Answer:
The final size is approximately equal to the initial size due to a very small relative increase of 
in its size
Solution:
As per the question:
The energy of the proton beam, E = 250 GeV =
Distance covered by photon, d = 1 km = 1000 m
Mass of proton, 
The initial size of the wave packet, 
Now,
This is relativistic in nature
The rest mass energy associated with the proton is given by:
This energy of proton is 
Thus the speed of the proton, v
Now, the time taken to cover 1 km = 1000 m of the distance:
T = 
T = 
Now, in accordance to the dispersion factor;
Thus the increase in wave packet's width is relatively quite small.
Hence, we can say that:
where
= final width
Answer:
(a). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².
(b). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².
Explanation:
Given that,
Electric field 
Electric field 
When the radius of curvature is greatest, the electric field at the surface will be smaller.
Where the radius of curvature is greatest
(a). We need to calculate the local surface charge density at the point on the surface
Using formula of charge density
Put the value into the formula
The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².
(b). We need to calculate the local surface charge density at the point on the surface where the radius of curvature of the surface is smallest
Using formula of charge density
Put the value into the formula
The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².
Hence, (a). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².
(b). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².
Change in Momentum = mv - mu.
u = 0, v = 10 m/s. Note ball accelerated from rest, so initial velocity = 0. u =0
Change in Momentum = mv - mu = 3*10 - 3*0 =30.
Change in Momentum = 30 kgm/s.
George's speed per second is 76.92 m/s
George's velocity in meters per second was 38.46 m/s (North)
<u>Explanation:</u>
When dividing the total distance covered by object by time, we get the value for average speed.
<u>Given:</u>
t = 13 seconds
Calculate the distance without consideration of motion’s direction. So, the distance walks 750 m north first, and then turn around walks 250 m south, so the total distance covered is
d = 750 + 250 = 1000 meters
To find the rate per second, simply divide the distance by the time.
In given case, the students walks 750 m north first, and then turn around walk 250 m south. The displacement is the distance in a straight line between the initial and final position: therefore, in this case, the displacement is
d = 750 (north) - 250 (south) = 500 m (north)
The time taken is t = 13 s
So, the average velocity is
(North)
And the direction is north (the same as the displacement).
Answer:
F = F2 - F1 where F is the net force upwards
F = 5 - (.5 * 9.8) = (5 - 4.9) N = .1 N
Since there is a net force in the upwards direction the rocket should experience an upwards acceleration.
