The echo is weaker than the original sound since the echo has a smaller amplitude than the original sound because sound spreads and its intensity decreases with distance. For example, when the sound from a speaker is heard and is compared to its echo, the echo heard has a lower amplitude than the actual sound.

5 0

3 years ago

A bullet is fired horizontally with initial velocity of 800 m/s as a target located 200 mfrom the rifle.(a).How much time is req

Aneli [31]

Answer:

t = 0.25 seconds

Explanation:

Given that,

Initial speed of a bullet, v = 800 m/s

Distance from the target is 200 m

We need to find the time required for the bullet to reach the target. Time is simply calculated by the definition of velocity i.e.

$t=\dfrac{d}{v}\\\\t=\dfrac{200\ m}{800\ m/s}\\\\t=0.25\ s$

So, it will take 0.25 seconds to reach the target.

3 0

3 years ago

A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac

uysha [10]

Answer:

a) v = 1.1 m/s

b) A = 0.315 m

c) v = 1.1 m/s A= 0.15 m

Explanation:

In any travelling wave, there exists a fixed relationship between the propagation speed, the wavelength and the frequency, as follows:

$v = \lambda * f (1)$

If the wave crests are spaced a horizontal distance of 5.7 m apart, this means that the wavelength of the wave is just the same, i.e., 5.70 m.
Regarding the frequency, we know that the frequency is just the inverse of the period, i.e., the time needed to complete one oscillation.

If it takes a time of 2.60 s to go from the highest point to the lowest, the time needed to complete an oscillation (the period T) will be just double of this time:
⇒ T = 2.60 s * 2 = 5.20 s (2)
Since we have now T, we can find the frequency f as follows:

$f = \frac{1}{T} = \frac{1}{5.20s} = 0.19 Hz (3)$

Replacing f and λ in (1) we get:

$v = \lambda * f = 5.70 m * 0.19 Hz = 1.10 m/s (4)$

The amplitude of the wave is just the amount that the water aparts from its equilibrium level, which is just the half of the distance between its highest point and the lowest one, as follows:

$A = \frac{0.630m}{2} = 0.315 m (5)$

Part a) will not be affected by the new amplitude, because we have showed that the speed is independent of the amplitude, so v can be written as follows:

v = 1.10 m/s (6)

Part b) will change , due to the amplitude changes. If the total vertical distance traveled by the boat is 0.30 m, by the same token as explained in b), the new amplitude will be just half of this, as follows:

$A = \frac{0.30m}{2} = 0.15 m (7)$

5 0

3 years ago