Answer:
37.125 m
Explanation:
Using the equation of motion
s=ut+0.5at^{2} where s is distance, u is initial velocity, t is time and a is acceleration
<u>Distance during acceleration</u>
Acceleration, a=\frac {V_{final}-V_{initial}}{t} where V_{final} is final velocity and V_{initial} is initial velocity.
Substituting 0.0 m/s for initial velocity and 4.5 m/s for final velocity, acceleration will be
a=\frac {4.5 m/s-0 m/s}{4.5 s}=1 m/s^{2}
Then substituting u for 0 m/s, t for 4.5 s and a for 1 m/s^{2} into the equation of motion
s=0*4.5+ 0.5*1*4.5^{2}=0+10.125
=10.125 m
<u>Distance at a constant speed</u>
At a constant speed, there's no acceleration and since speed=distance/time then distance is speed*time
Distance=4.5 m/s*6 s=27 m
<u>Total distance</u>
Total=27+10.125=37.125 m
Answer:
30ms
Explanation:
you need to multiple the 10ms by 3s which gives you 30ms
Answer:
480J
Explanation:
Using the formula:
Delta U = Q - W
Q:Heat (J)
Delta U: Changes in internal Energy (J)
W:Work (J)
We can plug in the give numbers, Q and W.
Delta U = 658J - 178J = 480J
Answer:
(a) 135 kV
(b) The charge chould be moved to infinity
Explanation:
(a)
The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by
where 
Difference in potential between the points is
![kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}](https://tex.z-dn.net/?f=kQ%5Cleft%5B-%5Cdfrac%7B1%7D%7B0.2%5Ctext%7B%20m%7D%7D%20-%5Cleft%28%20-%5Cdfrac%7B1%7D%7B0.1%5Ctext%7B%20m%7D%7D%5Cright%29%5Cright%5D%20%3D%20%5Cdfrac%7BkQ%7D%7B0.2%5Ctext%7B%20m%7D%7D%20%3D%20%5Cdfrac%7B9%5Ctimes10%5E9%5Ctext%7B%20F%2Fm%7D%5Ctimes3%5Ctimes10%5E%7B-6%7D%5Ctext%7B%20C%7D%7D%7B0.2%5Ctext%7B%20m%7D%7D)
(b)
If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.
![270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]](https://tex.z-dn.net/?f=270%5Ctimes10%5E3%20%3D%20kQ%5Cleft%5B-%5Cdfrac%7B1%7D%7Bx%7D-%5Cleft%28-%5Cdfrac%7B1%7D%7B0.1%5Ctext%7B%20m%7D%7D%5Cright%29%5Cright%5D)
The charge chould be moved to infinity
Answer:
16.9000000000000001 J
Explanation:
From the given information:
Let the initial kinetic energy from point A be 
= 1.9000000000000001 J
and the final kinetic energy from point B be 
= ???
The charge particle Q = 6 mC = 6 × 10⁻³ C
The change in the electric potential from point B to A;
i.e. V_B - V_A = -2.5 × 10³ V
According to the work-energy theorem:
-Q × ΔV = ΔK
