₂ = 6m₁ = 6m
v₁i = 4 m/s
v₂i = 2 m/s
v₁f = ((m₁ – m₂) / (m₁ + m₂)) v₁i + ((2m₂) / (m₁ + m₂)) v₂i
v₁f = ((m – 6m) / (m + 6m)) * (4) + ((2*6m) / (m + 6m)) * (2)
v₁f = 0.5714 m/s (→)
v₂f = ((2m₁) / (m₁ + m₂)) v₁i + ((m₂ – m₁) / (m₁ + m₂)) v₂i
v₂f = ((2m) / (m + 6m)) * (4) + ((6m -m) / (m + 6m)) * (2)
v₂f = 2.5714 m/s (→)
e = - (v₁f - v₂f) / (v₁i - v₂i) ⇒ e = - (0.5714 - 2.5714) / (4 - 2) = 1
It was a perfectly elastic collision.
