What force is required to stretched a steel wire of cross sectional area 1 cm² to double its length (Y=2×10¹¹ Nm⁻²)?

Physics
What force is required to stretched a steel wire of cross sectional area 1 cm² to double its length (Y=2×10¹¹ Nm⁻²)?

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Answer

\dfrac{5}{3}

Q20

Step 1: The formula for the angle of minimum deviation δm\delta_m for a prism is

δm=(μ1)A\delta_m = (\mu - 1) A

where μ\mu is the refractive index of the prism material and AA is the prism angle.

Step 2: Given A=60A = 60^\circ and δm=40\delta_m = 40^\circ, substitute the values.

40=(μ1)×6040^\circ = (\mu - 1) \times 60^\circ

Step 3: Solve for μ1\mu - 1.

μ1=4060=23\mu - 1 = \frac{40^\circ}{60^\circ} = \frac{2}{3}

Step 4: Find μ\mu.

μ=1+23=53\mu = 1 + \frac{2}{3} = \frac{5}{3}

\dfrac{5{3}}

Q22

Step 1: Coulomb's law for the electrostatic force FF between two point charges qq separated by distance rr in vacuum is

F=kq2r2F = k \frac{q^2}{r^2}

where k=9×109Nm2/C2k = 9 \times 10^9 \, \mathrm{N \cdot m^2 / C^2}.

Step 2: Given F=10NF = 10 \, \mathrm{N}, r=2mr = 2 \, \mathrm{m}, substitute the values.

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Q20 Step 1: The formula for the angle of minimum deviation _m for a prism is _m = ( - 1) A where is the refractive index of the prism material and A is the prism angle.

What force is required to stretched a steel wire of cross sectional area 1 cm² to double its length (Y=2×10¹¹ Nm⁻²)?
Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

ScanSolve AI Answer
Q20 Step 1: The formula for the angle of minimum deviation _m for a prism is _m = ( - 1) A where is the refractive index of the prism material and A is the prism angle. Step 2: Given A = 60^ and _m = 40^, substitute the values. 40^ = ( - 1) × 60^ Step 3: Solve for - 1. 1 = (40^)/(60^) = (2)/(3) Step 4: Find . = 1 + (2)/(3) = (5)/(3) (5)/(3) Q22 Step 1: Coulomb's law for the electrostatic force F between two point charges q separated by distance r in vacuum is F = k (q^2)/(r^2) where k = 9 × 10^9 \, N · m^2 / C^2. Step 2: Given F = 10 \, N, r = 2 \, m, substitute the values. 10 = (9 ×