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3 years ago

Which kind of mechanical wave needs a source of energy to produce it?

Physics

2 answers:

Over [174]3 years ago

6 0

I think the answer would be T<span>ransverse, Longitudinal, & Surface waves.

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marusya05 [52]3 years ago

3 0

The answer is surface wave, transverse wave, and longitudinal wave!

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Using monochromatic light of 410 nm wavelength, a single thin slit forms a diffraction pattern, with the first minimum at an ang

VLD [36.1K]

Answer:

c. 307 nm

Explanation:

angular position of first dark fringe = λ / d , λ is wavelength and d is width of slit .

(40 x π ) / 180 = 410 / d

angular position of second dark fringe = 2 x λ / d , λ is wavelength and d is width of slit .

(60 x π ) / 180 = 2 x λ / d

Dividing these equations

60 / 40 = 2 x λ / 410

λ = 307.5 nm.

5 0

3 years ago

Part 2: Design your own circuit (40 points)

Mrac [35]

Answer:

"YOUR OWN!"

Explanation:

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2 years ago

How is energy transferred through a generator that produces electric current? A.from armature to commutator to brushes to turbin

vaieri [72.5K]

Armature to commutator to brushes to turbine

7 0

2 years ago

Read 2 more answers

When you let the air go out of a balloon the balloon moves or shoots like a rocket. Would this happen if there was no surroundin

Pie

Answer:

The balloon would still move like a rocket

Explanation:

The principle of work of this system is the Newton's third law of motion, which states that:

"When an object A exerts a force on an object B (action), object B exerts an equal and opposite force (reaction) on object A"

In this problem, we can identify the balloon as object A and the air inside the balloon as object B. As the air goes out from the balloon, the balloon exerts a force (backward) on the air, and as a result of Newton's 3rd law, the air exerts an equal and opposite force (forward) on the balloon, making it moving forward.

This mechanism is not affected by the presence or absence of surrounding air: in fact, this mechanism also works in free space, where there is no air (and in fact, rockets also moves in space using this system, despite the absence of air).

3 0

3 years ago

A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she

lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

$f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L}$

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}} }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = 1.3225 \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz = 740.6 \ Hz$

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, $L_{new}$ = 2/3·L

The value of the fundamental frequency will then be given as follows;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \times \dfrac{2 \times L}{3} } =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz = 840 \ Hz$

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

6 0

2 years ago