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Vika [28.1K]
3 years ago
6

The range of the tides is affected by the relative positions of the sun and the moon. During the new moon and full moon, the hig

hest high tides and the lowest low tides occur. During the first and third quarters of the lunar month, the lowest high tides and the highest low tides occur. Throughout the month, the tidal range gradually increases and decreases between the minimum and maximum values of the range.
- Sketch a graph of a tidal range as a function of time, showing what you think the shape of a tidal cycle might look like for one month. On your graph, indicate where you think each....
Physics
1 answer:
Paladinen [302]3 years ago
4 0
<span>Daily the tides change 4 times a day, so high low high low, you can imagine it as Earth's daily rotation beneath the Moon, but the other side has a same tide since all water wants to be in balance. 

The Moon goes in a month around the Earth, and when she is New or Full (mistake in question!), together with the Sun all three are in one line and results in an extra attraction, which means extra high tides and extra low tides for that day. 

So in your graph you surely have 4 up and downs every day, and let's say halfway the month draw the ups and downs higher and deeper, and again at the end of the month.</span>marlies <span>· 3 years ago</span>
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Answer:

a

The total charge is q = 1.6*10^{-4}C

b

The change in potential energy is \Delta U = -4*10^{-2}J

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The height is  h= 8.2cm

Explanation:

From the question we are told that the

       The magnitude of electric field is E = 5000N/C

        The number of proton is  N_p = 1*10^{15}

        The distance between the plates is  d = 5cm = \frac{5}{100} = 0.05m

The total charge can be mathematically represented as

            q = N_p * e

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      Substituting values

           q = 1 *10^{15} * 1.6*10^{-19} = 1.6 *10^{-4}C

The change in potential energy is mathematically represented ads

         \Delta U = -(qE)d

where the negative sign shows that the work done by the electric force is against the electric field

                Substituting values

         \Delta U = - 1.6*10^{-4} * 5000 * 0.05

                = -4*10^{-2}J

 

   The mass of the bead is given as 0.05kg

  The change in potential due to gravity is mathematically given as

                    \Delta U = -mgh

the negative sign is  due to the fact that the height is decreasing

And g =9.8m/s^2

                  Making h the subject

        h = \frac{\Delta U}{mg}

Substituting values

                 h = \frac{4*10^{-2}}{0.05 * 9.8}

                   =0.081m = 0.082 *100 = 8.2cm

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