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3 years ago

How did rutherford's model of the atom differ from thomson's model?

1 answer:

7 0

Thompson thought atoms were made of a + nearly uniform medium (except for occasional electron)
Rutherford thought atoms had a central + nucleus with electrons orbiting around it.

So nuclear model vs non-nuclear model

You might be interested in

Which planet has a moon with many sulfur volcanoes?

Murljashka [212]

Jupiter i hope it is right answer

6 0

3 years ago

What is the function of the switch in an electric circuit?

aliya0001 [1]

Answer:

A switch opens or closes the electrical circuit, turning the flow of electricity on or off.

Explanation:

3 0

3 years ago

Read 2 more answers

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

3 years ago

A ball has a mass of 0.046kg. Calculate the change in gravitational potential energy when the ball is lifted through a vertical

loris [4]

Answer:

PE=0.92414J and KE=0.28175J

Explanation:

Gravitational potential energy=mass*gravity*height

PE=mgh

Data,

M=0.046kg

H=2.05m

g=9.8m/s^2

PE=0.046kg * 9.8m/s^2 * 2.05m

PE =0.92414J

KE=1/2mv^2

M=0.046kg

V=3.5m/s

KE=[(0.046kg)*(3.5m/s)^2]\2

KE=0.28175J

3 0

3 years ago

How much potential energy does a 40-N medicine ball gain when it is lifted 5 m?

777dan777 [17]

We know, Potential Energy = Force * Height
Here, F = 40 N
h = 5 m

Substitute their values,
U = 40 * 5
U = 200 J

In short, Your Answer would be Option A

Hope this helps!

6 0

3 years ago