Answer: 60
Explanation: if u hit it it will have impact and the impact added 10n
Answer:
z = 0.8 (approx)
Explanation:
given,
Amplitude of 1 GHz incident wave in air = 20 V/m
Water has,
μr = 1
at 1 GHz, r = 80 and σ = 1 S/m.
depth of water when amplitude is down to 1 μV/m
Intrinsic impedance of air = 120 π Ω
Intrinsic impedance of water = ![\dfrac{120\pi}{\epsilon_r}](https://tex.z-dn.net/?f=%5Cdfrac%7B120%5Cpi%7D%7B%5Cepsilon_r%7D)
Using equation to solve the problem
![E(z) = E_0 e^{-\alpha\ z}](https://tex.z-dn.net/?f=E%28z%29%20%3D%20E_0%20e%5E%7B-%5Calpha%5C%20z%7D)
E(z) is the amplitude under water at z depth
E_o is the amplitude of wave on the surface of water
z is the depth under water
![\alpha = \dfrac{\sigma}{2}\sqrt{\dfrac{(120\pi)^2}{\Epsilon_r}}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cdfrac%7B%5Csigma%7D%7B2%7D%5Csqrt%7B%5Cdfrac%7B%28120%5Cpi%29%5E2%7D%7B%5CEpsilon_r%7D%7D)
![\alpha = \dfrac{1}{2}\sqrt{\dfrac{(120\pi)^2}{80}}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%5Csqrt%7B%5Cdfrac%7B%28120%5Cpi%29%5E2%7D%7B80%7D%7D)
![\alpha =21.07\ Np/m](https://tex.z-dn.net/?f=%5Calpha%20%3D21.07%5C%20Np%2Fm)
now ,
![1 \times 10^{-6} = 20 e^{-21.07\times z}](https://tex.z-dn.net/?f=1%20%5Ctimes%2010%5E%7B-6%7D%20%3D%2020%20e%5E%7B-21.07%5Ctimes%20z%7D)
![e^{21.07\times z}= 20\times 10^{6}](https://tex.z-dn.net/?f=e%5E%7B21.07%5Ctimes%20z%7D%3D%2020%5Ctimes%2010%5E%7B6%7D)
taking ln both side
21.07 x z = 16.81
z = 0.797
z = 0.8 (approx)
Answer:
8.95 m/s
Explanation:
(3.5-1.0) x 2 = gt²
t = √5/9.8 = 0.714 sec
= X/t = 4/ 0.714 = 5.602 m/s
vertical velocity means it has to rise 1.0 m against gravity in 0.714s
= gt = 9.8 x 0.714 = 6.99 m/s
V = √
²+
² = √6.99²+5.602² = 8.95 m/s
Answer:
Your answer should be B.
If your internal energy or <em><u>heat</u></em> of a system <em><u>decreases</u></em> then you know that you are taking that energy out and doing <em><u>work.</u></em>