D. You’d need to experiment!
Sedimentary rock changes when it is exposed to extreme heat and pressure, but it can be caused by tectonic plate movement ( for heat ) and when the collide, you get the pressure.
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Answer:
Price=$7×10⁷
Explanation:
Step 1: Estimate the volume of the pile,
Step 2: Multiply it by the density to get its mass
Step 2: Then multiply the mass by the price per gram to get the total price
So
The average pile dimensions are=45.7×45×172.7
![V=3.6*10^{5}cm^{3}\\ m_{g}=V_{p}=3.6*10^{5}*19.3=7*10^{6}g\\](https://tex.z-dn.net/?f=V%3D3.6%2A10%5E%7B5%7Dcm%5E%7B3%7D%5C%5C%20%20m_%7Bg%7D%3DV_%7Bp%7D%3D3.6%2A10%5E%7B5%7D%2A19.3%3D7%2A10%5E%7B6%7Dg%5C%5C)
Price=m×$10
Price=(7×10⁶)×$10
Price=$7×10⁷
The starting angle θθ of a pendulum does not affect its period for θ<<1θ<<1. At higher angles, however, the period TT increases with increasing θθ.
The relation between TT and θθ can be derived by solving the equation of motion of the simple pendulum (from F=ma)
−gsinθ=lθ¨−gainθ=lθ¨
For small angles, θ≪1,θ≪1, and hence sinθ≈θsinθ≈θ. Hence,
θ¨=−glθθ¨=−glθ
This second-order differential equation can be solved to get θ=θ0cos(ωt),ω=gl−−√θ=θ0cos(ωt),ω=gl. The period is thus T=2πω=2πlg−−√T=2πω=2πlg, which is independent of the starting angle θ0θ0.
For large angles, however, the above derivation is invalid. Without going into the derivation, the general expression of the period is T=2πlg−−√(1+θ2016+...)T=2πlg(1+θ0216+...). At large angles, the θ2016θ0216 term starts to grow big and cause
To solve this problem we will apply the concepts related to load balancing. We will begin by defining what charges are acting inside and which charges are placed outside.
PART A)
The charge of the conducting shell is distributed only on its external surface. The point charge induces a negative charge on the inner surface of the conducting shell:
. This is the total charge on the inner surface of the conducting shell.
PART B)
The positive charge (of the same value) on the external surface of the conducting shell is:
![Q_{ext}=+Q_1=1.9*10^{-6} C](https://tex.z-dn.net/?f=Q_%7Bext%7D%3D%2BQ_1%3D1.9%2A10%5E%7B-6%7D%20C)
The driver's net load is distributed through its outer surface. When inducing the new load, the total external load will be given by,
![Q_{ext, Total}=Q_2+Q_{ext}](https://tex.z-dn.net/?f=Q_%7Bext%2C%20Total%7D%3DQ_2%2BQ_%7Bext%7D)
![Q_{ext, Total}=1.9+3.8](https://tex.z-dn.net/?f=Q_%7Bext%2C%20Total%7D%3D1.9%2B3.8)
![Q_{ext, Total}=5.7 \mu C](https://tex.z-dn.net/?f=Q_%7Bext%2C%20Total%7D%3D5.7%20%5Cmu%20C)