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spayn [35]
3 years ago
11

Where are metalloids found on the periodic table

Physics
1 answer:
Vsevolod [243]3 years ago
7 0

Answer:

to the right of the post transition metals and to the left of the non metals.

b, si, Ge, as, sb, te,po,at.

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I don't know the answer but I think that it is yes I just don't want to get it wrong
mrs_skeptik [129]
Yes your answer should be true.
8 0
3 years ago
Read 2 more answers
A metal detector used in airports is actually a large coil of wire carrying a small current. Explain how it detects a gun, even
Alexus [3.1K]

<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>

8 0
3 years ago
Two sinusoidal waves, which are identical except for a phase shift, travel along in the same direction. The wave equation of the
Y_Kistochka [10]

Answer:

two sinusoidal waves, which are identical except for a phase shift, travel along in the same direction. The wave equation of the resultant wave is yR (x, t) = 0.70 m sin⎛ ⎝3.00 m−1 x − 6.28 s−1 t + π/16 rad⎞ ⎠ . What are the angular frequency, wave number, amplitude, and phase shift of the individual waves?

ω = 6.28 s − 1 ,

k = 3.00 m− 1 ,

φ = π rad,

A R = 2 A cos (φ 2 ) ,

A = 0.37 m

Explanation:

y1 ( x , t ) = A sin( k x − ω t +φ ) ,

y 2 ( x , t ) = A sin ( k x − ω t ) .

from the principle of superposition which states that when two or more waves combine, there resultant wave is the algebriac sum of the individual waves

y1 ( x , t ) = A sin( k x − ω t +φ ) ,   is generaL form of thw wave eqaution

A=amplitude

k=angular wave number

ω=angular frequency

φ =phase constant

k=2π/lambda

ω=2π/T

yR (x, t) = 0.70 m sin{3.00 m−1 x − 6.28 s−1 t + π/16 rad}....................*

two waves superposed to give the above, assuming they are moving in the +x direction

y1 ( x , t ) = A sin( k x − ω t +φ ) , .....................1

y 2 ( x , t ) = A sin ( k x − ω t ) ...........................2

adding the two equation will give

A sin( k x − ω t +φ )+A sin ( k x − ω t ) .................3

A( sin( k x − ω t +φ )+ sin ( k x − ω t ) ),......................4

similar to the following trigonometry identity

sina+sinb=2cos(a-b)/2sin(a+b)/2

let a= ( k x − ω t

b=k x − ω t +φ )

y(x,t)=2Acos(φ/2)sin(k x − ω t +φ/2)

k=3m^-1

lambda=2π/k=2.09m

ω=6.28= T=2π/6.28

T=1s

φ/2=π/16

φ=π/8rad

amplitude

2Acos(φ/2)=0.70 m

A=0.7/2cos(π/8)

A=0.37 m

6 0
4 years ago
What is the frequency of a rubber band that vibrates back and forth 50 times in 1 second
ch4aika [34]

If it makes 50 complete back-and-forth wiggles in 1 second, then its frequency is <em>50 Hz</em>.

4 0
4 years ago
Which of the following statements are true for electric field lines? Check all that apply. Check all that apply. Electric field
Scilla [17]

Answer:

Electric field lines point away from positive charges and toward negative charges. <u>True</u>

Electric field lines are continuous; they do not have a beginning or an ending.<u> False</u>

Electric field lines can never intersect. <u>True</u>

Electric field lines are close together in regions of space where the magnitude the electric field is weak and are father apart where it is strong. <u>False</u>

At every point in space, the electric field vector at that point is tangent to the electric field line through that point.<u> True</u>

Explanation:

Electric field lines point away from positive charges and toward negative charges. Always the field lines go to negative charges and leave from positive charges.

Electric field lines are continuous; they do not have a beginning or an ending.<u> False  </u>

Because the field lines starts at positive charges and ends in negative charges.

Electric field lines can never intersect. <u>True</u>

It can not intercept the field lines because in that point the the field would have two directions. Besides, in that point the real value of the field should be found adding both field lines.

Electric field lines are close together in regions of space where the magnitude the electric field is weak and are father apart where it is strong. <u>False</u>

This fact is opposite to that so the regions of space where the magnitude the electric field is weak the lines are father apart and where the field is strong  the lines are close together.

At every point in space, the electric field vector at that point is tangent to the electric field line through that point.<u> True</u>

This statement correspond to the definition of the field line.

5 0
3 years ago
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