I believe it is C i am not sure
Hi! I believe your answer is decreasing. <u>An inclined plane makes work easier by decreasing the amount of effort force needed, but increases the distance</u>. I hope this helps you! Good luck and have a great day. ❤️✨
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:
For cast iron we have
![h = 0.92 m](https://tex.z-dn.net/?f=h%20%3D%200.92%20m)
For copper
![h = 1.05 m](https://tex.z-dn.net/?f=h%20%3D%201.05%20m)
For Lead
![h = 1.23 m](https://tex.z-dn.net/?f=h%20%3D%201.23%20m)
For Zinc
![h = 2.43 m](https://tex.z-dn.net/?f=h%20%3D%202.43%20m)
Explanation:
As we know that final speed of the block is calculated by work energy theorem
![W_f + W_g = \frac{1}{2}mv^2](https://tex.z-dn.net/?f=W_f%20%2B%20W_g%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
now we have
![-\mu_k mg cos\theta(\frac{h}{sin\theta}) + mgh = \frac{1}{2}mv^2](https://tex.z-dn.net/?f=-%5Cmu_k%20mg%20cos%5Ctheta%28%5Cfrac%7Bh%7D%7Bsin%5Ctheta%7D%29%20%2B%20mgh%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
now we have
![v^2 = 2gh - 2\mu_k g h cot\theta](https://tex.z-dn.net/?f=v%5E2%20%3D%202gh%20-%202%5Cmu_k%20g%20h%20cot%5Ctheta)
![v = \sqrt{2gh(1 - \mu_k cot\theta)}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B2gh%281%20-%20%5Cmu_k%20cot%5Ctheta%29%7D)
For cast iron we have
![4 = \sqrt{2(9.81)(h)(1 - 0.15cot52)}](https://tex.z-dn.net/?f=4%20%3D%20%5Csqrt%7B2%289.81%29%28h%29%281%20-%200.15cot52%29%7D)
![h = 0.92 m](https://tex.z-dn.net/?f=h%20%3D%200.92%20m)
For copper
![4 = \sqrt{2(9.81)(h)(1 - 0.29cot52)}](https://tex.z-dn.net/?f=4%20%3D%20%5Csqrt%7B2%289.81%29%28h%29%281%20-%200.29cot52%29%7D)
![h = 1.05 m](https://tex.z-dn.net/?f=h%20%3D%201.05%20m)
For Lead
![4 = \sqrt{2(9.81)(h)(1 - 0.43cot52)}](https://tex.z-dn.net/?f=4%20%3D%20%5Csqrt%7B2%289.81%29%28h%29%281%20-%200.43cot52%29%7D)
![h = 1.23 m](https://tex.z-dn.net/?f=h%20%3D%201.23%20m)
For Zinc
![4 = \sqrt{2(9.81)(h)(1 - 0.85cot52)}](https://tex.z-dn.net/?f=4%20%3D%20%5Csqrt%7B2%289.81%29%28h%29%281%20-%200.85cot52%29%7D)
![h = 2.43 m](https://tex.z-dn.net/?f=h%20%3D%202.43%20m)
Answer:
2.464 cm above the water surface
Explanation:
Recall that for the cube to float, means that the volume of water displaced weights the same as the weight of the block.
We calculate the weight of the block multiplying its density (0.78 gr/cm^3) times its volume (11.2^3 cm^3):
weight of the block = 0.78 * 11.2^3 gr
Now the displaced water will have a volume equal to the base of the cube (11.2 cm^2) times the part of the cube (x) that is under water. Recall as well that the density of water is 1 gr/cm^3.
So the weight of the volume of water displaced is:
weight of water = 1 * 11.2^2 * x
we make both weight expressions equal each other for the floating requirement:
0.78 * 11.2^3 = 11.2^2 * x
then x = 0.78 * 11.2 cm = 8.736 cm
This "x" is the portion of the cube under water. Then to estimate what is left of the cube above water, we subtract it from the cube's height (11.2 cm) as follows:
11.2 cm - 8.736 cm = 2.464 cm