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:
Explanation:
Aceleration is a change on the velocity of the object in a given time.
For this case: the initial velocity is
and the final velocity is :
so, the change in velocity is:
and the change in time , according to the problem:
So, the aceleration is:
Answer:
(a). The initial velocity is 28.58m/s
(b). The speed when touching the ground is 33.3m/s.
Explanation:
The equations governing the position of the projectile are
where 
is the initial velocity.
(a).
When the projectile hits the 50m mark, 
; therefore,
solving for 
we get:
Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that
which gives
(b).
The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,
the vertical component of the velocity is
which gives a speed 
of
Answer:
Distance between peak height (vertically) of projectile and mountain height = (2975.2 - 1800) = 1175.2 m
Distance between where the projectile lands and ship B = (3188.8 - 3110) = 8.8 m
Explanation:
Given the velocity and angle of shot of the projectile, one can calculate the range and maximum height attained by the projectile.
H = (v₀² Sin²θ)/2g
v₀ = initial velocity of projectile = 2.50 × 10² m/s = 250 m/s
θ = 75°, g = 9.8 m/s²
H = 250² (Sin² 75)/(2 × 9.8) = 2975.2 m
Range of projectile
R = v₀² (sin2θ)/g
R = 250² (sin2×75)/9.8
R = 250² (sin 150)/9.8 = 3188.8 m
Height of mountain = 1.80 × 10³ = 1800 m
Maximum height of projectile = 2975.2 m
Distance between peak height (vertically) of projectile and mountain height = 2975.2 - 1800 = 1175.2 m
Distance of ship B from ship A = 2.5 × 10³ + 6.1 × 10² = 2500 + 610 = 3110 m
Range of projectile = 3188.8 m
Distance between where the projectile lands and ship B = 3188.8 - 3110 = 8.8 m
Answer:
, 
Explanation:
The jet is flying at constant velocity: this means that its acceleration is zero, so the net force acting on the jet is also zero.
Therefore, we can write:
where
is the thrust force generated by each engine of the jet
is the drag force
Solving for Fd,
The velocity of the jet is
So, the rate at which the drag force does work (which is the power) is
and substituting
we find
Converting into horsepower,
