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3 years ago

Which atomic model was proposed as a result of j. J. Thomson’s work?

Physics

1 answer:

EleoNora [17]3 years ago

4 0

<h2>Answer: The "raising pudding" atomic model</h2>

Explanation:

<u>During the 19th century the accepted atomic model, was Dalton's atomic model</u>, which postulated the atom was an <u>"individible and indestructible mass".</u>

However, at the end of 19th century J.J. Thomson began experimenting with cathode ray tubes and found out that atoms contain small subatomic particles with a negative charge (later called <u>electrons</u>). This meant the atom was not indivisible as Dalton proposed. So, Thomson developed a new atomic model.

Taking into consideration that at that time there was still no evidence of the atom nucleus, Thomson thought the electrons (with negative charge) were immersed in the atom of positive charge that counteracted the negative charge of the electrons. <u>Just like the raisins embedded in a pudding or bread. </u>

That is why this model was called the <u>raisin pudding atomic model.</u>

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A large pizza is cut into 8 even slices. A person orders 4 large pizzas from a restaurant. How many total slices of pizza did th

Nikitich [7]

Answer:

<h2>32 slices</h2>

Explanation:

Step one:

given data

we are told that 1 large pizza can be cut into 8 even slices

Required

we want to find how many slices are there in 4 large pizzas

Step two:

so if 1 pizza has 8 slices

4 pizza will have x

cross multiply we have

x= 8*4

x=32 slices

7 0

3 years ago

A 10 kg frictionless cart is pushed at a constant force of 5.0 N for a distance of 10 m. The work done on the cart is 50J.

kykrilka [37]

1) Final kinetic energy of the cart: B) 50 J

2) Final speed of the cart: C) 3.2 m/s

3) Height reached along the ramp: A) 0.5 m

Explanation:

We can solve this part of the problem by using the work-energy theorem, which states that the work done on an object is equal to the kinetic energy gained by the object itself. Mathematically:

$W=K_f - K_i$

where

W is the work done

$K_f$ is the final kinetic energy

$K_i$ is the initial kinetic energy

In this problem, the work done on the cart is

W = 50 J

And assuming it starts from rest, its initial kinetic energy is zero:

$K_i = 0$

Therefore, the final kinetic energy is:

$K_f = K_i + W=0+50=50 J$

The kinetic energy of an object is the energy possessed by an object due to its motion; it is calculated as

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

For the cart in this problem, we have:

K = 50 J is its final kinetic energy

m = 10 kg is the cart

Therefore, solving the formula for v, we find its speed:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(50)}{10}}=3.2 m/s$

We can think this problem in terms of conservation of energy. In fact, as the cart rolls up the ramp, its kinetic energy is converted into gravitational potential energy, which is given by

$PE=mgh$