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4 years ago

Which letter represents a wavelet

Physics

2 answers:

Dennis_Churaev [7]4 years ago

7 0

The answer is C - Y, I just got it right on Edge.

Blizzard [7]4 years ago

6 0

A/c/d. that is the anser hope it helps

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Two 2.0cm×2.0cm metal electrodes are spaced 1.0 mm apart and connected by wires to the terminals of a 9.0 v battery. What is the

Fantom [35]

Answer:

The charge on each electrode is

$q=0.354 x10^{-12}$

Explanation:

The charge to find the equation is development

V=9v

$d=1.0 mm\\A=2cm*2cm=0.02m*0.02m\\A=4x10^{-4} \\E_{o} =8.85x10^{-12} (\frac{C^{2} }{N*m^{2} } )$

$q=C*V$

$C=\frac{E_{o}*A }{d} \\C=\frac{8.85x10^{-12} *4x10^{-4} }{0.01}=0.354 \\$

$q=0.354*9v\\q=3.186x10^{-12} C$

3 0

4 years ago

A driver can exceed the postage maximum speed limit in a work zone T or F?

PIT_PIT [208]

That statement is physically and grammatically true but legally false.

6 0

4 years ago

A circular horizontal surface having an area of 1020.5 mm2that is completely immersed in a fluid experiences a uniform force of

Bingel [31]

Answer:

Faya vei Abhikesh brass hahahahahaha

Explanation:

3 0

3 years ago

A general expression for an electromagnetic plane wave can be written as E Asin( k-r-cot) + B cos( k . r-ot) or E Dsin( k . r-do

solmaris [256]

Answer: $D = \sqrt{A^{2}+B^{2 }}$ and $\aplha = arctan(\frac{B}{A} )$

Explanation:

Hi! Since the notation is a little bit messed up, I am going to suppose that

$E = A sin(kr-\omega t)+B cos(kr-\omega t)$ --- (1)

and :

$E = D sin(kr-\omega t +\alpha)$ --- (2)

Here we are going to use a trigonometric identity of the sine of the sum of two angles, namely:

$sin(a+b)=cos(b)sin(a)+sin(b)cos(a)$ --- (3)

Lets set:

$a = kr - \omega t\\b = \alpha$

So now (2) becomes:

$E = Dcos( \alpha ) sin(kr - \omega t) + Dsin(\alpha)cos(kr - \omega t))$ --- (4)

Now (1) and (4) must be equal, and in particular we must have the following identities:

$Dcos(\alpha) = A\\Dsin(\alpha) = B$ --- (5)

If we square these two identities and sum them we got:

$D^{2} (cos(\alpha)^{2} +sin(\alpha)^{2}) = A^{2} +B^{2}$

And since:

$cos(a)^{2} +sin(a)^{2} =1$

We got the first solution:

$D = \sqrt{A^{2}+B^{2 }}$

For the second part we must divide the identies (5)

We got:

$\frac{sin(\alpha)}{cos(\alpha)}=\frac{B}{A}$

And since:

$\frac{sin(\alpha)}{cos(\alpha)}=tan(\alpha)$

We use the inverse of the tan function:

$\alpha =arctan(\frac{B}{A} )$

Greetings!

3 0

4 years ago

A 150-newton force, applied to a wooden crate at an angle of 30° above the horizontal, causesthe crate to travel at constant vel

omeli [17]

Answer:

Explanation:

A component of 150 N in vertical direction will reduce the magnitude of reaction force.

reaction force exerted by the floor

= mg - 150 sin 30

where m is mass of the crate .

the magnitude of the horizontal component of the 150-newton force

150 cos30

= 130 N

This force tries to pull the crate in forward direction with acceleration but it has no acceleration . It is so because frictional force of equal magnitude acts on it in opposite direction which makes the net force acting on it equal to zero.

Hence frictional force is equal to 150 cos 30.

= 130 N .

3 0

4 years ago