Santa doesn't want to push a 25.0 kg wooden box across a wooden floor with a uk of 0.20 at
1 answer:
Answer:
49 N
Explanation:
In order to move the box at constant speed, the acceleration of the box must be zero (a=0): this means, according to Newton's second law,
F = ma
that the net force acting on the box, F, must be zero as well.
Here there are two forces acting on the box in the horizontal direction while it is moving:
- The force of push applied by the guy, F
- The frictional force,
For an object moving on a flat surface, the frictional force is given by
where
is the coefficient of friction
m is the mass of the box
g is the acceleration of gravity
So the equation of the forces becomes
And substituting:
We find the force that must be applied by the guy:
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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.
Answer:
Magnitude of magnetic field is 1.29 x 10⁻⁴ T
Explanation:
Given :
Current flowing through the wire, I = 16.9 A
Length of wire. L = 0.69 m
Magnetic force experienced by the wire, F = 1.5 x 10⁻³ N
Consider B be the applied magnetic field.
The relation to determine the magnetic force experienced by current carrying wire is:
F = ILBsinθ
Here θ is the angle between magnetic field and current carrying wire.
According to the problem, the magnetic field and current carrying wire are perpendicular to each other, that means θ = 90⁰. So, the above equation becomes:
F = ILB
Substitute the suitable values in the above equation.
B = 1.29 x 10⁻⁴ T