Solution :
Energy of photon, E = 6.7 eV
E =
joule
Kinetic energy, 


Kinetic energy at high speeds


r - 1 = 7130
r = 7130 + 1
r = 7131


![$v^2=C^2\left[1-\left(\frac{1}{7131}\right)^2\right]$](https://tex.z-dn.net/?f=%24v%5E2%3DC%5E2%5Cleft%5B1-%5Cleft%28%5Cfrac%7B1%7D%7B7131%7D%5Cright%29%5E2%5Cright%5D%24)

Δ = 1 - 0.99999999017
= 0.00000000933
Relative mass, 

kg
Answer:
Yes, it's true. Computers do work that way. It's experienced by one of the authors of the book how computers work.
Explanation:
Answer:
Q/4πε0 [1/R - 1/√R2+d2]
Explanation:
Q/4πε0 [1/R - 1/√R2+d2] is the answer
explanation is attached.
toppr
1. Velocity at which the packet reaches the ground: 121.2 m/s
The motion of the packet is a uniformly accelerated motion, with constant acceleration
directed downward, initial vertical position
, and initial vertical velocity
. We can use the following SUVAT equation to find the final velocity of the packet after travelling for d=750 m:

substituting, we find

2. height at which the packet has half this velocity: 562.6 m
We need to find the heigth at which the packet has a velocity of

In order to do that, we use again the same SUVAT equation substituting
with this value, so that we find the new distance d that the packet travelled from the helicopter to reach this velocity:

Which means that the heigth of the packet was

the Orbital Velocity is the velocity sufficient to cause a natural or artificial satellite to remain in orbit. Inertia of the moving body tends to make it move on in a straight line, while gravitational force tends to pull it down. The orbital path, elliptical or circular, representing a balance between gravity and inertia, and it follows a rue that states that the more massive the body at the centre of attraction is, the higher is the orbital velocity for a particular altitude or distance.