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sashaice [31]
1 year ago
10

Calculate the force which will produce an extension of 0.30mm in a steel wire with a length of 4.0m and a cross section area of

2.0 x 10^(-6) m^2Youngs modulus of steel is 2.1 x 10 ^11 Pa

Physics
1 answer:
Anna [14]1 year ago
7 0

Given data:

* The extension of the steel wire is 0.3 mm.

* The length of the wire is 4 m.

* The area of cross section of wire is,

A=2\times10^{-6}m^2

* The young modulus of the steel is,

Y=2.1\times10^{11}\text{ Pa}

Solution:

The young modulus of the steel in terms of the force and extension is,

Y=\frac{F\times l}{A\times dl}

where F is the force acting on the steel wire,, l is the original length of the wire, dl is the extension of the wire, and A is the area,

Substituting the known values,

\begin{gathered} 2.1\times10^{11}=\frac{F\times4}{2\times10^{-6}\times0.3\times10^{-3}} \\ F=0.315\times10^2\text{ N} \\ F=31.5\text{ N} \end{gathered}

Thus, the force which produce the extension of 0.3 mm of the steel wire is 31.5 N.

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When the projectile reaches at the maximum height it travels only along the horizontal and thus it has only horizontal velocity at that instant.

Thus, the velocity of teh projectile at maximum height is same as horizontal component of velocity that meas 50 m/s.

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For the wave in this problem,

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v=(15500)(0.20)=3100 m/s

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What must be the distance in meters between point charge q1 = 23.5 µc and point charge q2 = -64.2 µc for the electrostatic force
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A string with a mass density of 3 * 10^-3 kg/m is under a tension of 380 N and is fixed at both ends. One of its resonance frequ
Delvig [45]

Answer:

(a) the fundamental frequency of this string is 65 Hz

(b) the harmonics of the given frequencies are third and fourth respectively.

(c) the length of the string is 2.74 m

Explanation:

Given;

mass density of the string, μ = 3 x 10⁻³ kg/m

tension of the string, T = 380 N

resonating frequencies, 195 Hz and 260 N

For the given resonant frequencies;

195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3

(c) From any of the equations, solve for Length of the string (L);

195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m

(a) the fundamental frequency is calculated as;

f_o = \frac{1}{2l} \sqrt{\frac{T}{\mu} } \\\\f_o = \frac{1}{2\times 2.74} \sqrt{\frac{380}{3\times 10^{-3} } }\\\\f_o =  65 \ Hz

(b) harmonics of the given frequencies;

the first harmonic (n = 1) = f₀ = 65 Hz

the second harmonic (n = 2) = 2f₀ = 130 Hz

the third harmonic (n = 3) = 3f₀ = 195 Hz

the fourth harmonic (n = 4) = 4f₀ = 260 Hz

Thus, the harmonics of the given frequencies are third and fourth respectively.

7 0
2 years ago
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