A will be the fastest and c the slowest because of the dip it has a is a straight line fastest way to get from a to b is a straight line b is the second fastest and d is last
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
From the starting depth to the surface, the vertical distance is 35 ft.
From the surface to the peak of the jump, the vertical distance is 27 ft.
From the peak of the jump to the surface, the vertical distance is 27 ft.
From the surface to the ending depth, the vertical distance is 18 ft.
Then the total vertical distance is ...
35 ft + 27 ft + 27 ft + 18 ft = 107 ft
Answer:
Mechanical advantage = 4
Explanation:
Given the following data;
Distance of effort, de = 8m
Distance of ramp, dr = 2m
To find the mechanical advantage;
Mechanical advantage = de/dr
Substituting into the equation, we have;
Mechanical advantage = 8/2
Mechanical advantage = 4
The different reflections of light through two separate mediums causes the bending of wave fronts associated with light rays. The reflection and refraction is caused by the medium associated with its light rays.
