Answer:
Based on its mass, the sun's gravitational attraction to the Earth is more than 177 times greater than that of the moon to the Earth.
Answer:W = 1.23×10^-6BTU
Explanation: Work = Surface tension × (A1 - A2)
W= Surface tension × 3.142 ×(D1^2 - D2^2)
Where A1= Initial surface area
A2= final surface area
Given:
D1=0.5 inches , D2= 3 inches
D1= 0.5 × (1ft/12inches)
D1= 0.0417 ft
D2= 3 ×(1ft/12inches)
D2= 0.25ft
Surface tension = 0.005lb ft^-1
W = [(0.25)^2 - (0.0417)^2]
W = 954 ×10^6lbf ft × ( 1BTU/778lbf ft)
W = 1.23×10^-6BTU
Explanation:
At point B, the velocity speed of the train is as follows.
![\nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})](https://tex.z-dn.net/?f=%5Cnu%5E%7B2%7D_%7BB%7D%20%3D%20%5Cnu%5E%7B2%7D_%7BA%7D%20%2B%202a_%7Bt%7D%20%28s_%7BB%7D%20-%20s_%7BA%7D%29)
= ![(30)^{2} + 2(-0.25(412 - 0))](https://tex.z-dn.net/?f=%2830%29%5E%7B2%7D%20%2B%202%28-0.25%28412%20-%200%29%29)
= 26.34 m/s
Now, we will calculate the first derivative of the equation of train.
y = ![200 e^{\frac{x}{1000}}](https://tex.z-dn.net/?f=200%20e%5E%7B%5Cfrac%7Bx%7D%7B1000%7D%7D)
![\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D%200.2%20e%5E%7B%5Cfrac%7Bx%7D%7B1000%7D%7D)
Now, second derivative of the train is calculated as follows.
Radius of curvature of the train is as follows.
![\rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}](https://tex.z-dn.net/?f=%5Crho%20%3D%20%5Cfrac%7B%5B1%20%2B%20%28%5Cfrac%7Bdy%7D%7Bdx%7D%29%5E%7B2%7D%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%7B%5Cfrac%7Bd%5E%7B2%7Dy%7D%7Bdx%5E%7B2%7D%7D%7D)
= ![\frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B1%20%2B%200.2e%5E%7B%5Cfrac%7B400%7D%7B1000%7D%7D%5E%7B2%7D%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%7B0.2%2810%5E%7B-3%7D%29e%5E%7B%5Cfrac%7B400%7D%7B1000%7D%7D%7D)
= 3808.96 m
Now, we will calculate the normal component of the train as follows.
![a_{n} = \frac{\nu^{2}_{B}}{\rho}](https://tex.z-dn.net/?f=a_%7Bn%7D%20%3D%20%5Cfrac%7B%5Cnu%5E%7B2%7D_%7BB%7D%7D%7B%5Crho%7D)
= ![\frac{(26.34)^{2}}{3808.96}](https://tex.z-dn.net/?f=%5Cfrac%7B%2826.34%29%5E%7B2%7D%7D%7B3808.96%7D)
= 0.1822 ![m/s^{2}](https://tex.z-dn.net/?f=m%2Fs%5E%7B2%7D)
The magnitude of acceleration of train is calculated as follows.
a = ![\sqrt{(a_{t})^{2} + (a_{n})^{2}}](https://tex.z-dn.net/?f=%5Csqrt%7B%28a_%7Bt%7D%29%5E%7B2%7D%20%2B%20%28a_%7Bn%7D%29%5E%7B2%7D%7D)
= ![\sqrt{(-0.25)^{2} + (0.1822)^{2}}](https://tex.z-dn.net/?f=%5Csqrt%7B%28-0.25%29%5E%7B2%7D%20%2B%20%280.1822%29%5E%7B2%7D%7D)
= ![0.309 m/s^{2}](https://tex.z-dn.net/?f=0.309%20m%2Fs%5E%7B2%7D)
Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is
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Hi, this sounds like a chemistry question:
If you wanted to separate sand from iron fillings for example, using tweezers would be a great tool to do this, depending on the size of the iron fillings.