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3 years ago

If asked to describe a wave, you could say waves are the

2 answers:

7 0

Wave is a disturbance that transfer energy from one point to another without permanent deformation of the transferring medium.

serious [3.7K]3 years ago

3 0

Good evening Carolina

You could say waves are the continuous transmission of energy from one location to the next.

I hope that's help:)

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Do geochemists need to have a knowledge of physical science? Explain your<br> answer. please help:))

saul85 [17]

No they don’t lol I hate school

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3 years ago

I need a little help with this

Mandarinka [93]

Answer:

truck 1 has the most velocity

Explanation:

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3 years ago

What is the maximum value of the magnetic field at a distance of 2.5 m from a light bulb that radiates 100 W of single-frequency

Anvisha [2.4K]

Answer:

$1.04\times 10^{-7}$ T

Explanation:

$IP$ = Power of the bulb = 100 W

$r$ = distance from the bulb = 2.5 m

$I$ = Intensity of light at the location

Intensity of the light at the location is given as

$I = \frac{P}{4\pi r^{2}}$

$I = \frac{100}{4(3.14) (2.5)^{2}}$

$I$ = 1.28 W/m²

$B_{o}$ = maximum magnetic field

Intensity is given as

$I = \frac{B_{o}^{2}c}{2\mu _{o}}$

$1.28 = \frac{B_{o}^{2}(3\times 10^{8})}{2(12.56\times 10^{-7})}$

$B_{o} = 1.04\times 10^{-7}$ T

7 0

3 years ago

What is the action force of a flying bird?

kirill [66]

It is called the reaction force of a bird flying.

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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

3 years ago