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Tcecarenko [31]
3 years ago
10

A chair of weight 105 N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F = 4

2.0 N directed at an angle of 43.0 ∘ below the horizontal and the chair slides along the floor.
Using Newton's laws, calculate n , the magnitude of the normal force that the floor exerts on the chair.
Physics
1 answer:
Nana76 [90]3 years ago
4 0

Answer:

  133.6 N

Explanation:

The vertical component of the force pushing on the chair is ...

  Fp = (42.0 N)sin(43°) ≈ 28.6 N

There is no acceleration of the chair in the direction normal to the floor, so the net force on it is zero. The floor counters the downward pushing force, together with the weight of the chair, with a force of ...

  105 N +28.6 N = 133.6 N

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Hope this helps :)

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A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in
koban [17]

Answer:

a) The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

b) The electric force is 2.0\times 10^{-6} newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

E = \frac{F_{e}}{q} (1)

Where:

E - Electric field, measured in newtons per coulomb.

F_{e} - Electric force, measured in newtons.

q - Electric charge, measured in coulombs.

If we know that F_{e} = 4.0\times 10^{-6}\,N and q = 2.0\times 10^{-8}\,C, then the electric field at that point is:

E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}

E = 2.0\times 10^{2}\,\frac{N}{C}

The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

b) If we know that E = 2.0\times 10^{2}\,\frac{N}{C} and q = 1.0\times 10^{-8}\,C, then the electric force is:

F_{e} = E\cdot q

F_{e} = \left(2.0\times 10^{2}\,\frac{N}{C} \right)\cdot (1.0\times 10^{-8}\,C)

F_{e} = 2.0\times 10^{-6}\,N

The electric force is 2.0\times 10^{-6} newtons.

7 0
3 years ago
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