To solve the problem we will apply the concepts related to the Intensity as a function of the power and the area, as well as the electric field as a function of the current, the speed of light and the permeability in free space, as shown below.
The intensity of the wave at the receiver is




The amplitude of electric field at the receiver is


The amplitude of induced emf by this signal between the ends of the receiving antenna is


Here,
I = Current
= Permeability at free space
c = Light speed
d = Distance
Replacing,


Thus, the amplitude of induced emf by this signal between the ends of the receiving antenna is 0.0543V
<span>IDK the answer but I think it was "Ptolemy" whose geocentric model of the solar system was accepted for 1,400 years, and was embraced by the the Church until Galileo called it into question.</span>
Answer:
KE=800,000
Explanation:
The formula for kinetic energy is KE=1/2mv^2 or Kinetic Energy= 0.5*mass*velocity^2
so 1000 is the mass and 40 is the velocity
KE=0.5*1000*40^2
KE=0.5*1,000*1,600
KE=800,000 Joules
Answer:
The object will travel 675 m during that time.
Explanation:
A body moves with constant acceleration motion or uniformly accelerated rectilinear motion (u.a.r.m) when the path is a straight line, but the velocity is not necessarily constant because there is an acceleration.
In other words, a body performs a u.a.r.m when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases uniformly.
In this case, the position is calculated using the expression:
x = xo + vo*t + ½*a*t²
where:
- x0 is the initial position.
- v0 is the initial velocity.
- a is the acceleration.
- t is the time interval in which the motion is studied.
In this case:
- x0= 0
- v0= 0 because the object is initially stationary
- a= 6

- t= 15 s
Replacing:
x= 0 + 0*15 s + ½*6
*(15s)²
Solving:
x=½*6
*(15s)²
x=½*6
*225 s²
x= 675 m
<u><em>
The object will travel 675 m during that time.</em></u>
Answer:
The appropriate solution is "2.78 mm".
Explanation:
Given:

or,



or,

As we know,
Fringe width is:
⇒ 
hence,
Separation between second and third bright fringes will be:
⇒ 


or,
