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GarryVolchara [31]
3 years ago
11

A steel piano wire is 0.7 m long and has a mass of 5 g. It is stretched with a tension of 500 N. What is the speed of transverse

waves on the wire? To reduce the wave speed by a factor of 2 without changing the tension, what mass of copper wire would have to be wrapped around the wire?
Physics
1 answer:
anygoal [31]3 years ago
6 0

Answer:

a.264.6m/s

b.4995g

Explanation:

Data given,

Tension T=500N,

length of wire=0.7m,

mass,m=5g=0.005kg

a.To determine the speed of the transverse wave, we use the equation below

V=\sqrt{\frac{FL}{m} } \\F=tension, L=length, m=mass\\

if we insert values as given in the question , we arrive at

V=\sqrt{\frac{500*0.7}{0.005}}\\ V=264.6m/s

b. To reduce he wave speed by a factor of 2, the new wave speed will become 132.3m/s.

From the equation, if mass,m becomes subject of formula, we have

m=\frac{V^{2}}{FL}\\ m=\frac{132.3^{2}}{500*0.7}\\m=50kg\\m=5000g

Hence to reduce the speed by a factor of 2, (5000-5)g=4995g mass of copper wire would have to be wrapped around the wire.

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Complete Question

The complete question is  shown on the first uploaded image (reference for Photobucket )

Answer:

The  electric field is  E = -1.3 *10^{-4} \ N/C

Explanation:

 From the question we are told that

    The linear charge density on the inner conductor is  \lambda _i  =  -26.8 nC/m  =  -26.8 *10^{-9} C/m

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Now this position we are considering is within the outer conductor so the electric field  at this point is due to the inner conductor (This is because the charges on the conductor a taken to be on the surface of the conductor according to Gauss Law )

Generally according to Gauss Law

         E (2 \pi r l) =  \frac{ \lambda_i }{\epsilon_o}

=>    E =  \frac{\lambda _i }{2 \pi *  \epsilon_o * r}

substituting values  

       E =  \frac{ -26 *10^{-9} }{2 * 3.142 *  8.85 *10^{-12} * 0.0373}

       E = -1.3 *10^{-4} \ N/C

The  negative  sign tell us that the direction of the electric field is radially inwards

    =>   |E| = 1.3 *10^{-4} \ N/C

5 0
4 years ago
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Delicious77 [7]

The new acceleration is 108 m/s^2

Explanation:

We can answer this problem by applying Newton's second law, which states that:

F=ma

where

F is the net force on an object

m is the mass of the object

a is its acceleration

The equation can be rewritten as

a=\frac{F}{m}

In this problem, the initial acceleration is

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Later:

- The net force is tripled: F'=3F

- The mass is halved: m'=\frac{m}{2}

Therefore, the new acceleration is:

a'=\frac{F'}{m'}=\frac{3F}{m/2}=6\frac{F}{m}=6a

which means that the new acceleration is 6 times the original acceleration, therefore

a'=6(18)=108 m/s^2

Learn more about acceleration:

brainly.com/question/9527152

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3 years ago
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AlladinOne [14]

Answer:

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ivann1987 [24]
I believe it would be A. (:
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A container has 79000cm3 of mercury. Find the mass of mercury if the density is 13.6g/cm3
ZanzabumX [31]

Explanation:

Density is mass divided by volume:

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Given V = 79000 cm³ and D = 13.6 g/cm³:

13.6 g/cm³ = M / (79000 cm³)

M = 1,074,400 g

M = 1,074 kg

Round as needed.

6 0
3 years ago
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