A student pulls a rope attached to a box of books and moves the box down the hall. The student pulls with a force of 185 N and a
1 answer:
You might be interested in
The force exerted by a magnetic field on a wire carrying current is:
where I is the current, L the length of the wire, B the magnetic field intensity, and
the angle between the wire and the direction of B.
In our problem, the force is F=0.20 N. The current is I=1.40 A, while the length of the wire is L=35.0 cm=0.35 m. The angle between the wire and the magnetic field is
, so we can re-arrange the formula and substitute the numbers to find B:
where I is the current, L the length of the wire, B the magnetic field intensity, and
In our problem, the force is F=0.20 N. The current is I=1.40 A, while the length of the wire is L=35.0 cm=0.35 m. The angle between the wire and the magnetic field is
Answer:
1keff=1k1+1k2
see further explanation
Explanation:for clarification
Show that the effective force constant of a series combination is given by 1keff=1k1+1k2. (Hint: For a given force, the total distance stretched by the equivalent single spring is the sum of the distances stretched by the springs in combination. Also, each spring must exert the same force. Do you see why?
From Hooke's law , we know that the force exerted on an elastic object is directly proportional to the extension provided that the elastic limit is not exceeded.
Now the spring is in series combination
Fe
F=ke
k=f/e.........*
where k is the force constant or the constant of proportionality
k=f/e
............................1
also for effective force constant
divide all through by extension
1) Total force is
Ft=F1+F2
Ft=k1e1+k2e2
F = k(e1+e2) 2)
Since force on the 2 springs is the same, so
k1e1=k2e2
e1=F/k1 and e2=F/k2,
and e1+e2=F/keq
Substituting e1 and e2, you get
1/keq=1/k1+1/k2
Hint: For a given force, the total distance stretched by the equivalent single spring is the sum of the distances stretched by the springs in combination.