The force needed to stretch the steel wire by 1% is 25,140 N.
The given parameters include;
- diameter of the steel, d = 4 mm
- the radius of the wire, r = 2mm = 0.002 m
- original length of the wire, L₁
- final length of the wire, L₂ = 1.01 x L₁ (increase of 1% = 101%)
- extension of the wire e = L₂ - L₁ = 1.01L₁ - L₁ = 0.01L₁
- the Youngs modulus of steel, E = 200 Gpa
The area of the steel wire is calculated as follows;

The force needed to stretch the wire is calculated from Youngs modulus of elasticity given as;


Thus, the force needed to stretch the steel wire by 1% is 25,140 N.
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Answer:
period of oscillations is 0.695 second
Explanation:
given data
mass m = 0.350 kg
spring stretches x = 12 cm = 0.12 m
to find out
period of oscillations
solution
we know here that force
force = k × x .........1
so force = mg = 0.35 (9.8) = 3.43 N
3.43 = k × 0.12
k = 28.58 N/m
so period of oscillations is
period of oscillations = 2π ×
................2
put here value
period of oscillations = 2π ×
period of oscillations = 0.6953
so period of oscillations is 0.695 second
Answer:
gas is dioatomic
T_f = 330.0 K

Explanation:
Part 1
below equation is used to determine the type Gas by determining
value

where V_i and V_f is initial and final volume respectively
and P_i and P_f are initial and final pressure


\gamma = 1.38
therefore gas is dioatomic
Part 2
final temperature in adiabatic process is given as
](https://tex.z-dn.net/?f=T_f%20%3D%20T_i%2A%5B%5Cfrac%7Bv_i%7D%7BV_f%7D%5D%28%5E%5Cgamma-1%29)
substituing value to get final temperature
![T_f = 260*[\frac{151}{80.6}]^ {(1.38-1)}](https://tex.z-dn.net/?f=T_f%20%3D%20260%2A%5B%5Cfrac%7B151%7D%7B80.6%7D%5D%5E%20%7B%281.38-1%29%7D)
T_f = 330.0 K
Part 3
determine number of moles by using following formula



Answer:
Explanation:
Givens
Vi = 10 m/s
Vf = 40 m/s
a = 3 m/s^2
Formula
a = (vf - vi) /t Substitute the givens into this formuls
Solution
3 = (40 - 10) / t Multiply both sides by t
3*t = t(40 - 10)/t Combine. Cancel t's on the right
3*t = 30 Divide by 3
3t/3 = 30 / 3
Answer: t = 10 seconds.