Clever problem.
We know that the beat frequency is the DIFFERENCE between the frequencies of the two tuning forks. So if Fork-A is 256 Hz and the beat is 6 Hz, then Fork-B has to be EITHER 250 Hz OR 262 Hz. But which one is it ?
Well, loading Fork-B with wax increases its mass and makes it vibrate SLOWER, and when that happens, the beat drops to 5 Hz. That means that when Fork-B slowed down, its frequency got CLOSER to the frequency of Fork-A ... their DIFFERENCE dropped from 6 Hz to 5 Hz.
If slowing down Fork-B pushed it CLOSER to the frequency of Fork-A, then its natural frequency must be ABOVE Fork-A.
The natural frequency of Fork-B, after it gets cleaned up and returns to its normal condition, is 262 Hz. While it was loaded with wax, it was 261 Hz.
Answer:
Explanation:
a ) Momentum of first cart = mass x velocity
= 3 x 4.6 =+13.8 kg m /s
Momentum of second cart = 1.3 x - 1.9 = - 2.47 kg m /s
Total momentum = 13.8 - 2.47
= +11.33 kg m /s
b )
Let the velocity of first cart be v at the moment when second cart was at rest
total momentum = 3 x v + 0 = 3 v
Applying conservation of momentum law
3 v = +11.33
v = +3.77 m /s
Answer:
a. 4
Explanation:
Hi there!
The equation of kinetic energy (KE) is the following:
KE = 1/2 · m · v²
Where:
m = mass of the car.
v = speed of the car.
Let´s see how would be the equation if the velocity is doubled (2 · v)
KE2 = 1/2 · m · (2 · v)²
Distributing the exponent:
KE2 = 1/2 · m · 2² · v²
KE2 = 1/2 · m · 4 · v²
KE2 = 4 (1/2 · m · v²)
KE2 = 4KE
Doubling the velocity increased the kinetic energy by 4.
Answer:
Acceleration, 
Explanation:
Initial speed of the skater, u = 8.4 m/s
Final speed of the skater, v = 6.5 m/s
It hits a 5.7 m wide patch of rough ice, s = 5.7 m
We need to find the acceleration on the rough ice. The third equation of motion gives the relationship between the speed and the distance covered. Mathematically, it is given by :
So, the acceleration on the rough ice 
and negative sign shows deceleration.
