Answer:
4 m/s
Explanation:
T = Tension
m = Mass of string
Velocity of wave in string is given by
For first cable
![v_1=\sqrt{\dfrac{T}{m}}=8](https://tex.z-dn.net/?f=v_1%3D%5Csqrt%7B%5Cdfrac%7BT%7D%7Bm%7D%7D%3D8)
For second cable
![v_2=\sqrt{\dfrac{T}{4m}}\\\Rightarrow v_2=\dfrac{1}{2}\sqrt{\dfrac{T}{m}}\\\Rightarrow v_2=\dfrac{1}{2}\times 8\\\Rightarrow v_2=4\ m/s](https://tex.z-dn.net/?f=v_2%3D%5Csqrt%7B%5Cdfrac%7BT%7D%7B4m%7D%7D%5C%5C%5CRightarrow%20v_2%3D%5Cdfrac%7B1%7D%7B2%7D%5Csqrt%7B%5Cdfrac%7BT%7D%7Bm%7D%7D%5C%5C%5CRightarrow%20v_2%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%208%5C%5C%5CRightarrow%20v_2%3D4%5C%20m%2Fs)
The speed of wave in cable two is 4 m/s
Answer:
![\boxed{ V_{2}= 0.03189 m^3}](https://tex.z-dn.net/?f=%5Cboxed%7B%20V_%7B2%7D%3D%200.03189%20m%5E3%7D)
Explanation:
According to Charles Law
=> ![\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}](https://tex.z-dn.net/?f=%5Cfrac%7BV_%7B1%7D%7D%7BT_%7B1%7D%7D%20%3D%20%5Cfrac%7BV_%7B2%7D%7D%7BT_%7B2%7D%7D)
Where
= 0.0279 m³,
= 280 K and
= 320 K
=> ![\frac{0.0279}{280} = \frac{V_{2}}{320}](https://tex.z-dn.net/?f=%5Cfrac%7B0.0279%7D%7B280%7D%20%3D%20%5Cfrac%7BV_%7B2%7D%7D%7B320%7D)
=>
= 0.03189 m³
No, light can't cause the butterfly's wing's to glow. Explanation: Butterflies contain wings which are of different colors for attractiveness and also to show warning alarm for it's predator.
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Answer:
The dog catches up with the man 6.1714m later.
Explanation:
The first thing to take into account is the speed formula. It is
, where v is speed, d is distance and t is time. From this formula, we can get the distance formula by finding d, it is ![d=v\cdot t](https://tex.z-dn.net/?f=d%3Dv%5Ccdot%20t)
Now, the distance equation for the man would be:
![d_{man}=v_{man}\cdot t=1.6\cdot t](https://tex.z-dn.net/?f=d_%7Bman%7D%3Dv_%7Bman%7D%5Ccdot%20t%3D1.6%5Ccdot%20t)
The distance equation for the dog would be obtained by the same way with just a little detail. The dog takes off running 1.8s after the man did. So, in the equation we must subtract 1.8 from t.
![d_{dog}=v_{dog}\cdot (t-1.8)=3\cdot (t-1.8)](https://tex.z-dn.net/?f=d_%7Bdog%7D%3Dv_%7Bdog%7D%5Ccdot%20%28t-1.8%29%3D3%5Ccdot%20%28t-1.8%29)
For a better understanding, at t=1.8 the dog must be in d=0. Let's verify:
![d_{dog}=v_{dog}\cdot (1.8-1.8)=3\cdot (0)=0](https://tex.z-dn.net/?f=d_%7Bdog%7D%3Dv_%7Bdog%7D%5Ccdot%20%281.8-1.8%29%3D3%5Ccdot%20%280%29%3D0)
Now, for finding how far they have each traveled when the dog catches up with the man we must match the equations of each one.
![d_{man}=d_{dog}](https://tex.z-dn.net/?f=d_%7Bman%7D%3Dd_%7Bdog%7D)
![1.6\cdot t=3\cdot (t-1.8)](https://tex.z-dn.net/?f=1.6%5Ccdot%20t%3D3%5Ccdot%20%28t-1.8%29)
![1.6\cdot t=3\cdot t-5.4](https://tex.z-dn.net/?f=1.6%5Ccdot%20t%3D3%5Ccdot%20t-5.4)
![1.4\cdot t=5.4](https://tex.z-dn.net/?f=1.4%5Ccdot%20t%3D5.4)
![t=\frac{5.4}{1.4}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B5.4%7D%7B1.4%7D)
![t=3.8571s](https://tex.z-dn.net/?f=t%3D3.8571s)
The result obtained previously means that the dog catches up with the man 3.8571s after the man started running.
That value is used in the man's distance equation.
![d_{man}=1.6\cdot t=1.6\cdot (3.8571)](https://tex.z-dn.net/?f=d_%7Bman%7D%3D1.6%5Ccdot%20t%3D1.6%5Ccdot%20%283.8571%29)
![d_{man}=6.1714m](https://tex.z-dn.net/?f=d_%7Bman%7D%3D6.1714m)
Finally, the dog catches up with the man 6.1714m later.