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2 years ago

14

HURRY UP PLZZZ Two identical waves are traveling toward each other in the same medium. One has a positive amplitude, meaning tha

t its peaks only go upward, and the other wave has a negative amplitude, with peaks going downward.
What will occur when the waves peak at the same place at the same time?

constructive interference
destructive interference
a reflecting wave
wave refraction

2 answers:

Illusion [34]2 years ago

6 0

The correct answer is Destructive Interference.

puteri [66]2 years ago

3 0

The answer is destructive interference

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Two point charges of values +3.4 and +6.6 μc are separated by 0.10 m. what is the electrical potential at the point midway betwe

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To solve this problem, we should remember that:

Energy = Force x Distance

Since we are talking about charges, therefore we make use of Coulumb’s law for the electrical force between the two charges:

F = k q1 q2 / d^2

Where,

k = Coulumb’s constant = 9 x 10^9 N m^2/ c^2

q = charge

d = distance between the charges

Plugging back into the energy equation:

E = (k q1 q2 / d^2) * d

E = k q1 q2 / d

Solving for E using the given values:

E = (9 x 10^9 N m^2/ c^2) (3.4 E -6 c) (6.6 E -6 c) / 0.10 m

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2 years ago

A 1.0-m-long copper wire of diameter 0.10 cm carries a current of 50.0 A to the east. Suppose we apply to this wire a magnetic f

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Answer:

The classification of that same issue in question is characterized below.

Explanation:

The given values are:

Current, I = 50.0 A

Diameter, d = 0.10 cm

(a)...

As we know,

⇒ Magnetic force = Copper wire's weight

So,

⇒ $B\times I\times L=M\times g$

On putting the estimated values, we get

⇒ $B\times 50\times 1=7.037\times 10^{-3}\times 9.81$

⇒ $50B=69.03297\times 10^{-3}$

⇒ $B=1.38\times 10^{-3} \ T$

(b)...

As we know,

⇒ $m=\delta\times L\times \frac{\pi \ d^2}{4}$

⇒ $=8960\times 1\times \frac{\pi \ (0.001)^2}{4}$

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2 years ago

An older camera has a lens with a focal length of 60mm and uses 34-mm-wide film to record its images. Using this camera, a photo

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Answer:

24.71 mm

Explanation:

Distance is proportional to focal length, so

d∝f

which means

$\frac{d'_1}{d'_2}=\frac{f_1}{f_2}$

Magnification of first lens

$M_2=-\frac{d'_1}{d_1}$

and

$M_2=\frac{h'_1}{h_1}$

Similarly, magnification of second lens

$M_2=-\frac{d'_2}{d_1}$

and

$M_2=\frac{h'_2}{h_1}$

From the above equations we get

$\frac{M_1}{M_2}=\frac{d'_1}{d_2'}$

and

$\frac{M_1}{M_2}=\frac{h'_1}{h_2'}$

which means,

$\frac{d'_1}{d_2'}=\frac{h'_1}{h_2'}$

and

$\frac{d'_1}{d_2'}=\frac{f_1}{f_2}$

So, we get

$\frac{f_1}{f_2}=\frac{h'_1}{h_2'}\\\Rightarrow f_2=f_1\times\frac{h_2'}{h'_1}\\\Rightarrow f_2=60\times\frac{14}{34}=24.71\ mm$

∴ Focal length should this camera's lens is 24.71 mm

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2 years ago

A teacher sets up a stand carrying a convex lens of focal length 15 cm at 20.5 cm mark on the optical bench. She asks the studen

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We get the clearest image if there is no magnification. When we have no magnification the image and real object have the same size.
If we look at the diagram that I attached we can see that:
$\frac{h_i}{h_0}=\frac{d_i}{d_0}$
Two triangles that I marked are similar and from this we get:
$\frac{h_i}{h_0}=\frac{d_i-f}{f}$
The image and the object must have the same height so we get:
$\frac{h_i}{h_0}=\frac{d_i-f}{f};h_i=h_0\\
1=\frac{d_i-f}{f}\\
d_i=2f$
This tells how far the screen should be from the lens.
The position of the screen on the optical bench is:
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1 year ago

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ddd [48]

Answer:

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<em>During the day the land is warmer than the sea so the sea breeze blows and during the night the water bodies are warmer than the land so the land breeze blows.</em>

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2 years ago