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aev [14]
3 years ago
6

There have been different models of the atom over time. How has the competition between these models affected our understanding

of the atom?
Physics
2 answers:
Reil [10]3 years ago
7 0

Answer:

Generally, these models have shed more light on the field of atom by describing the composition of atoms and how the atom functions . The competition of these models have been able to modify atomic model to it actual description, modification and functionality.      

Explanation:

There have been different models of atoms overtime . The competition between the various models of atom have been able to shed more light on atom and create more theory on the subject of atoms. The well known models of atom are Dalton's atomic model, Thomson's model of the atom , Ernest Rutherford model of atom, Niels Bohr's model and Schrödinger's model of atoms.

Dalton stated that atoms are indivisible and he was able to realize atom of a particular element different from others . That means atoms of the same element are identical. J.J Thompson was able to recognized electrons in atoms  he drew a plum pudding model  of an atom. It shows electron revolves around a positive charge through out a spherical cloud. Rutherford was able to detect positive charge of an atom found in the nucleus of an atom. Bohr's model modified Rutherford's model . He was able to recognized that the electron move or orbit around the nucleus. Schrödinger's  believes electrons moves around the nucleus in cloud and has an uncertain position.

Generally, these models have shed more light on the field of atom by describing the composition of atoms and how the atom functions . The competition of these models have been able to modify atomic model to it actual description, modification and functionality.      

Umnica [9.8K]3 years ago
4 0
Each time it gets changed we understand more about it
You might be interested in
An external resistor with resistance R is connected to a battery that has emf ε and internal resistance r. Let P be the electric
ELEN [110]

Answer:

a. 0 W b. ε²/R c. at R = r maximum power = ε²/4r d. For R = 2.00 Ω, P = 227.56 W. For R = 4.00 Ω, P = 256 W. For R = 6.00 Ω, P = 245.76 W

Explanation:

Here is the complete question

An external resistor with resistance R is connected to a battery that has emf ε and internal resistance r. Let P be the electrical power output of the source. By conservation of energy, P is equal to the power consumed by R. What is the value of P in the limit that R is (a) very small; (b) very large? (c) Show that the power output of the battery is a maximum when R = r . What is this maximum P in terms of ε and r? (d) A battery has ε= 64.0 V and r=4.00Ω. What is the power output of this battery when it is connected to a resistor R, for R=2.00Ω, R=4.00Ω, and R=6.00Ω? Are your results consistent with the general result that you derived in part (b)?

Solution

The power P consumed by external resistor R is P = I²R since current, I = ε/(R + r), and ε = e.m.f and r = internal resistance

P = ε²R/(R + r)²

a. when R is very small , R = 0 and P = ε²R/(R + r)² = ε² × 0/(0 + r)² = 0/r² = 0

b. When R is large, R >> r and R + r ⇒ R.

So, P = ε²R/(R + r)² = ε²R/R² = ε²/R

c. For maximum output, we differentiate P with respect to R

So dP/dR = d[ε²R/(R + r)²]/dr = -2ε²R/(R + r)³ + ε²/(R + r)². We then equate the expression to zero

dP/dR = 0

-2ε²R/(R + r)³ + ε²/(R + r)² = 0

-2ε²R/(R + r)³ =  -ε²/(R + r)²

cancelling out the common variables

2R =  R + r

2R - R = R = r

So for maximum power, R = r

So when R = r, P = ε²R/(R + r)² = ε²r/(r + r)² = ε²r/(2r)² = ε²/4r

d. ε = 64.0 V, r = 4.00 Ω

when R = 2.00 Ω, P = ε²R/(R + r)² = 64² × 2/(2 + 4)² = 227.56 W

when R = 4.00 Ω, P = ε²R/(R + r)² = 64² × 4/(4 + 4)² = 256 W

when R = 6.00 Ω, P = ε²R/(R + r)² = 64² × 6/(6 + 4)² = 245.76 W

The results are consistent with the results in part b

8 0
3 years ago
A 10kg block is Pulled along a horizontal
Leviafan [203]

Answer:

Explanation:

I hate these kinds of problems,  luckily I can't understand how much the kinetic friction is for this ,  the words are all mixed around.  and don't read well.  Maybe this went through a translator program?   My suggestion draw the free body diagram.   so you can see where the forces are, and how they are acting.   getting the free body diagram right.. usually makes these problems pretty straight forward.   just do the steps and you get the answer.

6 0
3 years ago
Read 2 more answers
I am planning to make two dives. The first dive is to 60 feet for 45 minutes, and the second dive is to 60 feet for 35 minutes.
Tpy6a [65]

Answer:

1:04-1:10 hours

Explanation:

You'll need a <em>Recreational dive planner</em> table, I annexed a copy, now you'll follow the next steps:

  1. In the first part of your table, you'll look for the distance row (in feet) of your first dive, for this specific exercise you'll find 60, once you locate it you'll go down that column until you reach the time you'll dive, in this case, 45 (minutes) or the closest value (47).
  2. You'll check and keep the letter in that 47 row (S) for future use.
  3. Now you have to go to the second part of your table and look for the distance column, in feet, of your second dive. We find 60 and then going right in the blue row, we'll look for the time (35) or its closest value (36).
  4. Finally, we have to check the letter for 36 minutes (F) and we'll make it met with the letter S in the first portion of your tables. This will give us an interval of time, 1:04-1:10 in this case.

I hope you find this information useful and interesting! Good luck!

4 0
3 years ago
A shot putter heaves a 7.26kg shot with final velocity of 7.50m/s what is kinetic energy?
ivanzaharov [21]

Answer: KE=.5mv^2

KE=.5(7.62)(7.5)^2

KE=.5(7.62)(56.25)

KE=.5(428.625)

KE=214.3125 J of KE

Explanation:

8 0
3 years ago
To practice Problem-Solving Strategy 27.1: Magnetic Forces. A particle with mass 1.81×10−3 kgkg and a charge of 1.22×10−8 CC has
Roman55 [17]

Answer:

The magnitude of the acceleration  is a = 0.33 m/s^2

The direction is - \r k i.e the negative direction of the z-axis

Explanation:

 From  the question we are that

       The mass of the particle m = 1.8*10^{-3} kg

         The charge on the particle is q = 1.22*10^{-8}C

         The velocity is \= v = (3.0*10^4 m/s ) j

        The the magnetic field is  \= B = (1.63T)\r i + (0.980T) \r j

The charge experienced  a force which is mathematically represented as

         

                    F = q (\= v * \= B)

    Substituting value

         F = 1.22*10^{-8} (( 3*10^4 ) \r j \ \ X \ \  ( 1.63 \r i  + 0.980 \r j )T)

            = 1.22 *10^{-8} ((3*10^4 * 1.63)(\r j \ \  X \ \  \r i) + (3*10^4 * 0.980) (\r j \ \ X \ \  \r  j))

            = (-5.966*10^4 N) \r k

Note :

           i \ \ X \ \ j = k \\\\j \ \  X  \ \ k = i\\\\k  \ \ X \ \ i = j\\\\j \ \ X \ \ i = -k \\\\k \ \  X  \ \ j = -i\\\\i  \ \ X \ \ k = - j\\

Now force is also mathematically represented as

        F = ma

Making a the subject

      a = \frac{F}{m}

   Substituting values

     a =\frac{(-5.966*10^4) \r k}{1.81*10^{-3}}

        = (-0.33m/s^2)\r k

        = 0.33m/s^2 * (- \r k)

6 0
3 years ago
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