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3 years ago

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Any kind of wave spreads out after passing through a small enough gap in a barrier. This phenomenon is known as _________. a) di

ffraction b) double-slit c) interference d) refraction e) antireflection

2 answers:

fenix001 [56]3 years ago

7 0

the answer is a diffraction

bulgar [2K]3 years ago

6 0

Answer:

a) diffraction

Explanation:

Diffraction occurs when waves pass through small openings, around obstacles or sharp edges.When an opaque object is between the point of light and a screen, the border between the shaded and illuminated regions on the screen is not defined. A careful inspection of the scrubber shows that a small amount of light is diverted to the shaded region. The region outside the shadow contains bright and dark altered bands, where the intensity of the first band is brighter than the region of uniform illumination.

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A crate filled with delicious junk food rests on a horizontal floor. Only gravity and the support force of the floor act on it,

stealth61 [152]

The words "... as shown ..." tell us that there's a picture that goes along
with this question, and you decided not to share it. That's sad and
disappointing, but I think the question can be answered without seeing
the picture.

The net force on the crate is zero. Evidence for this is that fact that
the crate is just sitting there. If the net force on an object is not zero,
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3 years ago

Cold butter placed on hot corn melts because of convection.<br> agree <br> disagree<br> it depends

frozen [14]

Explanation:

Cold butter placed on hot corn melts because of convection. Butter placed on hot corn melts because it touches the hot corn. Most of the heat is transferred through conduction

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3 years ago

An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at 12.8 m/s2 . At t1 the rocket e

Simora [160]

Answer:4.39 s

Explanation:

Given

initial velocity $u=0$

acceleration $a=12.8 m/s^2$

velocity acquired by sled in $t_1$ time

$v=0+at$

$v=12.8t_1$

distance traveled by sled in $t_1 s$

$v^2-u^2=2as$

$(12.8t_1)^2-0=2\times 12.8\times s_1$

$s_1=6.4\cdot t_1^2$

distance traveled in $t_2$ time with velocity $v=12.8t_1$

$s_2=v\times t_2$

$s_2=12.8\times t_1\times t_2$

$s_2=12.8\cdot t_1\cdot t_2$

$s_1+s_2=5.37\times 10^3$

$6.4t_1^2+12.8t_1t_2=5370$ ----1

$t_1+t_2=97.7 s$

$t_2=97.7-t_1$

substitute the value of $t_2$ in 1

we get

$6.4t_1^2-1250.56t_1+5370=0$

thus $t_1=\frac{1250.56-1194.33}{12.8}=4.39 s$

$t_1=4.39 s$

5 0

3 years ago