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emmasim [6.3K]
3 years ago
5

Describe J.J Thomson's plum pudding model,and explain how it compares to the modern day atomic theory.​

Physics
2 answers:
juin [17]3 years ago
8 0

The correct answer to this open question is the following.

The J.J Thomson's plum pudding model establishes that when he researched the atom, he found that electrons are positively charged. Because he observed that electrons were embedded like plums. Electrons are negative. However, compared to the modern-day atomic theory, scientist Ernest Rutherford established that the space between an atom is empty, the nucleus of the atom is positive, and negative electrons have fixed motion paths.

natita [175]3 years ago
4 0

Answer:

J J Thomson's plum pudding model indicated that <em>the negative electrons represented the raisins in the pudding and the dough contained the positive charge.</em>

Explanation:

Thomson discovered electron in 1897 and demonstrated that cathode rays were negatively charged. In addition, he also studied positively charged particles in neon gas. Thomson realized that the accepted model of an atom did not account for negatively or positively charged particles. Therefore, he proposed a model of the atom which he likened to plum pudding.  The model indicate that

<em>The negative electrons represented the raisins in the pudding and the dough contained the positive charge. </em>

Thomson's model of the atom did explain some of the electrical properties of the atom due to the electrons, but failed to recognize the positive charges in the atom as particles.

<em>The current model indicate the presence of the  a positively charged center of the atom that is denser than the rest of it called the nucleus. This dense center is made up of positively charged protons and neutrally charged neutrons. Around the outside of the nucleus the electrons are organized on rings. These electrons are arranged in  a certain pattern that is the same for all atoms.</em>

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Explanation:

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3 years ago
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2 years ago
A 1300 kg steel beam is supported by two ropes. (Figure
Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

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R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0

where R_1,R_2 are the magnitudes of the tensions in ropes 1 and 2, respectively;

  • the net vertical force acting on the beam is

R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0

where m=1300\,\rm kg and g=9.8\frac{\rm m}{\mathrm s^2}.

Eliminating R_2, we have

\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)

R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2

R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2

-R_1 \sin(50^\circ) = -\dfrac{mg}2

R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}

Solve for R_2.

\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0

\dfrac{R_2}2 = -mg\cot(110^\circ)

R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}

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