What keeps the electrons from leaving the atom in the thompson "raisin pudding" model of the atom?
2 answers:
The electrons keep from leaving the atom in the Thompson "raisin pudding" model of the atom because
a. they are embedded in a spherical cloud of a positive charge.
<h3>Further explanation</h3>John Dalton put forward the first theory about atoms which are particles that cannot be divided again
This is where research on atomic particles begins.
JJ Thomson conducted a cathode ray tube experiment
In this experiment, the cathode ray beam is deflected by a magnetic field. So it was concluded by JJ Thompson that the cathode ray has the same electric charge as the hydrogen atom in the electrolysis of water. And these negatively charged cathode ray particles are the basic particles that must be present in every atom
So JJ Thompson proposes an atomic model such as raisin bread or better known as JJ Thompson's raisin pudding model of the atom
JJ Thompson estimates that the atom consists of a cloud of negative charges in a sphere of positive charges
Atoms have neutral charges because positive and negative charges have the same amount
This atomic model can also be likened to a watermelon where the red edible part of a watermelon is part of a positive charge and black seeds filling the watermelon is a negative charge
Also described as a pudding because the model looks like dry fruits embedded in a sphere of positive charge
The existence of the nucleus in the next atom was discovered by Rutherford through experiments with firing alpha rays on a thin gold plate (gold foil)
And followed by the Bohr atomic model, which states that electrons circulate the nucleus on a particular path, so that electrons are not attracted to the nucleus
<h3>Learn more</h3>Bohr's model of the atom
Bohr's and Rutherford's model of the atom
Rutherford performed the gold foil experiment
Keywords: "raisin pudding" model of the atom, the gold foil experiment, a watermelon, cathode ray
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Answer:
Explanation:
Given
Minute hand length =16 cm
Time at a quarter after the hour to half past i.e. 1 hr 45 min
Angle covered by minute hand in 1 hr is 360 and in 45 minutes 270
(c)For the next half hour
Effectively it has covered 2 revolution and a quarter
angle turned
(f)Hour after that
After an hour it again comes back to its original position thus displacement is same =25.136
Angle turned will also be same i.e.
The resultant static friction force is equal to 20 N to the left.
Why?
I'm assuming that you forgot to write the question of the exercise, so, I will try to complete it:
"A 50-n crate sits on a horizontal floor where the coefficient of static friction between the crate and the floor is 0.50 . A 20-n force is applied to the crate acting to the right. What is the resulting static friction force acting on the crate?"
So, if we are going to calculate the resulting static friction force, it means that there is no movement, we must remember that the friction coefficient will give us the maximum force before the crate starts to move.
We can calculate the static friction force by using the following formula:
Since the crate is not moving (static), the static friction force acting on the crate will be equal to the applied force.
Calculating we have:
Hence, the static friction force is equal to 20 N to the left (since the applied force is acting to the right)
So,
Since the static friction force is equal to the applied force, the crate does not start to move.
Have a nice day!