A cable is 100 m long and has a cross-section area of 1 mm2. A 1000-N force is applied to stretch the cable. If Young’s modulus for the cable is 1.0x1011 N/m2
, how far does it stretch?
A. 10. m
B. 1.0 m
C. 0.10 m
D. 0.01 m
E. 0.001 m
To find the answer, we need to use Hooke's Law, which states that the extension of an elastic material is directly proportional to the force applied to it. The equation for Hooke's Law is:
F = k * ΔL
Where:
F is the force applied to stretch the cable,
k is the spring constant (Young's modulus), and
ΔL is the change in length (or extension) of the cable.
We can rearrange the equation to solve for ΔL:
ΔL = F / k
In this case, the force applied is 1000 N, and Young's modulus is 1.0x10^11 N/m^2. Plugging in these values, we get:
ΔL = 1000 N / (1.0x10^11 N/m^2)
Now we can calculate the answer.
ΔL = 1.0x10^-8 m
The cable stretches by 1.0x10^-8 meters.
Since all the answer options are in meters, we need to convert this scientific notation to decimal notation:
1.0x10^-8 = 0.00000001 m
So, the cable stretches by 0.00000001 m.
Among the given options, the closest value is option E. Therefore, the correct answer is:
E. 0.001 m