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:
0.002833 sec
Explanation:
Speed of light in vacuum is 
Given distance = 850 km = 850×1000=850000 m
We have to calculate the time that light take to travel the distance 850 km
Time 
So the time taken by light to travel 850 km is 0.002833 sec
The answer is up . tylrhscjwizn
Answer:
I think it's 2 the photo is hard to tell what they are exactly talking about.
Answer and Explanation: To know how much tape he will need, we have to calculate the perimeter of each parallelogram-shaped stripe.
Perimeter is the sum of all the sides of a figure.
For a parallelogram:
P = 2*length + 2*width
So, we need to determine width and length of the stripe.
Width is 3 inches. Length is the hypotenuse of the right triangle, whose sides are 6 and 18 inches. Then, length is


h = 19 in
Perimeter of the first stripe is
P = (2*19) + (2*3)
P = 44 inches
The hazard sign has 3 stripes. So total perimeter is
44 + 44 + 44
132 inches
To outline the parallelogram-shaped stripes, Charles need a total of 132 inches of tape. Since one roll has 144 inches, he will have enough tape to finish the job.