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Sliva [168]
3 years ago
14

A copper telephone wire has essentially

Physics
1 answer:
Lunna [17]3 years ago
5 0

Answer:

128.21 m

Explanation:

The following data were obtained from the question:

Initial temperature (θ₁) = 4 °C

Final temperature (θ₂) = 43 °C

Change in length (ΔL) = 8.5 cm

Coefficient of linear expansion (α) = 17×10¯⁶ K¯¹)

Original length (L₁) =.?

The original length can be obtained as follow:

α = ΔL / L₁(θ₂ – θ₁)

17×10¯⁶ = 8.5 / L₁(43 – 4)

17×10¯⁶ = 8.5 / L₁(39)

17×10¯⁶ = 8.5 / 39L₁

Cross multiply

17×10¯⁶ × 39L₁ = 8.5

6.63×10¯⁴ L₁ = 8.5

Divide both side by 6.63×10¯⁴

L₁ = 8.5 / 6.63×10¯⁴

L₁ = 12820.51 cm

Finally, we shall convert 12820.51 cm to metre (m). This can be obtained as follow:

100 cm = 1 m

Therefore,

12820.51 cm = 12820.51 cm × 1 m / 100 cm

12820.51 cm = 128.21 m

Thus, the original length of the wire is 128.21 m

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MariettaO [177]

1 kg ball can have more kinetic energy than a 100 kg ball as increase in velocity is having greater impact on K.E than increase in mass.

<u>Explanation</u>:

We know kinetic energy can be judged or calculated by two parameters only which is mass and velocity. As kinetic energy is directly proportional to the (velocity)^2 and increase in velocity leads to greater effect on translational Kinetic Energy. Here formula of Kinetic Energy suggests that doubling the mass will double its K.E but doubling velocity will quadruple its velocity:

\text { Kinetic Energy }=\frac{1}{2} m v^{2}

Better understood from numerical example as given:

If a man A having weight 50 kg run with speed 5 m/s and another man B having 100 kg weight run with 2.5 m / s. Which man will have more K.E?

This can be solved as follows:

\text { Kinetic Energy of } \mathrm{A}=\frac{1}{2} 50 \times 5^{2}=625 \mathrm{J}

\text { Kinetic Energy bf } \mathrm{B}=\frac{1}{2} 100 \times 2.5^{2}=312.5 \mathrm{J}

It shows that man A will have more K.E.

Hence 1 kg ball can have more K.E than 100 kg ball by doubling velocity.

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3 years ago
12. An organ pipe that is 1.75 m long and open at both ends produces sound of
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Answer:

354 m/s

Explanation:

For the second overtune (Third harmonic) of an open pipe,

λ = 2L/3................................ Equation 1

Where L = Length of the open pipe, λ = Wave length.

Given: L = 1.75 m.

Substitute into equation 1

λ = 2(1.75)/3

λ = 1.17 m.

From the question,

V = λf.......................... Equation 2

V = speed of sound in the room, f = frequency

Given: f = 303 Hz.

Substitute into equation 2

V = 1.17(303)

V = 353.5

V ≈ 354 m/s

Hence the right answer is 354 m/s

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I really need help.
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A mxiture of n2 and H2 has mole fraction of 0.4 and 0.6 respectively. Determine the density of the mixture at one bar and 0 c.
SOVA2 [1]

Answer:

The density of the mixture is 0.55kg/m^3

Explanation:

P = 1bar = 100kN/m^2, T = 0°C = 273K, n = 0.4+0.6 = 1mole

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