<u>Metal detectors work by transmitting an electromagnetic field from the search coil into the ground. Any metal objects (targets) within the electromagnetic field will become energised and retransmit an electromagnetic field of their own. The detector’s search coil receives the retransmitted field and alerts the user by producing a target response. metal detectors are capable of discriminating between different target types and can be set to ignore unwanted targets.
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1. Search Coil
The detector’s search coil transmits the electromagnetic field into the ground and receives the return electromagnetic field from a target.
2. Transmit Electromagnetic Field (visual representation only - blue)
The transmit electromagnetic field energises targets to enable them to be detected.
3. Target
A target is any metal object that can be detected by a metal detector. In this example, the detected target is treasure, which is a good (accepted) target.
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Answer:
0.07°C
Explanation:
<u>solution:</u>
the speed of a sound in water is<u>:</u>
v(T)=1480+4(T-4°C)
<u>at 4°C the travel time is:</u>
t(4◦C) = (
7600 × 103 m
)
/ (1480 m/s) = 5202.7 s
<u>5°C, the travel time is:</u>
t(5◦C) = (
7600 × 103 m
)
/ (1484 m/s) = 5188.7 s
<u>one degree C corresponds to a ∆t of 14 s so temperature difference is:</u>
ΔT=1 s/14 s=0.07◦C
Explanation:
Let h is the height of the plane above ground. x is the horizontal distance between the ground and the airport. Let s(t) is the distance between the plane and the airport. So,
...........(1)
Given, h = 4, x = 40 and s(t) = -20 mph
Differentiate equation (1) wrt t


When x = 40, 



So, the speed of the airplane is 241.14 m/s. Hence, this is the required solution.
Answer:
Hz
Explanation:
We know that
1 cm = 0.01 m
= Length of the human ear canal = 2.5 cm = 0.025 m
= Speed of sound = 340 ms⁻¹
= First resonant frequency
The human ear canal behaves as a closed pipe and for a closed pipe, nth resonant frequency is given as

for first resonant frequency, we have n = 1
Inserting the values


Hz
Answer:
4363.3231 feets²
Explanation:
Given that :
Distance, r = 50 ft
θ = 200°
The arc length of area covered :
Arc length = θ/360° * πr²
Arc length = (200/360) * 50 ft ^2 * π
Arc length = 0.5555555 * 2500 * π
Arc length = 4363.3231 feets²