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zhuklara [117]
3 years ago
7

The waves in a ripple tank are moving toward a barrier gap that is 1/2 the wavelength. A and B are equidistant from the gap. Sel

ect the correct statements about the diffraction that occurs.
1. Both A and B will encounter waves.
2. Only A will encounter waves.
3. Only B will encounter waves.
4. Waves will reach B more quickly than A.
5. Waves will reach A more quickly than B.
6. Waves will reach A and B at the same time.
Physics
2 answers:
kotegsom [21]3 years ago
6 0

Answer:  The correct statements about the diffraction are 1 and 6.

Explanation:    

Diffraction is the spreading of the wave from the gap in the barrier. If the wavelength increases then the diffraction also increases or it decreases with decrease in wavelength.

In the given problem, the waves in a ripple tank are moving toward a barrier gap that is 1/2 the wavelength. A and B are equidistant from the gap.

In the case of the waves in a ripple tank case, the diffraction is observed when the barrier is placed in a ripple tank. The path of water waves are observed when the they encounter the obstacles.  

The waves transmit energy in all direction then the waves will reach A and B at the same time.

Therefore, both A and B will encounter waves. Waves will reach A and B at the same time.

Romashka-Z-Leto [24]3 years ago
3 0
Diffraction is the spreading of a wave from a gap.

The statements that are true about diffraction in the case above are;

<span>1. Both A and B will encounter waves. and
</span><span>6. Waves will reach A and B at the same time.</span>
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A wave with a period of 0.008 second has a frequency of
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The frequency of a wave is the reciprocal of its period.

A period of 0.008 sec means a frequency of

         1 / 0.008 sec  =  125 per sec .  (125 Hz)

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Consider a 4-m-long, 4-m-wide, and 1.5-m-high above-the-ground swimming pool that is filled with water to the rim. (a) Determine
pashok25 [27]

Answer:

44100 N

Explanation:

Each wall will have dimension of 4 m x 1.5 m

Whole force will act on central point of wall situated at a depth of 1.5 /2 = .75m

pressure at CM = h d g , h = .75 , d ( density of water = 10³ )

pressure at CM = .75 x 10³ x 9.8

= 7350 N / m²

Total force on each wall

= pressure x area

= 7350 x 4 x 1.5

= 44100 N Ans

b ) If h = 1.5 x 2 = 3

Pressure = hdg

1.5 x 10³ x 9.8

= 14700 N / m²

Force

= pressure x area

14700 x 3 x 4

= 176400 N

Which is 4 times 44100 N

So force will quadruple.

It is so because both area and height have become twice.

3 0
3 years ago
What is tan(16°)?<br><br> A. 0.96<br> B.0.16<br> C.0.39<br> D.0.29
Alex Ar [27]

Answer:

D. 0.29

Explanation:

5 0
3 years ago
9. A radioisotope has a half-life of 4.50 min and an initial decay rate of 8400 Bq. What will be
Akimi4 [234]

Answer:

525 Bq

Explanation:

The decay rate is directly proportional to the amount of radioisotope, so we can use the half-life equation:

A = A₀ (½)^(t / T)

A is the final amount

A₀ is the initial amount,

t is the time,

T is the half life

A = (8400 Bq) (½)^(18.0 min / 4.50 min)

A = (8400 Bq) (½)^4

A = (8400 Bq) (1/16)

A = 525 Bq

8 0
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An organ pipe open at both ends is 1.5 m long. A second organ pipe that is closed at one end and open at the other is 0.75 m lon
telo118 [61]

Answer:

A. 110Hz,220Hz, 330 Hz

Explanation:

for organ open at open both ends;

the length of the organ for the fundamental frequency, L = A---->N + N----->A

A---->N  = λ /4 and N----->A = λ /4

L = λ /4 + λ /4 = λ /2

L = \frac{\lambda}{2} \\\\\lambda = 2L

λ  = 2 x 1.5m = 3.0 m

Wave equation is given by;

V = Fλ

Where;

V is the speed of sound

F is the frequency of the wave

F = V/ λ

F₀ = V / 2L

Where;

F₀  is the fundamental frequency

F₀ = 330 / 2(1.5)

F₀ = 330 / 3

F₀ = 110 Hz

the length of the organ for the first overtone, L = A---->N + N----->A + A----->N +  N----->A

L = 4λ /4

L = λ

λ = 1.5 m

F₁ = 330 / 1.5

F₁ = 220 Hz

Thus, F₁ = 2F₀

For open organ at one end

the length of the organ for the fundamental frequency, L = N------A

L = λ /4

λ = 4L

F₀ = V/4L

F₀ = 330 / (4 x 0.75)

F₀ = 110 Hz

the length of the organ for the first overtone, L = N-----N + N-----A

L = λ/2 + λ / 4

L = 3λ /4

F₁ = 3F₀

F₁ = 3 x 110

F₁ = 330 Hz

Thus the fundamental frequency for both organs is 110 Hz,

The first overtone for the organ open at both ends is 220 Hz

The first overtone for the organ open at one end is 330 Hz

The correct option is "A. 110Hz,220Hz, 330 Hz"

6 0
3 years ago
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