Answer:
(a) 62.69 nJ/m^3
(b) 1015.22 μJ/m^3
Explanation:
Electric field, E = 119 V/m
Magnetic field, B = 5.050 x 10^-5 T
(a) Energy density of electric field = 
= 6.269 x 10^-8 J/m^3 = 62.69 nJ/m^3
(b) energy density of magnetic field = 
= 1.01522 x 10^-3 J/m^3 = 1015.22 μJ/m^3
Let us situate this on the x axis, and let our uniform line of charge be positioned on the interval <span>(−L,0]</span> for some large number L. The voltage V as a function of x on the interval <span>(0,∞)</span> is given by integrating the contributions from each bit of charge. Let the charge density be λ. Thus, for an infinitesimal length element <span>d<span>x′</span></span>, we have <span>λ=<span><span>dq</span><span>d<span>x′</span></span></span></span>.<span>V(x)=<span>1/<span>4π<span>ϵ0</span></span></span><span>∫line</span><span><span>dq/</span>r</span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>∫<span>−L</span>0</span><span><span>d<span>x/</span></span><span>x−<span>x′</span></span></span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>(ln|x+L|−ln|x|)</span></span>
Given:
u(initial velocity)=0
a=5.54m/s^2
v(final velocity)=2 m/s
v=u +at
Where v is the final velocity.
u is the initial velocity
a is the acceleration.
t is the time
2=0+5.54t
t=2/5.54
t=0.36 sec
Answer:
After one half of the battery's useful life.
Explanation:
Batteries of the emergency locator transmitter (ELT) must be replaced or recharged after one half of the battery's useful life because if it is exposed to the high temperature for a long period of time such as the air plane parked in the sun will result in the deterioration of battery which may makes the transmitter out of order before the expiry date of the battery. So it will be safe to do that after the use of one half of the battery's life.
