Some examples of insulators are plastic, glass and rubber.
Answer: Dalton’s model
Explanation:
In the attached image we can see four atomic models labeled with four letters:
W represents the current and accepeted atomic model: a nucleus with an electron cloud, where the orbit and position of the electrons around the nucleus is defined by specific regions (associated with specific energy levels) where there is a greater probability of finding the electron at any given moment. It is important to note this model was improved by the works in quantum physics done by Louis de Broglie and Erwin Schrodinger.
X represents Rutherford's model (This model was proposed after Thomson's model). Ernest Rutherford conducted a series of experiments in order to corroborate Thomson's atomic model. However the results of the experiment led him to find out there is a concentration of charge in the atom's core (which was later called nucleus) surrounded by electrons. This lead to a new atomic model, in which the atom has a positive charged nucleus surrounded by negative charged particles that move similar to the orbit of the planet around the Sun.
Y represents Thomson's model, also called the <em>plum pudding</em> model. This scientific found out that atoms contain small subatomic particles with a negative charge (later called electrons). However, taking into consideration that at that time there was still no evidence of the atom nucleus, Thomson thought the electrons were immersed in the atom of positive charge that counteracted the negative charge of the electrons. Just like the raisins embedded in a pudding or bread.
Z represents Bohr's model. This model was proposed by the danish physicist Niels Bohr after Rutherford's model. In fact, this model was Rutherford's model with the following addition: electrons orbit the nucleus (like planets around the sun) in specific orbits at different energy levels around the nucleus.
So, the only missing model is <u>Dalton's model</u>, which was the first atomic model: the atom represented as a solid, indestructible and indivisible mass. An idea that was already accepted by that time since the ancient Greeks.
<span>(a) 12.02 m/s
(b) 52.2 meters
This problem is an example of integral calculus. You've been given an acceleration vector which is usually known as the 2nd derivative. From that you need to calculate the velocity function (1st derivative) and position (actual function) by successively calculating the anti-derivative. So:
A(t) = 6.30 - 2.20t
V(t) = 6.30t - 1.10t^2 + C
We now have a velocity function, but need to determine C. Since we've been given the velocity at t = 0, that's fairly trivial.
V(t) = 6.30t - 1.10t^2 + C
3 = 6.30*0 - 1.10*0^2 + C
3 = 0 + 0 + C
3 = C
So the entire velocity function is:
V(t) = 6.30t - 1.10t^2 + 3
V(t) = -1.10t^2 + 6.30t + 3
Now for the location function which is the anti-derivative of the velocity function.
V(t) = -1.10t^2 + 6.30t + 3
L(t) = -0.366666667t^3 + 3.15t^2 + 3t + C
Now we need to calculate C. And once again, we've been given the location for t = 0, so
L(t) = -0.366666667t^3 + 3.15t^2 + 3t + C
7.3 = -0.366666667*0^3 + 3.15*0^2 + 3*0 + C
7.3 = 0 + 0 + 0 + C
7.3 = C
L(t) = -0.366666667t^3 + 3.15t^2 + 3t + 7.3
Now that we have the functions, they are:
A(t) = 6.30 - 2.20t
V(t) = -1.10t^2 + 6.30t + 3
L(t) = -0.366666667t^3 + 3.15t^2 + 3t + 7.3
let's answer the questions.
(a) What is the maximum speed achieved by the cyclist?
This can only happen at those points that meet either of the following criteria.
1. The derivative is undefined for the point.
2. The value of the derivative is 0 for the point.
As it turns out, the 1st derivative of the velocity function is the acceleration function which we have. So
A(t) = 6.30 - 2.20t
0 = 6.30 - 2.20t
2.20t = 6.30
t = 2.863636364
So one of V(0), V(2.863636364), or V(6) will be the maximum value. Therefore:
V(0) = 3
V(2.863636364) = 12.0204545454545
V(6) = 1.2
So the maximum speed achieved is 12.02 m/s
(b) Total distance traveled?
L(0) = 7.3
L(6) = 59.5
Distance traveled = 59.5 m - 7.3 m = 52.2 meters</span>
Earth's land and ocean surface to the atmosphere.