Answer:
1.moon
2.Earth is the first planet with a satellite, or moon.
3.Mars, the fourth planet out, has two moons, but they are probably just a couple of big old rocks.
5.The incoming object was destroyed on impact, reduced to vapor, dust, and chunks. Large surviving parts were driven deep into the interior of Earth. A signicant portion of Earth was destroyed as well. The energy that resulted from the crash produced an explosion of unimaginable magnitude.
Explanation:
i did what i can
Answer:
option D
Explanation:
given,
length of the pipe, L = 0.96 m
Speed of sound,v = 345 m/s
Resonating frequency when both the end is open
n is the Harmonic number
2nd overtone = 3rd harmonic
so, here n = 3
now,
f = 540 Hz
The common resonant frequency of the string and the pipe is closest to 540 Hz.
the correct answer is option D
Explanation:
There are two components of a longitudinal sound wave which are compression and rarefaction. Similarly, there are two components of the transverse wave, the crest, and trough.
The crest of a wave is defined as the part that has a maximum value of displacement while the trough is defined as the part which corresponds to minimum displacement.
While compression is that space where the particles are close together while the rarefaction is that space where the particles are far apart from each other.
So, the refraction or the rarefied part of a longitudinal sound wave is analogous to a trough of a transverse wave.
Answer:
Force(Romeo moving) = 5,000 N
Explanation:
Given:
Mass of horse = 900 kg
Acceleration = 20 km/hr
Find:
Force(Romeo moving)
Computation:
Acceleration = 20 km/hr
Acceleration in m/s = 20 / 3.6 = 5.555556 m/s²
Force = m x a
Force(Romeo moving) = 900 x 5.555556
Force(Romeo moving) = 5,000 N
Answer:
The observed frequency by the pedestrian is 424 Hz.
Explanation:
Given;
frequency of the source, Fs = 400 Hz
speed of the car as it approaches the stationary observer, Vs = 20 m/s
Based on Doppler effect, as the car the approaches the stationary observer, the observed frequency will be higher than the transmitted (source) frequency because of decrease in distance between the car and the observer.
The observed frequency is calculated as;
![F_s = F_o [\frac{v}{v_s + v} ] \\\\](https://tex.z-dn.net/?f=F_s%20%3D%20F_o%20%5B%5Cfrac%7Bv%7D%7Bv_s%20%2B%20v%7D%20%5D%20%5C%5C%5C%5C)
where;
F₀ is the observed frequency
v is the speed of sound in air = 340 m/s
![F_s = F_o [\frac{v}{v_s + v} ] \\\\400 = F_o [\frac{340}{20 + 340} ] \\\\400 = F_o (0.9444) \\\\F_o = \frac{400}{0.9444} \\\\F_o = 423.55 \ Hz \\](https://tex.z-dn.net/?f=F_s%20%3D%20F_o%20%5B%5Cfrac%7Bv%7D%7Bv_s%20%2B%20v%7D%20%5D%20%5C%5C%5C%5C400%20%3D%20F_o%20%5B%5Cfrac%7B340%7D%7B20%20%2B%20340%7D%20%5D%20%5C%5C%5C%5C400%20%3D%20F_o%20%280.9444%29%20%5C%5C%5C%5CF_o%20%3D%20%5Cfrac%7B400%7D%7B0.9444%7D%20%5C%5C%5C%5CF_o%20%3D%20423.55%20%5C%20Hz%20%5C%5C)
F₀ ≅ 424 Hz.
Therefore, the observed frequency by the pedestrian is 424 Hz.
