Because they behave just like all the electromagnetic waves of the spectrum. Same equations, just shorter wavelengths and more energy.
Hope you get it :)
if rebecca stands in front of david at a distance of do and david perceives the position of rebecca at di, di will be +84 cm
<h3>What is focal length ?</h3>
How strongly light converges or diverges depends on an optical system's focal length, which is the inverse of optical power. A system with a positive focus length is said to converge light, whereas one with a negative focal length is said to diverge light.
focal length = +60 cm
magnification m = -0.40
focal length being positive an magnification negative.
given lens is a convex lens.
for a lens
m = di/do and 1/f = (1/di) - (1/do)di
= -0.4do1/f = (1/-0.4do) - 1/do do
= -210 cmdi = -0.4 * -210
di = +84 cm
To learn more about focal length go to - brainly.com/question/25779311
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Answer:
a)![t=5.5\mu s](https://tex.z-dn.net/?f=t%3D5.5%5Cmu%20s)
b)![\Delta t'=12.5\mu s](https://tex.z-dn.net/?f=%5CDelta%20t%27%3D12.5%5Cmu%20s)
Explanation:
a) Let's use the constant velocity equation:
![v=\frac{\Delta x}{\Delta t}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B%5CDelta%20x%7D%7B%5CDelta%20t%7D)
- v is the speed of the muon. 0.9*c
- c is the speed of light 3*10⁸ m/s
![t=\frac{\Delta x}{v}=\frac{1500}{0.9*(3*10^{8})}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B%5CDelta%20x%7D%7Bv%7D%3D%5Cfrac%7B1500%7D%7B0.9%2A%283%2A10%5E%7B8%7D%29%7D)
![t=5.5\mu s](https://tex.z-dn.net/?f=t%3D5.5%5Cmu%20s)
b) Here we need to use Lorentz factor because the speed of the muon is relativistic. Hence the time in the rest frame is the product of the Lorentz factor times the time in the inertial frame.
![\Delta t'=\gamma\Delta t](https://tex.z-dn.net/?f=%5CDelta%20t%27%3D%5Cgamma%5CDelta%20t)
![\gamma =\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}](https://tex.z-dn.net/?f=%5Cgamma%20%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B1-%5Cfrac%7Bv%5E%7B2%7D%7D%7Bc%5E%7B2%7D%7D%7D%7D)
v is the speed of muon (0.9c)
Therefore the time in the rest frame will be:
![\Delta t'=\frac{1}{\sqrt{1-\frac{(0.9c)^{2}}{c^{2}}}}\Delta t](https://tex.z-dn.net/?f=%5CDelta%20t%27%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B1-%5Cfrac%7B%280.9c%29%5E%7B2%7D%7D%7Bc%5E%7B2%7D%7D%7D%7D%5CDelta%20t)
![\Delta t'=\frac{1}{\sqrt{1-0.9^{2}}}\Delta t](https://tex.z-dn.net/?f=%5CDelta%20t%27%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B1-0.9%5E%7B2%7D%7D%7D%5CDelta%20t)
![\Delta t'=\frac{1}{0.44}\Delta t](https://tex.z-dn.net/?f=%5CDelta%20t%27%3D%5Cfrac%7B1%7D%7B0.44%7D%5CDelta%20t)
No we use the value of Δt calculated in a)
![\Delta t'=2.27*5.5=12.5\mu s](https://tex.z-dn.net/?f=%5CDelta%20t%27%3D2.27%2A5.5%3D12.5%5Cmu%20s)
I hope it helps you!
#16
If we put a resistor in circuit it will slow the speed of current
Let's check ohms law
![\\ \rm\Rrightarrow \dfrac{V}{I}=R](https://tex.z-dn.net/?f=%5C%5C%20%5Crm%5CRrightarrow%20%5Cdfrac%7BV%7D%7BI%7D%3DR)
- So if resistance is more current is less
#17
Again use ohms law
![\\ \rm\Rrightarrow V=IR](https://tex.z-dn.net/?f=%5C%5C%20%5Crm%5CRrightarrow%20V%3DIR)
![\\ \rm\Rrightarrow V\propto I](https://tex.z-dn.net/?f=%5C%5C%20%5Crm%5CRrightarrow%20V%5Cpropto%20I)
- Voltage must be increased
8 ∙ 10^-4 / 2 ∙ 10^2 = (8/2) ∙ ((10^-4)*(10^-2)) = <span>4 ∙ 10^-6</span>