A force of 42N is needed to start a box sliding across the floor. The weight of the box is 55N.
2 answers:
Explanation:
It is given that,
Force applied to the box, F = 42 N
Weight of the box, W = 55 N
(a) Let f is the force of the friction acting on the box. The force of friction always opposes the motion of an object. It always equal to the applied force as :
f = -F = -42 N
(b) The frictional force is kinetic in nature because when the force is applied to the box it starts to move.
(c) Let is the coefficient of friction. It is given by :
Hence, this is the required solution.
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1. increase the area of the plates
4. decrease the separation between the plates
5. insert a dielectric between the plates
Explanation:
The energy stored in a capacitor is given by
where
C is the capacitance
V is the potential difference across the capacitor
For a parallel-plate capacitor, the capacitance is given by
where
k is the dielectric constant of the material
is the vacuum permittivity
A is the area of the plates of the capacitor
d is the separation between the plates
So we can rewrite the energy stored in the capacitor as
Here the potential difference is kept constant, so the energy depends only on the dielectric constant of the medium, the area and on the distance between the plates. In particular:
- The energy is directly proportional to the area, so as the area increases, the energy will increase
- The energy is inversely proportional to the distance, so as the distance decreases, the energy will increase
- The energy increases if the value of k increases (that is, if a dielectric is put between the plates)
It follows that the correct options to increase the energy are:
1. increase the area of the plates
4. decrease the separation between the plates
5. insert a dielectric between the plates
Learn more about capacitors:
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Answer:
The speed decreases 75%.
Explanation:
- Since no friction present, assuming no external forces acting during the three collisions, total momentum must be conserved.
- For the first collission, only mass 1 is moving before it, so we can write the following equation:
- Since both masses are identical, and they stick together after the collision, we can express the final momentum as follows:
- From (1) and (2) we get:
- v₁ = v₀/2 (3)
- Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1 and 2 combined together before colliding with mass 3 is just v₁, so the initial momentum prior the second collision (p₁) can be expressed as follows:
- Since after the collision the three masses stick together, we can express this final momentum (p₂) as follows:
- From (4) and (5) we get:
- v₂ = v₀/3 (6)
- Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1, 2 and 3 combined together before colliding with mass 4 is just v₂, so the initial momentum prior the third collision (p₂) can be expressed as follows:
- Since after the collision the four masses stick together, we can express this final momentum (p₃) as follows:
- From (7) and (8) we get:
- v₃ = v₀/4
- This means that after the last collision, the speed will have been reduced to a 25% of the initial value, so it will have been reduced in a 75% regarding the initial value of v₀.