6. fuel is needed in space to power certain things like engines.
7. It couldn't, as the lowest speed acceptable to get into space is 7.9 km per second (28,440 km per hour). This is because to break into the barrier that no longer contains oxygen, you would need to be going at an extremely fast rate.
8. It involves Newton's 3rd law. The movement requires that same thrust to be thrown back, also.
9. There is no gravity. So, basically, you just float.
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.
Kinetic energy is transformed into Thermal energy
Answer:
The position for mass is ![x(t)=36.75sin(37.5t)](https://tex.z-dn.net/?f=x%28t%29%3D36.75sin%2837.5t%29)
Explanation:
Let x(t) donate the position of mass at time t,Then x satisfies the differential equation
![m\frac{d^2x}{dt^2} +kx=0\\here\\m=4kg\\k=30N/0.2m\\thus\\w^2=k/m=30N/0.2m*4kg=37.5](https://tex.z-dn.net/?f=m%5Cfrac%7Bd%5E2x%7D%7Bdt%5E2%7D%20%2Bkx%3D0%5C%5Chere%5C%5Cm%3D4kg%5C%5Ck%3D30N%2F0.2m%5C%5Cthus%5C%5Cw%5E2%3Dk%2Fm%3D30N%2F0.2m%2A4kg%3D37.5)
The general solution is
![(x)t=C_{1}cos(37 .5t)+C_{2}sin(37.5t)](https://tex.z-dn.net/?f=%28x%29t%3DC_%7B1%7Dcos%2837%20.5t%29%2BC_%7B2%7Dsin%2837.5t%29)
It follows
![x'(t)=-37.5C_{1}sin(37.5t)+37.5C_{2}cos(37.5t)\\now\\x(0)=0\\gives\\C_{1}=0\\ and\\x'(0)=1m/s\\gives\\C_{2} =36.75](https://tex.z-dn.net/?f=x%27%28t%29%3D-37.5C_%7B1%7Dsin%2837.5t%29%2B37.5C_%7B2%7Dcos%2837.5t%29%5C%5Cnow%5C%5Cx%280%29%3D0%5C%5Cgives%5C%5CC_%7B1%7D%3D0%5C%5C%20and%5C%5Cx%27%280%29%3D1m%2Fs%5C%5Cgives%5C%5CC_%7B2%7D%20%3D36.75)
thus the position of mass is
![x(t)=36.75sin(37.5t)](https://tex.z-dn.net/?f=x%28t%29%3D36.75sin%2837.5t%29)
Answer:
The velocity of the blood in the thinner arteries is 0.1 times that of the thicker artery.
Explanation:
To find the velocity of the blood we need to use the continuity equation:
(1)
<u>Where</u>:
n: is the number of branches
A: is the cross-sectional area
v: is the velocity
For artery 1, we have:
n₁ = 1, A₁ = 1 cm², v₁ = v
For the 20 arteries (2), we have:
n₂ = 20, A₂ = 0.5 cm², v₂ =?
By using equation (1):
![n_{1}A_{1}v_{1} = n_{2}A_{2}v_{2}](https://tex.z-dn.net/?f=%20n_%7B1%7DA_%7B1%7Dv_%7B1%7D%20%3D%20n_%7B2%7DA_%7B2%7Dv_%7B2%7D%20)
![1 cm^{2}*v = 20*0.5 cm^{2}*v_{2}](https://tex.z-dn.net/?f=%201%20cm%5E%7B2%7D%2Av%20%3D%2020%2A0.5%20cm%5E%7B2%7D%2Av_%7B2%7D%20)
Therefore, the velocity of the blood in the thinner arteries is 0.1 times that of the thicker artery.
I hope it helps you!