Answer:
The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Explanation:
Given that,
Amplitude = 0.08190 m
Frequency = 2.29 Hz
Wavelength = 1.87 m
(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave
Using formula of distance
![d=2A](https://tex.z-dn.net/?f=d%3D2A)
Where, d = distance
A = amplitude
Put the value into the formula
![d=2\times0.08190](https://tex.z-dn.net/?f=d%3D2%5Ctimes0.08190)
![d=0.1638\ m](https://tex.z-dn.net/?f=d%3D0.1638%5C%20m)
Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
I think its Coulomb's law<span>
</span>
<span>The moon is smaller and more dense than the Earth, and has less extreme temperature changes. The statement presented is True. In terms of temperature, since there is no atmosphere on the moon, then it has less extreme temperature changes. The moon can reach 253 Fahrenheit in the day and -387 Fahrenheit at night.</span>
Charles Law
Explanation:
Step 1:
It is given that the original volume of the gas is 250 ml at 300 K temperature and 1 atmosphere pressure. We need to find the volume of the same gas when the temperature is 350 K and 1 atmosphere pressure.
Step 2:
We observe that the gas pressure is the same in both the cases while the temperature is different. So we need a law that explains the volume change of a gas when temperature is changed, without any change to the pressure.
Step 3:
Charles law provides the relationship between the gas volume and temperature, at a given pressure
Step 4:
Hence we conclude that Charles law can be used.
Answer: ![1.13(10)^{3} Pa](https://tex.z-dn.net/?f=1.13%2810%29%5E%7B3%7D%20Pa)
Explanation:
This problem can be solved by the following equation:
![\Delta P=\frac{8 \eta L Q}{\pi r^{4}}](https://tex.z-dn.net/?f=%5CDelta%20P%3D%5Cfrac%7B8%20%5Ceta%20L%20Q%7D%7B%5Cpi%20r%5E%7B4%7D%7D)
Where:
is the pressure difference between the two ends of the pipe
is the viscosity of oil
is the length of the pipe
is the Rate of flow of the fluid
is the diameter of the pipe
is the radius of the pipe
Soving for
:
![\Delta P=\frac{8 (0.20 Pa.s)(2600 m)(0.0009 \frac{m^{3}}{s})}{\pi (0.18 m)^{4}}](https://tex.z-dn.net/?f=%5CDelta%20P%3D%5Cfrac%7B8%20%280.20%20Pa.s%29%282600%20m%29%280.0009%20%5Cfrac%7Bm%5E%7B3%7D%7D%7Bs%7D%29%7D%7B%5Cpi%20%280.18%20m%29%5E%7B4%7D%7D)
Finally:
![\Delta P=1135.26 Pa \approx 1.13(10)^{3} Pa](https://tex.z-dn.net/?f=%5CDelta%20P%3D1135.26%20Pa%20%5Capprox%201.13%2810%29%5E%7B3%7D%20Pa)