To solve this problem it is necessary to apply the concepts related to frequency and vibration of strings. Mathematically the frequency can be expressed as

Then the relation between two different frequencies with same wavelength would be


The beat frequency heard when the two strings are sounded simultaneously is



We have the velocity of the transverse waves in stretched string as


And,

Therefore the relation between the two is,


Finally substituting this value at the frequency beat equation we have


Therefore the beats per second are 11.92Hz
Answer: Yos
Explanation: Becouse i experimented that btw
Approximately 100-120 wolves
Answer:
(B) The speed is larger at A than at B.
Explanation:
Point B, the final point of the trajectory, has higher electric potential than point A, the initial point of the trajectory, so the electric potential energy of the charged particle increases, which means that its kinetic energy must be decreasing, thus the speed at B must be lower than the speed at A.
Answer:
correct option is b. 31.3 m/s
Explanation:
given data
artificial gravity a1 = 1 g
artificial gravity a2 = 2 g
diameter = 100 m
radius r= 50 m
speed v1 = 22.1 m/s
solution
As acceleration is ∝ v²
so we can say
.....................1
put here value
solve it
v2 =
× 22.1
v2 = 31.25 m/s
so correct option is b. 31.3 m/s