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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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7nadin3 [17]

Answer:

double replacement

Explanation:

sorry if im wrong

8 0
3 years ago
What is the momentum of a 2.0kg ball rolling at 6.0 m/s
irga5000 [103]
Momentum can be defined by the formula p=m*V (where m is mass and V is velocity) so if we plug in these numbers:

p = 2kg * 6m/s
p = 12 kgm/s
4 0
3 years ago
A charged capacitor is connected to an ideal inductor. At time t = 0, the charge on the capacitor is equal to 6.00 μC. At time t
Len [333]

Answer:

4.71\times 10^{-3}A

Explanation:

Q_{max} = Maximum charge stored by capacitor = 6 μC = 6 x 10⁻⁶ C

t  = time taken for charge on the capacitor to become zero = 2 ms = 2 x 10⁻³ s

Time period is given as

T = 4t

T = 4(2\times 10^{-3})

T = 8\times 10^{-3} s

Angular frequency is given as

w = \frac{2\pi }{T}

w = \frac{2(3.14) }{8\times 10^{-3}}

w =785 rad/s

Charge at any time is given as

Q(t) = Q_{max}Coswt

Taking derivative both side relative to "t"

\frac{\mathrm{d}Q(t) }{\mathrm{d} t} = \frac{\mathrm{d}(Q_{max}Coswt) }{\mathrm{d} t}

i(t)= -Q_{max} w Sinwt

Amplitude of the current is given as

i_{max}= Q_{max} w

i_{max}= (6\times 10^{-6}) (785)

i_{max}= 4.71\times 10^{-3}A

8 0
3 years ago
What laws of motion are demonstrated by a hammer pounding a nail into a board?
julia-pushkina [17]
A hammer pounding a nail into a board is an example of Newton’s Third law.

Newton’s third law states that for every action there is an equal and opposite reaction. Meaning, when you hit the hammer on the board the same amount of energy that is going into the board, is going into the hammer. Causing the hammer to bounce off the board.

Hope this helps!
6 0
3 years ago
PLZ HELP ME!!!!!!!!
mote1985 [20]

Answer:

1.     Frequency = speed / wavelength

here we know

speed = 343 m/s

wavelength = 0.77955 m

now frequency is given by  

f = \frac{343m/s}{ 0.77955m}

f = 440 Hz  

2.     Frequency = speed / wavelength

here we know

speed = 343 m/s

wavelength = 0.52028 m

now frequency is given by  

f = \frac{343m/s}{ 0.52028m}

f = 659.3 Hz

3.     Frequency = speed / wavelength

here we know

speed = 343 m/s

wavelength = 0.65552 m

now frequency is given by  

f = \frac{343m/s}{ 0.65552m}

f = 523.25 Hz  

4.     Wavelength = speed / Frequency

here we know

speed = 343 m/s

Frequency = 587.33 Hz

now wavelength is given by  

\lambda = \frac{343m/s}{587.33}

\lambda = 0.584 m    

5.     Wavelength = speed / Frequency

here we know

speed = 343 m/s

Frequency = 493.88 Hz

now wavelength is given by  

\lambda = \frac{343m/s}{493.88}

\lambda = 0.6945 m    

6.     Wavelength = speed / Frequency

here we know

speed = 343 m/s

Frequency = 698.46 Hz

now wavelength is given by  

\lambda = \frac{343m/s}{698.46}

\lambda = 0.491 m    

7.     Frequency = speed / wavelength

here we know

speed = 343 m/s

wavelength = 0.5840 m

now frequency is given by  

f = \frac{343m/s}{0.5840m}

f = 587.3 Hz  

8.     Frequency = speed / wavelength

here we know

speed = 343 m/s

wavelength = 0.4375 m

now frequency is given by  

f = \frac{343m/s}{0.4375m}

f = 784 Hz  

9.     Wavelength = speed / Frequency

here we know

speed = 343 m/s

Frequency = 783.99 Hz

now wavelength is given by  

\lambda = \frac{343m/s}{783.99}

\lambda = 0.4375 m    

10.  Wavelength = speed / Frequency

here we know

speed = 343 m/s

Frequency = 659.26 Hz

now wavelength is given by  

\lambda = \frac{343m/s}{659.26}

\lambda = 0.520 m    

4 0
3 years ago
Read 2 more answers
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