In preparing to shoot an arrow, an archer pulls a bow string back 0.424 m by exerting a force that increases uniformly from 0 to
1 answer:
Answer:
525.94N/m
Explanation:
According to Hooke's law, the extension or compression of an elastic material is proportional to an applied force provided that the elastic limit of the material is not exceeded.
where F is the applied force or load, k is the elastic constant or stiffness of the material and e is the extension.
In this problem, the bow string is assumed to behave like an elastic material that is stretched not beyond the elastic limit.
Given;
F = 223N;
e = 0.424m
k = ?
We make substitutions into equation (1) and then solve for k.
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Answer:
a) 2.063*10^-4
b) 1.75*10^-4
Explanation:
Given that: d= 1.628 mm = 1.628 x 10-3 I= 12 mA = 12.0 x 10-8 A The Cross-sectional area of the wire is:
a) <em>The Potential difference across a 2.00 in length of a 14-gauge copper </em>
<em> wire: </em>
L= 2.00 m
From Table Copper Resistivity = 1.72 x 10-8 S1 • m The Resistance of the Copper wire is:
=0.0165Ω
The Potential difference across the copper wire is:
V=IR
=2.063*10^-4
b) The Potential difference if the wire were made of Silver: From Table: Silver Resistivity p= 1.47 x 10-8 S1 • m
The Resistance of the Silver wire is:
=0.014Ω
The Potential difference across the Silver wire is:
V=IR
=1.75*10^-4