A car rolls down a ramp in a parking garage. The horizontal position of the car in meters over time is shown below. Graph of ver

Answer:

The final size is nearly the same as the initial size because the increase in size is remarkably small

Solution:

According to the problem:

The proton beam energy is E = 250 GeV =

Distance traveled by the photon, d = 1 km = 1000 m

Proton mass,

Initial size of the wave packet,

Now,

This operates under relativistic principles

The rest mass energy for the proton is expressed as:

This proton energy is

Thus, the speed of the proton, v

The time to cover 1 km = 1000 m of distance is calculated as:

T =

T =

According to the dispersion factor;

Thus, the widening of the wave packet is relatively minor.

Hence, we can conclude that:

where

= final width

As the parachutist is descending at a steady rate

we can conclude that

Acceleration indicates the change in velocity

given the constant velocity in this scenario

Thus, in this situation, we find the acceleration to be zero

It’s understood from Newton's second law

where a is equal to 0

Here, the force due to gravity

Hence, we can deduce

therefore

as such the upward force is counteracted by the downward force.

Answer: 592.37m

Explanation:

Person D is represented by the blue line.

The total displacement is calculated by subtracting the initial position from the final position. Starting at (0,0), the path consists of moving down two blocks, then right six blocks, followed by moving up four blocks, and finally left one block.

Here, I consider the positive direction of the x-axis to the right and the positive direction of the y-axis as upward.

Thus, the new coordinates will be, with B representing a block:

P =(6*B - 1*B, -2*B + 4*B) = (5*B, 2*B)

Given that B = 110m

P = (550m, 220m)

The displacement corresponds to the length of the vector, since the change from the initial position (0,0) to P is simply P:

P = √(550^2 + 220^2) = 592.37m