Answer:
a) 20 seconds
b) No.
Explanation:
t = Time taken for jet to stop
u = Initial velocity = 100 m/s (given in the question)
v = Final velocity = 0 (because the jet will stop at the end)
s = Displacement of the jet (Distance between the moment the jet touches the ground to the point the point it stops)
a = Acceleration = -5.00 m/s² (slowing down, so it is negative)
a) Equation of motion
The time required for the plane to slow down from the moment it touches the ground is 20 seconds.
The distance it requires for the jet to stop is 1000 m so in a small tropical island airport where the runway is 0.800 km long the plane would not be able to land. The runway needs to be atleast 1000 m long here the runway on the island is 1000-800 = 200 m short.
We will put the number of trips in the first column, the miles driven in the second column and gallons of fuel used in the third column.
8 7,680 1,010
7 9,940 1,330
12 14,640 1,790
12 13,920 2,050
What is the difference between<span> a</span>size declarator<span> and a </span>subscript<span>? The </span>size declarator<span> is ... When writing a function that accepts a two-dimensional </span>array<span> as an argument, which </span>size declarator<span> must you provide in the parameter </span>for<span> the</span>array<span>? The second size ...</span>
It is also tripled, there is a rule to everything, whatever you do to one thing, you do the exact thing to the other. Hope this solves it :)
Answer:
145 m
Explanation:
Given:
Wavelength (λ) = 2.9 m
we know,
c = f × λ
where,
c = speed of light ; 3.0 x 10⁸ m/s
f = frequency
thus,
substituting the values in the equation we get,
f = 1.03 x 10⁸Hz
Now,
The time period (T) = 
or
T = 
= 9.6 x 10⁻⁹ seconds
thus,
the time interval of one pulse = 100T = 9.6 x 10⁻⁷ s
Time between pulses = (100T×10) = 9.6 x 10⁻⁶ s
Now,
For radar to detect the object the pulse must hit the object and come back to the detector.
Hence, the shortest distance will be half the distance travelled by the pulse back and forth.
Distance = speed × time = 3 x 10^8 m/s × 9.6 x 10⁻⁷ s) = 290 m {Back and forth}
Thus, the minimum distance to target = 
= 145 m
