Answer:
The copper wire stretches 6.25 cm and the steel wire stretches 3.75 cm.
Explanation:
Young's modulus is defined as:
E = stress / strain
E = (F / A) / (dL / L)
E = (F L) / (A dL)
Solving for dL:
dL = (F L) / (A E)
The wires have the same force, length, and cross-sectional area. So:
dL₁ + dL₂ = (FL/A) (1/E₁ + 1/E₂)
Given that dL₁ + dL₂ = 0.10 m, E₁ = 20×10¹⁰ N/m², and E₂ = 12×10¹⁰ N/m²:
0.10 = (FL/A) (1/(20×10¹⁰) + 1/(12×10¹⁰))
FL/A = 0.75×10¹⁰ N/m
Solving for dL₁ and dL₂:
dL₁ = (FL/A) / E₁
dL₁ = (0.75×10¹⁰ N/m) / (20×10¹⁰ N/m²)
dL₁ = 0.0375 m
dL₂ = (FL/A) / E₂
dL₂ = (0.75×10¹⁰ N/m) / (12×10¹⁰ N/m²)
dL₂ = 0.0625 m
The copper wire stretches 6.25 cm and the steel wire stretches 3.75 cm.
W is the noble gas having atomic number 10 and configuration (2,8)
x, y and z are the s block elements. They have 2,2 and 1 electrons in their outermost shell respectively.
Atomic size decreases across the period and increases down the group. Element z is having more atomic size.
Hope it helps you!!!!
We can compare the two by their kinetic energies. The kinetic energy is the energy when an object
is in motion. It is expressed as the product of the mass of the object and the
square of the velocity divided by two. We assume a velocity of 1 m/s for this problem.<span>
KE = mv^2/2
KE1 = 10 (1)^2
KE1 = 10 J
KE2 = 1(1)^2
KE2 = 1 J
Therefore, c</span><span>ompared to the 10 kg ball, the 1 kg ball has lesser kinetic energy.</span>
<span>Let's first off calculate the kinetic energy using the formula 1/2MV^2. Where the mass, M, is 0.6Kg. And speed, V, is 2. Hence we have 1/2 * 0.6 * 2^2 = 1.2J. Since kinetic energy is energy due to motion; hence at point B the rubber has a KE of 1.2J and not 7.5J. So I would say that only the Mass and speed is actually true; While it's kinetic energy is not true.</span>
