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makkiz [27]
4 years ago
5

What is magnetism please define it

Physics
1 answer:
Scrat [10]4 years ago
4 0

Hello There!

It's a Property of matter where atoms in an object are aligned into domains

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Which would show an example of how physical changes are reversible?
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Answer:

Melting tin and then cooling it into a mold

Explanation:

When you melt something, and when it cools it returns to its physical state, therefore the physical changes are reversible. For example, take chocolate. When chocolate melts its liquid, then when it's in a colder situation it becomes solid again and so on, the changes are reversible.

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The Venus flytrap is known for which of these behaviors?
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Snapping a leaf shut around an insect, I think.
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Which of the following statements about asteroids is true? A. The asteroids in our solar system are spread out uniformly and are
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in the space between the orbits of Mars and Jupiter. (B)</span>
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A woman is standing in the ocean, and she notices that after a wave crest passes, five more crests pass in a time of 50.0 s. The
Masteriza [31]

Answer

given,

number of crest (N)= 5

time(t) = 50 s

distance between to successive crest = 32 m

a) Period

   T = \dfrac{t}{N}

   T = \dfrac{50}{5}

  T = 10\ s

b) frequency

   f = \dfrac{1}{T}

   f = \dfrac{1}{10}

   f =0.1\ Hz

c) wavelength

   distance between to consecutive crest is wavelength

    wavelength = 32 m

d) speed

         v = f λ

         v = 0.1 x 32

         v = 3.2 m/s

e) Amplitude

    We cannot determine amplitude because data is not given.

8 0
3 years ago
Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.340 mm wide. The diffraction pattern is observed
VARVARA [1.3K]

Answer:

a) The width of the central bright fringe is 1.41492×10^-2 m

b) The width of the first bright field on either side of the bright fringe is 7.075×10^-3 m

Explanation:

Let d be the width of the slit, s be the screen distance, m the order of diffraction. And x be the distance from the center to the edge of the central bright fringe and W be the width of the central bright fringe . And let y be the distance from the center of the central bright fringe to the edge of the first bright fringe on either side of the central bright fringe and w be the width of the first bright fringe. Let θ1 and θ2 respectively be the scattering angles of the central bright fringe and the first bright fringe.

a) According to Bragg's law

d×sinθ1 = m×λ

(0.340×10^-3)×sinθ1 = 1×(633×10^-9)

θ1 = 10.667×10^-2°

x = s×tanθ1

x = 3.80×tan(10.667×10^-2°)

x = 7.0741×10^-3 m

Now to get the width of the fringe you need to add x and x together

W = 2×x

W = 2×7.0741×10^-3

W= 1.41492×10^-2 m

b) For the second bright fringe

d×sinθ2 = m×λ

3.8×sinθ2 =  2×(633×10^-9)

θ2 = 21.33×10^-2°

y = s×tanθ2

y = 3.80×tan( 21.33×10^-2°)

y = 1.4149×10^-2 m

The width of the first bright fringe on either side of the central bright fringe is given by:

w = y - x

w = 7.075×10^-3 m

6 0
3 years ago
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