Answer: action forc roketorce
reaction force is engine fires
Subduction is, "<span>the sideways and downward movement of the edge of a plate of the earth's crust into the mantle beneath another plate." The basalt would most likely be swallowed up into the ground.
Hope this is what you were looking for! :)
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To solve this problem we will use the concepts related to Magnification. Magnification is the process of enlarging the apparent size, not physical size, of something. This enlargement is quantified by a calculated number also called "magnification".
The overall magnification of microscope is
![M = \frac{Nl}{f_ef_0}](https://tex.z-dn.net/?f=M%20%3D%20%5Cfrac%7BNl%7D%7Bf_ef_0%7D)
Where
N = Near point
l = distance between the object lens and eye lens
= Focal length
= Focal of eyepiece
Given that the minimum distance at which the eye is able to focus is about 25cm we have that N = 25cm
Replacing,
![M = \frac{25*10}{3*5}](https://tex.z-dn.net/?f=M%20%3D%20%5Cfrac%7B25%2A10%7D%7B3%2A5%7D)
![M = 16.67\approx 17\\](https://tex.z-dn.net/?f=M%20%3D%2016.67%5Capprox%2017%5C%5C)
Therefore the correct answer is C.
Answer:
a) t=24s
b) number of oscillations= 11
Explanation:
In case of a damped simple harmonic oscillator the equation of motion is
m(d²x/dt²)+b(dx/dt)+kx=0
Therefore on solving the above differential equation we get,
x(t)=A₀![e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)](https://tex.z-dn.net/?f=e%5E%7B%5Cfrac%7B-bt%7D%7B2m%7D%7Dcos%28w%27t%2B%5Cphi%29%3DA%28t%29cos%28w%27t%2B%5Cphi%29)
where A(t)=A₀
A₀ is the amplitude at t=0 and
is the angular frequency of damped SHM, which is given by,
![w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }](https://tex.z-dn.net/?f=w%27%3D%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D-%5Cfrac%7Bb%5E%7B2%7D%7D%7B4m%5E%7B2%7D%7D%20%7D)
Now coming to the problem,
Given: m=1.2 kg
k=9.8 N/m
b=210 g/s= 0.21 kg/s
A₀=13 cm
a) A(t)=A₀/8
⇒A₀
=A₀/8
⇒![e^{\frac{bt}{2m}}=8](https://tex.z-dn.net/?f=e%5E%7B%5Cfrac%7Bbt%7D%7B2m%7D%7D%3D8)
applying logarithm on both sides
⇒![\frac{bt}{2m}=ln(8)](https://tex.z-dn.net/?f=%5Cfrac%7Bbt%7D%7B2m%7D%3Dln%288%29)
⇒![t=\frac{2m*ln(8)}{b}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B2m%2Aln%288%29%7D%7Bb%7D)
substituting the values
![t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B2%2A1.2%2Aln%288%29%7D%7B0.21%7D%3D24s%28approx%29)
b) ![w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }](https://tex.z-dn.net/?f=w%27%3D%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D-%5Cfrac%7Bb%5E%7B2%7D%7D%7B4m%5E%7B2%7D%7D%20%7D)
![w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}](https://tex.z-dn.net/?f=w%27%3D%5Csqrt%7B%5Cfrac%7B9.8%7D%7B1.2%7D-%5Cfrac%7B0.21%5E%7B2%7D%7D%7B4%2A1.2%5E%7B2%7D%7D%7D%3D2.86s%5E%7B-1%7D)
, where
is time period of damped SHM
⇒![T'=\frac{2\pi}{2.86}=2.2s](https://tex.z-dn.net/?f=T%27%3D%5Cfrac%7B2%5Cpi%7D%7B2.86%7D%3D2.2s)
let
be number of oscillations made
then, ![nT'=t](https://tex.z-dn.net/?f=nT%27%3Dt)
⇒![n=\frac{24}{2.2}=11(approx)](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B24%7D%7B2.2%7D%3D11%28approx%29)