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denis-greek [22]
3 years ago
7

A steel wire of length 1.5 m and diameter 1 mm is joined to an aluminum wire of identical dimensions to make a composite wire of

length 3 m. Find the resulting change in the length of this composite wire if an object with a mass of 5 kg is hung vertically from one of its ends. (Neglect any effects the masses of the two wires have on the changes in their lengths.). Use 2 x 1011 N/m2 and 0.7 x 1011 N/m2 for the Young's modulus for steel and aluminum respectively.
Physics
1 answer:
Anarel [89]3 years ago
4 0

Answer:

1.805 mm

Explanation:

Extension in the steel wire = WL_{steel}/AE_{steel}

Extension in the aluminium wire = WL_{Al}/AE_{Al}

Total extension = W/A * (L_{steel}/E_{steel} + L_{Al}/E_{Al})

we have:

W = mg

W = 5 × 9.8

W = 49 N

Area A = π/4 × (0.001)²

= 7.85398 × 10 ⁻⁷ m²

Total extension = W/A * (L_{steel}/E_{steel} + L_{Al}/E_{Al})

Total extension = 49/ 7.85398 × 10 ⁻⁷ ( (1.5/ 200×10⁹) + 1.5/ 70×10⁹))

Total extension =  0.0018048

Total extension = 1.805 mm

Thus, the total extension = the resulting change in the length of this composite wire = 1.805 mm

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Answer:

The inner planets are closer to the Sun and are smaller and rockier. The outer planets are further away, larger and made up mostly of gas. The inner planets (in order of distance from the sun, closest to furthest) are Mercury, Venus , Earth and Mars.

Explanation:

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2 years ago
A 0.017-kg acorn falls from a position in an oak tree that is 18.5 meters above the ground.
Marianna [84]

Answer:

Part a)

Final speed of the corn is 19.05 m/s

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Explanation:

Part a)

As we know that the initial position of the corn is

h = 18.5 m

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4 0
4 years ago
Please help ASAP! Thank you!
Likurg_2 [28]

Answer:

Sound energy wave is transferred through a substance as a series of compressions.

Explanation:

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4 0
3 years ago
Two point charges, A and B, are separated by a distance of 22.0 cm . The magnitude of the charge on A is twice that of the charg
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Answer:

1.12×10⁻⁵ C and 2.24×10⁻⁵ C.

Explanation:

From coulomb's law,

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Let: B = q, the A = 2q.

Substituting these values into equation 1,

47 = 9.0×10⁹(q×2q)/0.22²

47 = 18×10⁹(q²)/0.0484

q² = (47×0.0484)/(18×10⁹)

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q = √( 1.26×10⁻¹⁰)

q = 1.12×10⁻⁵ C

The charge at point A = 2q = 2× 1.12×10⁻⁵  = 2.24×10⁻⁵ C.

Hence the charges are 1.12×10⁻⁵ C and 2.24×10⁻⁵ C.

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3 years ago
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3 0
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