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:
This distance is measured from the center of the earth r = 3.4 10⁸ m
Explanation:
The equation for gravitational attraction force is
F = G m1 m2 / r²
Where g is the universal gravitation constant, m are the masses of the body and r is the distance between them, remember that this force is always attractive
Let's write the sum of force on the ship and place the condition that is balanced
F1 -F2 = 0
F1 = F2
Let's write this equation for our case
G m Me / r² = G m Mm / (r'.)²
The distance r is measured from the center of the earth and the distance r' is measured from the center of the moon,
r' = 3.85 10⁸ m
Let's simplify and calculate the distance
Me / r² = Mm / / (3.85 108- r)²
Me / Mm (3.85 108- r)² = r²
√ 81.4 (3.85 108 -r) = r
√ 81.4 3.85 108 = r (1 + √ 81.4)
34.74 108 = r (10.02)
r = 34.74 10⁸ / 10.2
r = 3.4 10⁸ m
This distance is measured from the center of the earth
The two units are Inches of Mercury and Millibars
Answer:
X=92.49 m
Explanation:
Given that
u= 21 m/s
h= 97 m
Time taken to cover vertical distance h
h= 1/2 g t²
By putting the values
97 = 1/2 x 10 x t² ( g = 10 m/s²)
t= 4.4 s
The horizontal distance
X= u .t
X= 21 x 4.4
X=92.49 m
Answer:
a
Explanation:
If its not right choose D
