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3 years ago

How did Rutherford's experiments demonstrate that Thomson's model of the atom was incorrect?

2 answers:

8 0

Rutherford's Gold foil experiment involved the firing of atoms at a piece of gold foil only a few atoms thick. Rutherford expected that the particles would go straight through, but some were deflected. This experiment proved that atoms have a dense positively charged nucleus.

Answer: d. They showed that atoms have a dense nucleus.

mestny [16]3 years ago

8 0

Answer:

d) They showed that atoms have a dense nucleus.

Explanation:

Thomson's model predicted that the atom is composed of negatively charged electrons, which is surrounded by positively charged cloud. This was called Plum pudding model of the atom.

But Rutherford proved it to be incorrect . His gold foil experiment showed that the atom's central part is heavy and positively charged. The deflections of the alpha particles showed that the heavy central region is surrounded by negatively charged particles called electrons.

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A high-jumper, having just cleared the bar, lands on an air mattress and comes to rest. Had she landed directly on the hard grou

Hoochie [10]

Answer:

e. the air mattress exerts the same impulse, but a smaller net avg force, on the high-jumper than hard-ground.

Explanation:

This is according to the Newton's second law and energy conservation that the force exerted by the hard-ground is more than the force exerted by the mattress.

The hard ground stops the moving mass by its sudden reaction in the opposite direction of impact force whereas the mattress takes a longer time to stop the motion of same mass in a longer time leading to lesser average reaction force.

Mathematical expression for the Newton's second law of motion is given as:

$F=\frac{dp}{dt}$ ............................................(1)

where:

dp = change in momentum

dt = time taken to change the momentum

We know, momentum:

$p=m.v$

Now, equation (1) becomes:

$F=\frac{d(m.v)}{dt}$

∵mass is constant at speeds v << c (speed of light)

$\therefore F=m.\frac{dv}{dt}$

and, $\frac{dv}{dt} =a$

where: a = acceleration

$\Rightarrow F=m.a$

also

$F\propto \frac{1}{dt}$

so, more the time, lesser the force.

& Impulse:

$I=F.dt$

$I=m.a.dt$

$I=m.\frac{dv}{dt}.dt$

$I=m.dv=dp$

∵Initial velocity and final velocity(=0), of a certain mass is same irrespective of the stopping method.

So, the impulse in both the cases will be same.

4 0

3 years ago

Me Ayudan con este ejercicio por favor !!!

klasskru [66]

Tom llega primero a la superficie de el lago

6 0

3 years ago

Lori’s family is on a road trip. They split their drive into the five legs listed in the table. Find the average velocity for ea

PIT_PIT [208]

Average velocity is displacement divided by time elapsed; Δv/Δt

You will need to use the information in the table you are given. Subtract: (final velocity - initial velocity) and divide by (final time - initial time).

3 0

3 years ago

How long can you keep open beer in the fridge?

Gre4nikov [31]

You can keep it for as long as you want; you're the boss of your fridge.
But anyway, an open beer bottle can stay good in the fridge for up to 96 hours, probably more.

7 0

3 years ago

Read 2 more answers

What belongs in the center section

MissTica

Answer:

The second one I think

Explanation:

8 0

3 years ago