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>
-Slow down
-Speed up
-Turning
Hope this helps
The answer is 9.8 ms^-2, because there is only one force acting on the object so the acceleration will be numerically equal to the gravitational field strength.
Answer:
Explanation:
Given
Two masses 
and 
is released and there is tension T in the string
Suppose a is the acceleration of the system
Therefore from Diagram
For
------1
for m_2 body
-------2
From above two Equation it is said that Tension is greater than m_1g and less than m_2g
