Answer:
Wegener first thought of this idea by noticing that the different large landmasses of the Earth almost fit together like a jigsaw puzzle. The continental shelf of the Americas fits closely to Africa and Europe, and other continents showed the same trend. Wegner also analyzed both sides of the Atlantic Ocean for rock type, geological structures and fossils and noticed that there was a significant similarity between matching sides of the continents, especially in fossil plants.
The answer is C 300,000 kilometers per second
Answer:

Explanation:
given,
frequency of tuba.f = 64 Hz
Speed of train approaching, v = 8.50 m/s
beat frequency = ?
using Doppler's effect formula

v_s is the velocity of the source
v is the speed of sound, v = 340 m/s
now,

f' = 65.64 Hz
now, beat frequency is equal to



hence, beat frequency is equal to 1.64 Hz
There are 6 types of quarks.
Answer: C. 6
To solve this problem, it will be necessary to apply the concepts related to the fundamental resonance frequency in a closed organ pipe.
This is mathematically given as

For fundamental frequency n is 0, then,

When,
v = Velocity of sound
L = Length,
Rearranging to find the velocity,



Therefore the speed of sound in this gas is 416m/s