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Schach [20]
3 years ago
7

steel wire 8m long and 4mm in diameter is fixed to two rigid b. [1] supports. Calculate the increase in tension when the tempera

ture falls by 10°C. Given linear expansivity of steel, a 12x10 %/K, Young's modulus for steel, Y = 2x10¹¹ N/m².​
Physics
1 answer:
marin [14]3 years ago
7 0

Answer:

301.6 N

Explanation:

The length of the wire L₀ = 8 m and its diameter, d = 4 mm = 4 × 10⁻³ m. Since its temperature drops by  10°C, it will have a change in length ΔL = L₀αΔθ where α = linear expansivity of steel, a 12 × 10⁻⁶ /K, and Δθ = temperature change = -10°C = -10 K(negative since it is a drop)

So, the strain, ε = ΔL/L₀ = αΔθ = 12 × 10⁻⁶ /K × 10 K = 12 × 10⁻⁵

Now the Young's modulus of steel, Y = σ/ε where σ = stress = T/A where T = increase in tension in steel wire and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 4 × 10⁻³ m and ε = strain = 12 × 10⁻⁵

So, σ = Yε

Since Y = 2 × 10¹¹ N/m².

Substituting the values of the variables into the equation, we have

σ = Yε

σ = 2 × 10¹¹ N/m² × 12 × 10⁻⁵

σ = 24 × 10⁶ N/m²

Since σ = T/A

T = σA

T = σπd²/4

Substituting the values of the variables into the equation, we have

T = σπd²/4

T = 24 × 10⁶ N/m² × π × (4 × 10⁻³ m)²/4

T = 24 × 10⁶ N/m² × π × 16 × 10⁻⁶ m²/4

T = 24 × 10⁶ N/m² × π × 4 × 10⁻⁶ m²

T = 96 N × π

T = 301.59 N

T ≅ 301.6 N

So, the increase in tension in the steel wire is 301.6 N

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Answer:60 ohms

Explanation:

R1=30 ohms

R2=15 ohms

R3=15 ohms

Let the total resistance be R

R=R1 + R2 + R3

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R=60

Total resistance is 60 ohms

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A 23 kg body is moving through space in the positive direction of an x axis with a speed of 130 m/s when, due to an internal exp
babymother [125]

Answer:

a) Vx = 1088m/s

b) Vy = -162.93m/s

c) 5246745J

Explanation:

Mass of unbroken body = 23kg

Its velocity along +ve X-axis = 130m/s

Mass of first broken body, m1= 9.4kg

Its velocity along +ve X-axis = 130m/s

Nass of 2nd broken body, m2 = 6.1kg

Its velocity long-lived X - axis = -550m/s

Mass of 3rd broken body = ?

m3 = (23 - 9.4 - 6.1)kg

m3 = 7.5kg

Let velocity along the x-axis = Vx

Let the velocity along the x-axis = Vy

Applying law of conservation of momentum along x-axis

a) m1×0 + m2×(-550) + m3×(Vx) =M × 130

9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130

0 + (-5170) + 7.5Vx = 2990

2990 + 5170 = 7.5Vx

8160 = 7.5Vx

Vx = 8160/7.5

Vx = 1088m/s

b) Aplying conservation of momentum along the x-axis

(m1×130) + (m2 × 0) + (m3× Vy) = 0

(9.4 × 130) + (6.1 ×550) + 7.5Vy = 0

1222 + 0 + 7.5Vy = 0

1222 = -7.5Vy

Vy = 1222/(-7.5)

Vy = -262.93m/s

c) The energy released or change in KE is given by:

1/2[(m1v1^2) + (m2v2^2) +(m3Vx^2) ]= MV^2

Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)

Change in KE = 1/2[158860 + 1845250 + 8878080] - 1/2[388700]

Change in KE = 5441095 - 194350

Change in KE = 5246745J

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Answer:

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        D = 8.712 10⁶ + 1.21 10⁶

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