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Vaselesa [24]
3 years ago
10

A metal detector used in airports is actually a large coil of wire carrying a small current. Explain how it detects a gun, even

if it is wrapped in towels in a suitcase.
Physics
1 answer:
Alexus [3.1K]3 years ago
8 0

<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. </u>

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.

<em>hope this helps PLEASE MARK AS BRAINLIEST:)</em>

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It is customary to work in SI units.

Calculate the volume of the concrete.
V = 3.7*2.1*5.8 cm³ = 45.066 cm³ = 45.066 x 10 ⁻⁶ m³

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Which law of thermodynamics would be violated if heat were to spontaneously flow between two objects of equal temperature?
maks197457 [2]
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Is a magnetic reversal a threat to life on Earth ? Whay do you say?
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Water flows through a cast steel pipe (k = 50 W m.K, ε = 0.8) with an outer diameter of 104mm and 2 mm wall thickness. Calculate
masha68 [24]

Answer:

The heat loss per unit length is   \frac{Q}{L}   = 2981 W/m

Explanation:

From the question we are told that

     The outer diameter of the pipe is d = 104mm = \frac{104}{1000} = 0.104 m

     The thickness is  D = 2mm = \frac{2}{1000} = 0.002m  

      The temperature  of water is  T = 90^oC = 90 + 273 = 363K  

      The outside air temperature is T_a = -10^oC = -10 +273 = 263K

        The water side heat transfer coefficient is z_1 = 300 W/ m^2 \cdot K

       The  heat transfer coefficient is  z_2 = 20 W/m^2 \cdot K

The heat lost per unit length is mathematically represented as

           \frac{Q}{L}   = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1}  +  \frac{ln [\frac{d}{D} ]}{z_2}}

Substituting values

         \frac{Q}{L}   = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300}  +  \frac{ln [\frac{0.104}{0.002} ]}{20}}

           \frac{Q}{L}   = \frac{628}{0.2107}

           \frac{Q}{L}   = 2981 W/m

6 0
3 years ago
True or False<br> further the sun, the longer the shadow
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Answer: Yes the further the sun is away the longer the shadow is. At noon,the shadow is the shortest because its straight up above you. If this helps pls mark brainliest!

6 0
3 years ago
Read 2 more answers
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