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Sidana [21]
3 years ago
7

An 8 Newton wooden block slides across a horizontal wooden floor at constant velocity. What is the magi notice of the force of k

inetic friction between the block and the floor
Physics
1 answer:
kondaur [170]3 years ago
8 0

Answer:

The force of kinetic friction between the block and the floor is 2.4 N

Explanation:

The given parameters are;

The weight of the block = 8 Newtons

The velocity of the block = Constant velocity

Taking the kinetic friction for wood, \mu _k = 0.3, we have;

The normal reaction of the block on horizontal ground, N = The weight of the block = 8 N

The force of kinetic friction between , F_k = \mu _k × N

Therefore, we have;

The force of kinetic friction between the block and the floor, F_k = 0.3 × 8 N = 2.4 N.

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Complete Question

A  voltaic cell is  constructed with two Zn^{2+}- Zn electrodes. The  two cell compartment have  [Zn^{2+}] =  1.6 \ M and  [Zn^{2+}] =  2.00*10^{-2} \  M respectively.

What is the cell emf for the concentrations given? Express your answer using two significant figures

Answer:

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Explanation:

Generally from the question we are told that

   The  concentration of [Zn^{2+}] at the cathode is  [Zn^{2+}]_a =  1.6 \ M

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Generally the the cell emf for the concentration is mathematically represented as

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Generally the E^ois the standard emf of a cell, the value is  0 V

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