All categories

3 years ago

What are the differences between the plum pudding model, nuclear model and the atomic model

2 answers:

8 0

The plum pudding diagram models the protons and electrons in an atom and gives evidence of different charges in an atom. The nuclear model was made to prove that the subatomic particles were placed in different places in the atom such as the neutrons and protons being in the nucleus and the electrons on the outer shells. The atomic model, buy Dalton, was made to prove that the atom is always in constant motion and never still in one place.

HOPE THIS HELPED :)

kakasveta [241]3 years ago

7 0

The plum pudding model suggested that the electrons were dispersed throughout the atom (like chocolate chips in a cookie) and the space was positively charged so that in the end the atom was neutral. Today, people know that the electrons are in a "cloud" around the atom and the protons (and neutrons) are in the nucleus, at the center of the atom.

You might be interested in

Convert 55 miles into inches

frosja888 [35]

Answer:

3.485e+6 inches. hope this helps

6 0

3 years ago

Read 2 more answers

At what speed must the electron revolve round the nucleus of

EleoNora [17]

Explanation:

I think this is it, give it a try

6 0

3 years ago

If the spring constant is doubled, what value does the period have for a mass on a spring?

soldi70 [24.7K]

if spring constant is doubled, the mass on spring will be doubled as well. according to this formula, F=ke

k stands for spring constant and e stands for the length extended

4 0

3 years ago

Read 2 more answers

Please help with Physics Circuits!

Zigmanuir [339]

1) Let's start by calculating the equivalent resistance of the three resistors in parallel, $R_2, R_3, R_4$ :
$\frac{1}{R_{234}}= \frac{1}{R_2}+ \frac{1}{R_3}+ \frac{1}{R_4}= \frac{1}{4.5 \Omega}+ \frac{1}{1.3 \Omega}+ \frac{1}{6.3 \Omega}=1.15 \Omega^{-1}$
From which we find
$R_{234}= \frac{1}{1.15 \Omega^{-1}}=0.9 \Omega$

Now all the resistors are in series, so the equivalent resistance of the circuit is the sum of all the resistances:
$R_{eq}=R_1 + R_{234} = 5 \Omega + 0.9 \Omega = 5.9 \Omega$
So, the correct answer is D)

2) Let's start by calculating the equivalent resistance of the two resistors in parallel:
$\frac{1}{R_{23}} = \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{5 \Omega}+ \frac{1}{5 \Omega}= \frac{2}{5 \Omega}$
From which we find
$R_{23} = 2.5 \Omega$

And these are connected in series with a resistor of $10 \Omega$ , so the equivalent resistance of the circuit is
$R_{eq}=10 \Omega + 2.5 \Omega = 12.5 \Omega$

And by using Ohm's law we find the current in the circuit:
$I= \frac{V}{R_{eq}}= \frac{9 V}{12.5 \Omega}=0.72 A$
So, the correct answer is C).

3) Let' start by calculating the equivalent resistance of the two resistors in parallel:
$\frac{1}{R_{23}} = \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{5 \Omega}+ \frac{1}{5 \Omega}= \frac{2}{5 \Omega}$
From which we find
$R_{23} = 2.5 \Omega$
Then these are in series with all the other resistors, so the equivalent resistance of the circuit is
$R_{eq}=R_1 + R_{23}+R_4 = 5 \Omega + 2.5 \Omega + 5 \Omega =12.5 \Omega$

And by using Ohm's law we find the current flowing in the circuit:
$I= \frac{V}{R_{eq}}= \frac{12 V}{12.5 \Omega}=0.96 A$

And so the voltage read by the voltmeter V1 is the voltage drop across the resistor 2-3:
$V= I R_{23} = (0.96 A)(2.5 \Omega)=2.4 V$
So, the correct answer is D).

4) Again, let's start by calculating the equivalent resistance of the two resistors in parallel:
$\frac{1}{R_{23}} = \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{13 \Omega}+ \frac{1}{18 \Omega}=0.13 \Omega^{-1}$
From which we find
$R_{23} = 7.55 \Omega$

Now all the resistors are in series, so the equivalent resistance of the circuit is:
$R_{eq}= R_1 + R_{23}+R_4=8.5 \Omega+7.55 \Omega + 3.2 \Omega = 19.25 \Omega$

The current in the circuit is given by Ohm's law
$I= \frac{V}{R_{Eq}}= \frac{15 V}{19.25 \Omega}=0.78 A$

Now we can compare the voltage drops across the resistors. Resistor 1:
$V_1 = I R_1 = (0.78 A)(8.5 \Omega)=6.63 V$
Resistor 2 and resistor 3 are in parallel, so they have the same voltage drop:
$V_2 = V_3 = V_{23} = I R_{23} = (0.78 A)(7.55 \Omega)=5.89 V$
Resistor 4:
$V_4 = I R_4 = (0.78 A)(3.2 \Omega)=2.50 V$

So, the greatest voltage drop is on resistor 1, so the correct answer is D).

5) the figure shows a circuit with a resistor R and a capacitor C, so it is an example of RC circuit. Therefore, the correct answer is D).

6) The circuit is the same as part 4), so the calculations are exactly the same. Therefore, the power dissipated on resistor 3 is
$P_3 = I_3^2 R_3 = \frac{V_3^2}{R_3}= \frac{(5.89 V)^2}{18 \Omega}=2.0 W$
So, correct answer is B).

7) The circuit is the same as part 4), so we can use exactly the same calculation, and we immediately see that the resistor with lowest voltage drop was R4 (2.50 V), so the correct answer is B) R4.

5 0

3 years ago

Read 2 more answers

True or False? In an RL circuit, the larger the value of R (i.e. the resistance of the circuit), the longer it takes the circuit

Luba_88 [7]

Answer:

1.False

2.4 E

Explanation:

1.We have to find the statement is true or false.

We know that

Time constant, $\tau=\frac{L}{R}$

Time constant is inversely proportional to resistance R .It means when R increases then time constant decreases when R decrease then the time constant decreases.

Statement:The larger the value of R ,the longer it takes the circuit to reach its final steady- state value

It is false.

2.Energy stored in inductor,E= $\frac{1}{2}LI^2$

When I'=2I

Then, $E'=\frac{1}{2}L(2I)^2=\frac{1}{2}LI^2\times 4=4 E$

Hence,the new total energy will become 4 E.

3 0

3 years ago