Answer:
Earths shadow covering the moon would create a lunar eclipse.
Explanation:
because i just know
Amplitude is the pair of vertical buttons, so to speak. Compressions are the bunched up vertical lines with the purple arrows pointing left and right. Rarefactions are purple arrows pointing down. Wavelength is crest to crest purple buttons. Associated LH and RH pointing arrows.
Each successive graph is at a later time. You can see from these graphs how the amplitude of the total electric field changes, but the positions of the crests and troughs (called antinodes) and places of zero field (called nodes) never change.!!!!!!!!!!!!!!!!!
Answer:
The discharge rate is ![Q = 0.0192 \ m^3 /s](https://tex.z-dn.net/?f=Q%20%3D%200.0192%20%5C%20%20m%5E3%20%2Fs)
Explanation:
From the question we are told that
The diameter is ![d = 60 \ mm = 0.06 \ m](https://tex.z-dn.net/?f=d%20%3D%20%2060%20%5C%20mm%20%20%20%3D%20%200.06%20%5C%20m)
The head is ![h = 6 \ m](https://tex.z-dn.net/?f=h%20%20%3D%20%206%20%5C%20m)
The coefficient of contraction is ![Cc = 0.68](https://tex.z-dn.net/?f=Cc%20%20%3D%20%200.68)
The coefficient of velocity is ![Cv = 0.92](https://tex.z-dn.net/?f=Cv%20%20%3D%20%200.92)
The radius is mathematically evaluated as
![r = \frac{d}{2}](https://tex.z-dn.net/?f=r%20%3D%20%20%5Cfrac%7Bd%7D%7B2%7D)
substituting values
![r = \frac{ 0.06 }{2}](https://tex.z-dn.net/?f=r%20%3D%20%20%5Cfrac%7B%200.06%20%7D%7B2%7D)
![r = 0.03 \ m](https://tex.z-dn.net/?f=r%20%3D%20%200.03%20%5C%20m)
The area is mathematically represented as
![A = \pi r^2](https://tex.z-dn.net/?f=A%20%3D%20%20%5Cpi%20r%5E2)
substituting values
![A = 3.142 * (0.03)^2](https://tex.z-dn.net/?f=A%20%3D%20%203.142%20%2A%20%20%280.03%29%5E2)
![A = 0.00283 \ m^2](https://tex.z-dn.net/?f=A%20%3D%200.00283%20%5C%20m%5E2)
The discharge rate is mathematically represented as
![Q = Cv *Cc * A * \sqrt{ 2 * g * h}](https://tex.z-dn.net/?f=Q%20%3D%20%20Cv%20%2ACc%20%20%2A%20%20A%20%20%2A%20%20%5Csqrt%7B%202%20%2A%20g%20%2A%20%20h%7D)
substituting values
![Q = 0.68 * 0.92* 0.00283 * \sqrt{ 2 * 9.8 * 6}](https://tex.z-dn.net/?f=Q%20%3D%200.68%20%2A%20%200.92%2A%20%20%200.00283%20%20%2A%20%20%5Csqrt%7B%202%20%2A%209.8%20%2A%20%206%7D)
![Q = 0.0192 \ m^3 /s](https://tex.z-dn.net/?f=Q%20%3D%200.0192%20%5C%20%20m%5E3%20%2Fs)
Answer:
The work done by friction was ![-4.5\times10^{5}\ J](https://tex.z-dn.net/?f=-4.5%5Ctimes10%5E%7B5%7D%5C%20J)
Explanation:
Given that,
Mass of car = 1000 kg
Initial speed of car =108 km/h =30 m/s
When the car is stop by brakes.
Then, final speed of car will be zero.
We need to calculate the work done by friction
Using formula of work done
![W=\Delta KE](https://tex.z-dn.net/?f=W%3D%5CDelta%20KE)
![W=K.E_{f}-K.E_{i}](https://tex.z-dn.net/?f=W%3DK.E_%7Bf%7D-K.E_%7Bi%7D)
![W=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{f}^2](https://tex.z-dn.net/?f=W%3D%5Cdfrac%7B1%7D%7B2%7Dmv_%7Bf%7D%5E2-%5Cdfrac%7B1%7D%7B2%7Dmv_%7Bf%7D%5E2)
Put the value of m and v
![W=0-\dfrac{1}{2}\times1000\times(30)^2](https://tex.z-dn.net/?f=W%3D0-%5Cdfrac%7B1%7D%7B2%7D%5Ctimes1000%5Ctimes%2830%29%5E2)
![W=-450000 \ J](https://tex.z-dn.net/?f=W%3D-450000%0A%5C%20J)
![W=-4.5\times10^{5}\ J](https://tex.z-dn.net/?f=W%3D-4.5%5Ctimes10%5E%7B5%7D%5C%20J)
Hence, The work done by friction was ![-4.5\times10^{5}\ J](https://tex.z-dn.net/?f=-4.5%5Ctimes10%5E%7B5%7D%5C%20J)