Answer:
3×10^-12 C
Explanation:
The total of the three charges is ...
(-3 +8 +4)×10^-12 C = 9×10^-12 C
Assuming the charge is equally distributed between the balls when they are brought in contact, the charge on each ball will be ...
(9/3)×10^-12 C = 3×10^-12 C
The strength of electric field E is 17 N / C.
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<u>Explanation:</u>
Electric field strength is defined as the force per unit charge acting at a point in the given field. The equation for the strength of the electric field is given by
E = F / q
where E represents the electric field strength,
F represents the force in newton,
q represents the charge in coulomb.
Given the charge q = 0.30 coulombs
force F = 5.0 N
Electric field strength E = force / charge
= 5.0 / 0.30
E = 16.66 = 17 N / C.
Answer:
Explanation:
Given that,
Initial angular velocity is 0
ωo=0rad/s
It has angular velocity of 11rev/sec
ωi=11rev/sec
1rev=2πrad
Then, wi=11rev/sec ×2πrad
wi=22πrad/sec
And after 30 revolution
θ=30revolution
θ=30×2πrad
θ=60πrad
Final angular velocity is
ωf=18rev/sec
ωf=18×2πrad/sec
ωf=36πrad/sec
a. Angular acceleration(α)
Then, angular acceleration is given as
wf²=wi²+2αθ
(36π)²=(22π)²+2α×60π
(36π)²-(22π)²=120πα
Then, 120πα = 8014.119
α=8014.119/120π
α=21.26 rad/s²
Let. convert to revolution /sec²
α=21.26/2π
α=3.38rev/sec
b. Time Taken to complete 30revolution
θ=60πrad
∆θ= ½(wf+wi)•t
60π=½(36π+22π)t
60π×2=58πt
Then, t=120π/58π
t=2.07seconds
c. Time to reach 11rev/sec
wf=wo+αt
22π=0+21.26t
22π=21.26t
Then, t=22π/21.26
t=3.251seconds
d. Number of revolution to get to 11rev/s
∆θ= ½(wf+wo)•t
∆θ= ½(0+11)•3.251
∆θ= ½(11)•3.251
∆θ= 17.88rev.
Kinetic energy has nothing to do with anything other than motion of the particle.
When a particle with velocity v collides another particle(suppose it is at rest for simplication), assuming that there is perfectly elastic collision between them, the velocity of particle which was at rest becomes mv/M ( assuming mass of particle in motion to be m and at rest to be M) from convervation of linear momentum. And all this transfer of energy happens in a fraction of seconds which is not visible to naked eyes.
Hence 1st option is correct!