KE = 2000 J
Explanation:
KE = (1/2)mv^2
= (1/2)(0.100 kg)(200 m/s)^2
= 2000 J
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>
Answer:ummm ok
Explanation:I really don’t get it but ok
Answer:
k = 4422.35 KN/m
Explanation:
Given that
Frequency ,f= 29 Hz
m = 7.5 g
Natural frequency ω
ω = 2 π f
We also know that for spring mass system
ω ² m =k
k=Spring constant
So we can say that
( 2 π f)² = m k
By putting the values
(2 x π x 29)² = 7.5 x 10⁻³ k
33167.69 = 7.5 x 10⁻³ k
k=4422.35 x 10³ N/m
k = 4422.35 KN/m
Therefore spring constant will be 4422.35 KN/m
Given: Mass m = 60 Kg
Weight W = 96 N
Required: Acceleration due to gravity, g = ?
Formula: W = mg
g = W/g
g = 96 Kg.m/s²/60 Kg (note: this is the derive unit for Newton "N")
g = 1.6 m/s²
