The answer to this question is dropping it on a hard surface.
A) travel outside the necleus
Answer:
q=3.5*10^-4
Explanation:
<u>concept:</u>
The force acting on both charges is given by the coulomb law:
F=kq1q2/r^2
the centripetal force is given by:
Fc=mv^2/r
The kinetic energy is given by:
KE=1/2mv^2
<u>The tension force:</u>
<u><em>when the plane is uncharged </em></u>
T=mv^2/r
T=2(K.E)/r
T=2(50 J)/r
T=100/r
<u><em>when the plane is charged </em></u>
T+k*|q|^2/r^2=2(K.E)charged/r
100/r+k*|q|^2/r^2=2(53.5 J)/r
q=√(2r[53.5 J-50 J]/k) √= square root on whole
q=√2(2)(53.5 J-50 J)/8.99*10^9
q=3.5*10^-4
The product of √30 and √610 is 10√183.
√30 = √(2×3×5)
and √610 = √(2×5×61
Since 61 can't be factorised further.
Therefore, the value of √30×√610 is
= √(2×3×5×2×5×61)
= 2×5×√(3×61)
=10√183