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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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Answer:

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A point would eventually be reached where the gas agitation would lead to an explosion.

Overheated aerosol cans would explode.

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4 years ago
At what condition does a body become weightless at the equator?
vesna_86 [32]
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3 years ago
A toy car is tied to a post with string, causing it to drive in a tight circle. This is
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7 0
3 years ago
A blue shift in light from a star indicates what?
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3 years ago
Your starship lands on a mysterious planet. As chief scientist-engineer, you make the following measurements: A 2.50-kg stone th
olga55 [171]

Answer:

Part a)

M = 6.08 \times 10^{19} kg

Part b)

T = 4510 hours

Explanation:

As we know that stone is thrown upwards with speed

v_i = 12 m/s

Now it returns back to the surface of Earth after t = 6 s

so the displacement of the stone is zero

\Delta y = 0 = v t + \frac{1}{2}at^2

0 = 12 t - \frac{1}{2}g t^2

g = \frac{2(12)}{t}

g = 4 m/s^2

Part a)

Now we know that the circumference of the planet at the equator is of length

L = 2 \times 100 km

2\pi R = 2\times 10^5 m

R = 3.2 \times 10^4 m

Now we have formula of acceleration due to gravity as

g = \frac{GM}{R^2}

4 = \frac{6.67 \times 10^{-11} M}{(3.2 \times 10^4)^2}

M = 6.08 \times 10^{19} kg

Part b)

Time to complete one revolution around the planet is given as

T = 2\pi\sqrt{\frac{r^3}{GM}}

here we know that

r = distance from center of the planet

r = 3.2 \times 10^4 + 3\times 10^7 = 3.003 \times 10^7 m

now we have

T = 2\pi\sqrt{\frac{(3.003\times 10^7)^3}{(6.67 \times 10^{-11})(6.08\times 10^{19})}}

T = 4510 hours

7 0
3 years ago
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