Which magnetic property best describes a magnet’s ability to act at a distance?
2 answers:
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The magnitude of the magnetic field inside the solenoid is .
The given parameters;
- <em>length of the solenoid, L = 91 cm = 0.91 m</em>
- <em>radius of the solenoid, r = 1.5 cm = 0.015 m</em>
- <em>number of turns of the solenoid, N = 1300 </em>
- <em>current in the solenoid, I = 3.6 A</em>
The magnitude of the magnetic field inside the solenoid is calculated as;
where;
is the permeability of frees space = 4π x 10⁻⁷ T.m/A
Thus, the magnitude of the magnetic field inside the solenoid is .
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Answer:
the work done by the lawnmower is 236.14 J.
Explanation:
Given;
power exerted by the lawnmower engine, P = 19 hp
time in which the power was exerted, t = 1 minute = 60 s.
1 hp = 745.7 watts
The work done by the lawnmower is calculated as follows;
Therefore, the work done by the lawnmower is 236.14 J.
Answer:
a)W=8.333lbf.ft
b)W=0.0107 Btu.
Explanation:
<u>Complete question</u>
The force F required to compress a spring a distance x is given by F– F0 = kx where k is the spring constant and F0 is the preload. Determine the work required to compress a spring whose spring constant is k= 200 lbf/in a distance of one inch starting from its free length where F0 = 0 lbf. Express your answer in both lbf-ft and Btu.
Solution
Preload = F₀=0 lbf
Spring constant k= 200 lbf/in
Initial length of spring x₁=0
Final length of spring x₂= 1 in
At any point, the force during deflection of a spring is given by;
F= F₀× kx where F₀ initial force, k is spring constant and x is the deflection from original point of the spring.
Change to lbf.ft by dividing the value by 12 because 1ft=12 in
100/12 = 8.333 lbf.ft
work required to compress the spring, W=8.333lbf.ft
The work required to compress the spring in Btu will be;
1 Btu= 778 lbf.ft
?= 8.333 lbf.ft----------------cross multiply
(8.333*1)/ 778 =0.0107 Btu.