Answer:
120 m
Explanation:
Given:
wavelength 'λ' = 2.4m
pulse width 'τ'= 100T ('T' is the time of one oscillation)
The below inequality express the range of distances to an object that radar can detect
τc/2 < x < Tc/2 ---->eq(1)
Where, τc/2 is the shortest distance
First we'll calculate Frequency 'f' in order to determine time of one oscillation 'T'
f = c/λ (c= speed of light i.e 3 x
m/s)
f= 3 x
/ 2.4
f=1.25 x
hz.
As, T= 1/f
time of one oscillation T= 1/1.25 x
T= 8 x
s
It was given that pulse width 'τ'= 100T
τ= 100 x 8 x
=> 800 x
s
From eq(1), we can conclude that the shortest distance to an object that this radar can detect:
= τc/2 => (800 x
x 3 x
)/2
=120m
Answer:
Part 1 
Part 2 
Part 3 
Explanation:
Given
Number of protons 
Radius of nucleus 
Distance of the electrons 
Part 1
Electric field produced by just outside its surface

Part 2
Electric field produced by just outside its surface

Part 3
The net electric field inside a uniform shell of negative charge is zero because the electric flux lines cancel out each other
hence, the solution is
Part 1 
Part 2 
Part 3 
Answer:
20 Yards
Explanation:
|---20----|
| |
| 50 |50
|---D--->|
Start End
Total displacement(D) 20 yards (East).
Explanation:
The given value of P is as follows.
P = 1.06F + 22.18, 
or, P' = 1.06
As p' is defined and non-zero. Hence, only critical points are boundary points.
For F = 10, the value of P will be calculated as follows.
P = 
= 32.78
For F = 70, the value of P will be calculated as follows.
P = 
= 96.38
Therefore, the minimum value of P is 32.78 and maximum value of P is 96.38.
The answer to your question is B. <span>Analgesics. Hope that helps.</span>