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likoan [24]
4 years ago
13

a student drops an object from the top of a building which is 19.6 m high. How long does it take the object to fall to the groun

d?
Physics
2 answers:
zubka84 [21]4 years ago
8 0

Here's a formula that's simple and useful, and if you're really in
high school physics, I'd be surprised if you haven't see it before. 
This one is so simple and useful that I'd suggest memorizing it,
so it's always in your toolbox.

This formula tells how far an object travels in how much time,
when it's accelerating:

               Distance = (1/2 acceleration) x (Time²).

                           D = 1/2 A T²

For your student who dropped an object out of the window,

     Distance = 19.6 m
     Acceleration = gravity = 9.8 m/s²

                                              D = 1/2 G T²

                                          19.6 =   4.9   T²

Divide each side by 4.9 :       4  =           T²

Square root each side:           2  =          T

When an object is dropped in Earth gravity,
it takes  2  seconds to fall the first 19.6 meters.

Crazy boy [7]4 years ago
5 0
Earth's gravity is 9.807 m/s²

Using this we can divide 19.6/9.807 to get 1.999 seconds or two if you round.
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    Fe^{2+}  + 2e^-----> Fe \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \   E^0_{red}  = - 0.441 V

   

    Cd^{2+} + 2e^-  -----> Cd  \ \ \ \ \ \ \ \ \ \ \ \ \ \  E^0_{red} = -0.403V

A galvanic cell based on these half-reactions is set up under standard conditions where each solutions is 1.00 L and each electrode weighs exactly 100.0 g. How much will the Cd electrode weigh when the non-standard potential of the cell is 0.03305 V?

Answer:

The mass is M= 117.37g

Explanation:

The overall reaction  is as follows

             Cd^{2+} + Fe  Fe^{2+} + Cd

The reaction is this way because the potential  of Cd^{2+} \ reducing \  to \ Cd is higher than the potential  of Fe^{2+} \ reducing \  to \ Fe so the the Fe would be oxidized and Cd^{2+} would be reduced

  At equilibrium the rate constant of the reaction is

                Q = \frac{concentration \ of \ product  }{concentration \ of  reactant }

                      = \frac{[Fe^{2+}[Cd]]}{[Cd^{2+}][Fe]}

The Voltage of the cell E_{cell} = E_{Cd^{2+}/Cd } + E_{Fe^{2+} /Fe}

     Substituting the given values into the equation

                         E_{cell} = -0.403 -(-0.441)

                                 = 0.038V

The voltage of the cell at any point can be calculated using the equation

               E = E_{cell} - \frac{0.059}{n_e}  Q

Where n_e \ is \ the \ number\  of\  electron

Substituting for Q

           E = E_{cell} - \frac{0.059}{n_e} \frac{[Fe^{2+}[Cd]]}{[Cd^{2+}][Fe]}

 When E = 0.03305 V

            E = E_{cell} - \frac{0.59}{n_e} \frac{Fe^{2+}}{Cd^{2+}}

Since we are considering the Cd electrode the equation becomes

            E= E_{cell} - \frac{0.059}{n_e}  [\frac{1}{Cd^{2+}} ]

Substituting values and making [Cd^{2+}]  the subject

          [Cd^{2+}] =\frac{1}{e^{[\frac{0.03305- 0.038}{\frac{0.059 }{2} }] }}

                      = 0.8455M

Given from the question that the volume is 1 Liter

   The number of mole = concentration * volume

                                       = 0.8455 * 1

                                        = 0.8455 moles

At the standard state the concentration of Cd^{2+} is  =1 mole /L

  Hence the amount deposited on the Cd electrode would be

              =  Original amount - The calculated amount

              =   1 - 0.8455

              = 0.1545 moles

The mass deposited is mathematically represented as

             mass = mole * molar \ mass

The Molar mass of Cd = 112.41 g/mol

          Mass  = 0.1545 *112.41

                    = 17.37g

Hence the total mass of the electrode is = standard mass + calculated mass

            M= 100+ 17.37

            M= 117.37g

                               

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