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
Answer:
Distance, d = 0.1 m
It is given that,
Initial velocity of meson,
Finally, the meson is coming to rest v = 0
Acceleration of the meson, (opposite to initial velocity)
Using third equation of motion as :
s is the distance the meson travelled before coming to rest.
So,
s = 0.1 m
The meson will cover the distance of 0.1 m before coming to rest. Hence, this is the required solution.
It would be funny because . I will not be good
Answer:
v = 1.4 m /s
Explanation:
We shall apply law of conservation of mechanical energy
The kinetic energy of dart and block is converted into potential energy of both dart and block .
1 /2 (m+M) v² = ( m +M) gH
.5 x v² = 9.8 x .1
= v² = 1.96
v = 1.4
v = 1.4 m /s
Answer:
The distance represents the difference of the first position and last position of the body.
Explanation:
For example, if y axis represents the position axis, and the first position is 3, second 9 we can see that the distance is a (positive) projection of one position into another. 9-3=6
Hope this helps.
