Answer:
Please see below as the answer is self-explanatory.
Explanation:
- In order to have a destructive interference, the path difference between the sources of the sound, must be equal to an odd multiple of the semi-wavelength, as follows:
- The minimum posible value for this distance, is when n=1, as it can be seen here:
- In any traveling wave, there exists a fixed relationship between the wave speed, the frequency and the wavelength:
- Therefore, assuming that the speed of sound keeps constant, if the frequency is increased, in order to keep the right side of the expression above balanced, λ must be decreased.
- As the smallest separation that produces destructive interference is directly proportional to the wavelength, this means that this separation will decrease if the cellists produced a note with a higher frequency.
<span>The student's power rating is approximately 800W.
</span>
Power is the rate of doing work. Its SI unit is Watt. 1 W = 1J/sec
Given: <span>The work done to get to the top is 1200 Joules
</span>
Time = 1.5 seconds.
Power =
I believe the answer is C. It will maintain its state of motion
Answer:
a) t = 3.35[s]; b) t = 1.386[s]
Explanation:
We can solve this problem by dividing it into two parts, for the first 55 [m] and then the second part with the remaining 55 [m].
We will take the initial velocity as zero, as the problem does not mention that the Rock was thrown at initial velocity.
And using kinematics equations:
![v_{f}^{2}= v_{o}^{2}+2*g*y\\where:\\v_{o}=0\\g=gravity = 9.81[m/s^2]\\y=55 [m]\\v_{f}^{2}=0+2*9.81*55\\v_{f}=\sqrt{2*9.81*55} \\v_{f}=32.85[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%5E%7B2%7D%3D%20v_%7Bo%7D%5E%7B2%7D%2B2%2Ag%2Ay%5C%5Cwhere%3A%5C%5Cv_%7Bo%7D%3D0%5C%5Cg%3Dgravity%20%3D%209.81%5Bm%2Fs%5E2%5D%5C%5Cy%3D55%20%5Bm%5D%5C%5Cv_%7Bf%7D%5E%7B2%7D%3D0%2B2%2A9.81%2A55%5C%5Cv_%7Bf%7D%3D%5Csqrt%7B2%2A9.81%2A55%7D%20%5C%5Cv_%7Bf%7D%3D32.85%5Bm%2Fs%5D)
Now we can calculate the time:
![v_{f}=v_{o}+g*t\\t=\frac{v_{f}-v_{o}}{g}\\ t=\frac{32.85-0}{9.81}\\ t=3.35[s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3Dv_%7Bo%7D%2Bg%2At%5C%5Ct%3D%5Cfrac%7Bv_%7Bf%7D-v_%7Bo%7D%7D%7Bg%7D%5C%5C%20t%3D%5Cfrac%7B32.85-0%7D%7B9.81%7D%5C%5C%20t%3D3.35%5Bs%5D)
Now we can calculate the second time, but using as a initial velocity 32.85[m/s].
The final velocity will be:
![v_{f}^{2}= v_{o}^{2}+2*g*y\\v_{f}=\sqrt{v_{o}^{2}+2*g*y} \\v_{f}=\sqrt{32.85^{2}+2*9.81*55 } \\v_{f}=46.45[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%5E%7B2%7D%3D%20v_%7Bo%7D%5E%7B2%7D%2B2%2Ag%2Ay%5C%5Cv_%7Bf%7D%3D%5Csqrt%7Bv_%7Bo%7D%5E%7B2%7D%2B2%2Ag%2Ay%7D%20%5C%5Cv_%7Bf%7D%3D%5Csqrt%7B32.85%5E%7B2%7D%2B2%2A9.81%2A55%20%7D%20%5C%5Cv_%7Bf%7D%3D46.45%5Bm%2Fs%5D)
Now we can calculate the second time:
![t=\frac{46.45-32.85}{9.81} \\t= 1.386[s]](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B46.45-32.85%7D%7B9.81%7D%20%5C%5Ct%3D%201.386%5Bs%5D)
Note: The reason the second time is shorter even though it is the same distance is that the acceleration of gravity increases the speed of the rock more and more as it falls.