A=vf-vi/tf-ti
a= velocity final-velocity initial / time final-time initial
a= 6 m/s - 3 m/s divided by 8 s - 0s
a= 2 m/s / 8 m/s
a= 1/4 m/s^2
"increments of 8" means the major divisions are 0,8,16,24 ?
<span>x axis, calculate the moment arms from 0 </span>
<span>3x4, 2x12, 1x20 </span>
<span>from an arbitrary C </span>
<span>3(c-4) + 2(c-12) + (c-20) = 0 </span>
<span>3c - 12 + 2c -24 + c - 20 = 0 </span>
<span>6c = 56 </span>
<span>c = 9.33 </span>
<span>y axis </span>
<span>3x3, 1x12, 2x20 </span>
<span>3(c-4) + 1(c-12) +2 (c-20) = 0 </span>
<span>3c - 12 + c - 12 + 2c - 40 = 0 </span>
<span>6c = 64 </span>
<span>c = 10.67 </span>
<span>so center is x = 9.33, y = 10.67 </span>
<span>5.82 x 10-49 joules7.62 x 10-19 joules8.77 x 10-12 joules1.09 x 10-12<span> joules </span><span>answer is b</span></span>
Explanation:
It is given that,
Mass of the object, m = 0.8 g = 0.0008 kg
Electric field, E = 534 N/C
Distance, s = 12 m
Time, t = 1.2 s
We need to find the acceleration of the object. It can be solved as :
m a = q E.......(1)
m = mass of electron
a = acceleration
q = charge on electron
"a" can be calculated using second equation of motion as :
a = 16.67 m/s²
Now put the value of a in equation (1) as :
q = 0.0000249 C
or
Hence, this is the required solution.