Los Angeles lies on the Pacific plate, San Francisco lies on the North American plate, and the meeting point of the two cities is mathematically given as
T = 120 x 105 years
<h3>What is the meeting point of the two plates?</h3>
Generally, the equation for Distance is mathematically given as
D = Rate x Time
Therefore
T = D/R
T = (600 x 105) / 5
T = 120 x 105 years
In conclusion, the meeting point of the two plates will be
T = 120 x 105 years
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Answer:
The inductance of the inductor is 35.8 mH
Explanation:
Given that,
Voltage = 120-V
Frequency = 1000 Hz
Capacitor ![C= 2.00\mu F](https://tex.z-dn.net/?f=C%3D%202.00%5Cmu%20F)
Current = 0.680 A
We need to calculate the inductance of the inductor
Using formula of current
![I = \dfrac{V}{Z}](https://tex.z-dn.net/?f=I%20%3D%20%5Cdfrac%7BV%7D%7BZ%7D)
![Z=\sqrt{R^2+(L\omega-\dfrac{1}{C\omega})^2}](https://tex.z-dn.net/?f=Z%3D%5Csqrt%7BR%5E2%2B%28L%5Comega-%5Cdfrac%7B1%7D%7BC%5Comega%7D%29%5E2%7D)
Put the value of Z into the formula
![I=\dfrac{V}{\sqrt{R^2+(L\omega-\dfrac{1}{C\omega})^2}}](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7BV%7D%7B%5Csqrt%7BR%5E2%2B%28L%5Comega-%5Cdfrac%7B1%7D%7BC%5Comega%7D%29%5E2%7D%7D)
Put the value into the formula
![0.680=\dfrac{120}{\sqrt{(100)^2+(L\times2\pi\times1000-\dfrac{1}{2\times10^{-6}\times2\pi\times1000})^2}}](https://tex.z-dn.net/?f=0.680%3D%5Cdfrac%7B120%7D%7B%5Csqrt%7B%28100%29%5E2%2B%28L%5Ctimes2%5Cpi%5Ctimes1000-%5Cdfrac%7B1%7D%7B2%5Ctimes10%5E%7B-6%7D%5Ctimes2%5Cpi%5Ctimes1000%7D%29%5E2%7D%7D)
![L=35.8\ mH](https://tex.z-dn.net/?f=L%3D35.8%5C%20mH)
Hence, The inductance of the inductor is 35.8 mH
Explanation:
Given that,
Radius R= 2.00
Charge = 6.88 μC
Inner radius = 4.00 cm
Outer radius = 5.00 cm
Charge = -2.96 μC
We need to calculate the electric field
Using formula of electric field
![E=\dfrac{kq}{r^2}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7Bkq%7D%7Br%5E2%7D)
(a). For, r = 1.00 cm
Here, r<R
So, E = 0
The electric field does not exist inside the sphere.
(b). For, r = 3.00 cm
Here, r >R
The electric field is
![E=\dfrac{kq}{r^2}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7Bkq%7D%7Br%5E2%7D)
Put the value into the formula
![E=\dfrac{9\times10^{9}\times6.88\times10^{-6}}{(3.00\times10^{-2})^2}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7B9%5Ctimes10%5E%7B9%7D%5Ctimes6.88%5Ctimes10%5E%7B-6%7D%7D%7B%283.00%5Ctimes10%5E%7B-2%7D%29%5E2%7D)
![E=6.88\times10^{7}\ N/C](https://tex.z-dn.net/?f=E%3D6.88%5Ctimes10%5E%7B7%7D%5C%20N%2FC)
The electric field outside the solid conducting sphere and the direction is towards sphere.
(c). For, r = 4.50 cm
Here, r lies between R₁ and R₂.
So, E = 0
The electric field does not exist inside the conducting material
(d). For, r = 7.00 cm
The electric field is
![E=\dfrac{kq}{r^2}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7Bkq%7D%7Br%5E2%7D)
Put the value into the formula
![E=\dfrac{9\times10^{9}\times(-2.96\times10^{-6})}{(7.00\times10^{-2})^2}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7B9%5Ctimes10%5E%7B9%7D%5Ctimes%28-2.96%5Ctimes10%5E%7B-6%7D%29%7D%7B%287.00%5Ctimes10%5E%7B-2%7D%29%5E2%7D)
![E=5.43\times10^{6}\ N/C](https://tex.z-dn.net/?f=E%3D5.43%5Ctimes10%5E%7B6%7D%5C%20N%2FC)
The electric field outside the solid conducting sphere and direction is away of solid sphere.
Hence, This is the required solution.
![\huge\underline{\underline{\boxed{\mathbb {EXPLANATION}}}}](https://tex.z-dn.net/?f=%5Chuge%5Cunderline%7B%5Cunderline%7B%5Cboxed%7B%5Cmathbb%20%7BEXPLANATION%7D%7D%7D%7D)
The heat capacity is given by the expression:
![\longrightarrow \sf{\triangle Q= m \triangle C \triangle T}](https://tex.z-dn.net/?f=%5Clongrightarrow%20%5Csf%7B%5Ctriangle%20Q%3D%20m%20%5Ctriangle%20C%20%20%5Ctriangle%20%20%20T%7D)
![\longrightarrow \sf{Q= \: Heat}](https://tex.z-dn.net/?f=%5Clongrightarrow%20%5Csf%7BQ%3D%20%5C%3A%20Heat%7D)
![\longrightarrow \sf{M= \: Mass}](https://tex.z-dn.net/?f=%5Clongrightarrow%20%5Csf%7BM%3D%20%5C%3A%20Mass%7D)
![\longrightarrow \sf{C= \: Specific \: Heat}](https://tex.z-dn.net/?f=%5Clongrightarrow%20%5Csf%7BC%3D%20%5C%3A%20Specific%20%5C%3A%20Heat%7D)
![\longrightarrow \sf{T= \: Temperature}](https://tex.z-dn.net/?f=%5Clongrightarrow%20%5Csf%7BT%3D%20%5C%3A%20Temperature%7D)
![\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}](https://tex.z-dn.net/?f=%5Chuge%5Cunderline%7B%5Cunderline%7B%5Cboxed%7B%5Cmathbb%20%7BANSWER%3A%7D%7D%7D%7D)
When the
is measured in the calorimeter, we obtain a value, and since we know the mass of the material and we control the change in
, we can then determine the specific heat "C" by simply remplazing in the expression.
DIAMONDS are NOT rocks. Diamonds are minerals, rocks are made up of many different fragments of minerals.