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kherson [118]
3 years ago
6

-g With what tension must a rope with length 2.50 m and mass 0.120 kg be stretched for transverse waves of frequency 40.0 Hz to

have a wavelength of 0.750 m?
Physics
1 answer:
Alex17521 [72]3 years ago
8 0

Answer:

43.2 N

Explanation:

\lambda = Wavelength = 0.75 m

f = Frequency = 40 Hz

m = Mass of string = 0.12 kg

L = Length of string = 2.5 m

\mu = Linear density = \dfrac{m}{L}

Velocity of wave is given by

v=\lambda f\\\Rightarrow v=0.75\times 40\\\Rightarrow v=30\ m/s

The tension in string is given by

T=v^2\mu\\\Rightarrow T=30^2\times \dfrac{0.12}{2.5}\\\Rightarrow T=43.2\ N

The tension in the string is 43.2 N

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Melting butter is a physical change. Which best describes what is happening?
kati45 [8]

Answer:

Option B

Explanation:

The butter forms a new arrangement of atoms i.e from solid to liquid

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You and a friend each hold a lump of wet clay. Each lump has a mass of 30 grams. You each toss your lump of clay into the air, w
Vesna [10]

Answer:

\ \text{m/s}

Explanation:

u_1 = Velocity of one lump = 3x+3y-3z

u_2 = Velocity of the other lump = -4x+0y-4z

m = Mass of each lump = 30\ \text{g}

The collision is perfectly inelastic as the lumps stick to each other so we have the relation

mu_1+mu_2=(m+m)v\\\Rightarrow m(u_1+u_2)=2mv\\\Rightarrow v=\dfrac{u_1+u_2}{2}\\\Rightarrow v=\dfrac{3x+3y-3z-4x+0y-4z}{2}\\\Rightarrow v=-0.5x+1.5y-3.5z=\ \text{m/s}

The velocity of the stuck-together lump just after the collision is \ \text{m/s}.

4 0
3 years ago
You have to run 2.2 miles in track. How far is this in feet? Note: There are
Romashka [77]

Answer: B

Explanation: This can be easily done by inputting 5280 * 2.2.

3 0
3 years ago
Read 2 more answers
You are comparing two diffraction gratings using two different lasers: a green laser and a red laser. You do these two experimen
Nikolay [14]

Answer:

a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center

Explanation:

Let  n₁ and n₂ be no of lines per unit length  of grating A and B respectively.

λ₁ and λ₂ be wave lengths of green and red respectively , D be distance of screen and d₁ and d₂ be distance between two slits of grating A and B ,

Distance of first maxima for green light

= λ₁ D/ d₁

Distance of first maxima for red light

= λ₂ D/ d₂

Given that

λ₁ D/ d₁ = λ₂ D/ d₂

λ₁ / d₁ = λ₂ / d₂

λ₁ / λ₂  = d₁ / d₂

But

λ₁  <  λ₂

d₁ < d₂

Therefore no of lines per unit length of grating A will be more because

no of lines per unit length  ∝ 1 / d

If grating B is illuminated with green light first maxima will be at distance

λ₁ D/ d₂

As λ₁ < λ₂

λ₁ D/ d₂ < λ₂ D/ d₂

λ₁ D/ d₂ < 1 m

In this case position of first maxima will be less than 1 meter.

Option a is correct .

5 0
3 years ago
A steel piano wire, of length 1.150 m and mass 4.80 g is stretched under a tension of 580.0 N.
kaheart [24]

A steel piano wire, of length 1.150 m and mass of 4.80 g is stretched under a tension of 580.0 N.the speed of transverse waves on the wire would be  372.77 m/s

<h3>What is a sound wave?</h3>

It is a particular variety of mechanical waves made up of the disruption brought on by the movements of the energy. In an elastic medium like the air, a sound wave travels through compression and rarefaction.

For calculating the wave velocity of the sound waves generated from the piano can be calculated by the formula

V= √F/μ

where v is the wave velocity of the wave travel on the string

F is the tension in the string of piano

μ is the mass per unit length of the string

As given in question a steel piano wire, of length 1.150 m and mass of 4.80 g is stretched under a tension of 580.0 N.

The μ is the mass per unit length of the string would be

μ = 4.80/(1.150×1000)

μ = 0.0041739 kg/m

By substituting the respective values of the tension on the string and the density(mass per unit length) in the above formula of the wave velocity

V= √F/μ

V=√(580/0.0041739)

V =  372.77 m/s

Thus,  the speed of transverse waves on the wire comes out to be  372.77 m/s

Learn more about sound waves from here

brainly.com/question/11797560

#SPJ1

6 0
1 year ago
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