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2 years ago

5

Maya uses a tuning fork of frequency 250 Hz to produce transverese waves on a stretched string. Her friend justin measure the di

stance between the.two consecutive crest on the string as 5 cm. What was the velocity of the wave?

1 answer:

Goshia [24]2 years ago

6 0

Answer:

The velocity of the wave is 12.5 m/s

Explanation:

The given parameters are;

he frequency of the tuning fork, f = 250 Hz

The distance between successive crests of the wave formed, λ = 5 cm = 0.05 m

The velocity of a wave, v = f × λ

Where;

f = The frequency of the wave

λ = The wavelength of the wave - The distance between crests =

Substituting the known values gives;

v = 250 Hz × 0.05 m = 12.5 m/s

The velocity of the wave, v = 12.5 m/s.

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adell [148]

Answer:

It takes $\frac{1}{2}T$ to accelerate the object from rest to the speed v.

Explanation:

From Newton's second law:

$F=m\cdot a$ (1)

and the definition of acceleration,

$\displaystyle{a = \frac{\Delta v}{\Delta t}}$ (2)

we can solve this problem. Putting (2) in (1) we have:

$\displaystyle{F = m\cdot \frac{\Delta v}{\Delta t}}$ and solving for $\Delta t$ and considering the initial time as zero ( $t_0=0$ ) and the initial velocity also zero ( $v_0=0$ ) we have:

$\displaystyle{T=\frac{mv}{F}}$

Now, for a mass $m^*= 2m$ and the $F^*=4F$ we can wrtie the same equation:

$\displaystyle{T^*=\frac{m^*v}{F^*}}$ and substituting $m^*$ and $F^*$ :

$\displaystyle{T^*=\frac{2m\cdotv}{4F}=\frac{2}{4}T=\boxed{\frac{1}{2}T}}$

So now, it only takes half the time to accelerate the object from rest to the speed v

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3 years ago

A string is stretched and fixed at both ends, 200 cm apart. If the density of the string is 0.015 g/cm, and its tension is 600 N

Gre4nikov [31]

Answer:

<h2>f₀ = 158.12 Hertz</h2>

Explanation:

The fundamental frequency of the string f₀ is expressed as f₀ = V/4L where V is the speed experienced by the string.

$V = \sqrt{\frac{T}{\mu} }$ where T is the tension in the string and $\mu$ is the density of the string

Given T = 600N and $\mu$ = 0.015 g/cm = 0.0015kg/m

$V = \sqrt{\frac{600}{0.0015} }\\ \\V = \sqrt{400,000}\\ \\V = 632.46m/s$

The next is to get the length L of the string. Since the string is stretched and fixed at both ends, 200 cm apart, then the length of the string in metres is 2m.

L = 2m

Substituting the derived values into the formula f₀ = V/2L

f₀ = 632.46/2(2)

f₀ = 632.46/4

f₀ = 158.12 Hertz

Hence the fundamental frequency of the string is 158.12 Hertz

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3 years ago

A ball is falling from the second floor balcony to the floor below

aleksandrvk [35]

Answer:

KE increases and PE decreases

Explanation:

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2 years ago

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LiRa [457]

Answer:

R = 5.73 m

Explanation:

For an angle of rotation through 21 degree we know that

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$angle \times Radius = Arc$

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$R = \frac{2.1 \times 180}{21 \times \pi}$

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3 years ago

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Answer:

The answer is 13 however make sure if they ask for a certain measurement like meter answer it by saying 13 meters.

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All you do is divide 650J by 50N and you get a total of 13 (meters since I don't know what they want you to put it in).

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2 years ago