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kompoz [17]
3 years ago
14

A person pushes a 50kg box to the right with 100N of force at a constant velocity of 5m/s. Calculate the coefficient of friction

between the box and the surface.
Physics
1 answer:
Anna11 [10]3 years ago
8 0
The weight of the box is 50x9.8 = 490 N. The force of friction is 100N. F= μΝ so coefficient = 100/490 = 0.20
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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:

The value is   E =  0.06 V

Explanation:

Generally from the question we are told that

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

    The  concentration of [Zn^{2+}] at the anode is [Zn^{2+}]_c =  2.00*10^{-2} \  M

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

So

      E =  0  -  \frac{0.0591}{2}  * log[\frac{ 2.00*10^{-2}}{1.6} ]

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