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3 years ago

How does an atom change if all of its electrons are removed?

2 answers:

6 0

Because the transfer of electrons is called an ion, and also, they are being lost, they atom will become positively charged. The Ion will lose electrons and become a cation.

Answer: The atom becomes positively charged.

Tanzania [10]3 years ago

4 0

The correct answer to the question is : C) The atom becomes positively charged.

EXPLANATION :

An isolated atom is electrically neutral. It is so because the total number of electrons is equal to total number of protons.

As per the question, all the electrons from the atom are removed.

We know that electron is a negatively charged particle, and proton is a positively charged particle.

When all the electrons are removed from the atom, the atom contains only protons and neutrons. The neutron is a neutral nucleon. Hence, the atom will be positively charged due to the excess number of protons.

Hence, the correct answer is that atom will be positively charged.

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The parallel axis theorem relates Icm, the moment of inertia of an object about an axis passing through its center of mass, to I

erma4kov [3.2K]

Answer:

Part a)

$I_{end} = \frac{mL^2}{3}$

Part b)

$I_{edge} = \frac{2ma^2}{3}$

Explanation:

As we know that by parallel axis theorem we will have

$I_p = I_{cm} + Md^2$

Part a)

here we know that for a stick the moment of inertia for an axis passing through its COM is given as

$I = \frac{mL^2}{12}$

now if we need to find the inertia from its end then we will have

$I_{end} = I_{cm} + Md^2$

$I_{end} = \frac{mL^2}{12} + m(\frac{L}{2})^2$

$I_{end} = \frac{mL^2}{3}$

Part b)

here we know that for a cube the moment of inertia for an axis passing through its COM is given as

$I = \frac{ma^2}{6}$

now if we need to find the inertia about an axis passing through its edge

$I_{edge} = I_{cm} + Md^2$

$I_{edge} = \frac{ma^2}{6} + m(\frac{a}{\sqrt2})^2$

$I_{edge} = \frac{2ma^2}{3}$

7 0

3 years ago

A 74 KG student starting from rest slides down 11.8 m high water slide how fast is it going at the bottom of the slide?

-Dominant- [34]

the answer is going to be 873.2

8 0

3 years ago

Calculate the kinetic energy, in joules of a 1160-kg automobile moving at 19.0 m/s.

Rudiy27

Answer:

Explanation:

K.E=1/2 mv²

m=1160 kg

v=19.0 m/s

so k.e=1/2*1160*(19.0)²

K.E=1/2*1160*361

K.E=1/2*418760

K.E=209380=2.0*10^5 j

7 0

2 years ago

A can of sardines is made to move along an x axis from x = 0.47 m to x = 1.20 m by a force with a magnitude given by F = exp(–8x

sattari [20]

If the force were constant or increasing, we could guess that the speed of the sardines is increasing. Since the force is decreasing but staying in contact with the can, we know that the can is slowing down, so there must be friction involved.
Work is the integral of (force x distance) over the distance, which is just the area under the distance/force graph.
The integral of exp(-8x) dx that we need is (-1/8)exp(-8x) evaluated from 0.47 to 1.20 .

I get 0.00291 of a Joule ... seems like a very suspicious solution, but for an exponential integral at a cost of 5 measly points, what can you expect. On the other hand, it's not really too unreasonable. The force is only 0.023 Newton at the beginning, and 0.000067 newton at the end, and the distance is only about 0.7 meter, so there certainly isn't a lot of work going on. The main question we're left with after all of this is: Why sardines ? ?

6 0

2 years ago

Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the ot

almond37 [142]

Answer:

The kinetic coefficient of friction of the crate is 0.235.

Explanation:

As a first step, we need to construct a free body diagram for the crate, which is included below as attachment. Let supposed that forces exerted on the crate by both workers are in the positive direction. According to the Newton's First Law, a body is unable to change its state of motion when it is at rest or moves uniformly (at constant velocity). In consequence, magnitud of friction force must be equal to the sum of the two external forces. The equations of equilibrium of the crate are:

$\Sigma F_{x} = P+T-\mu_{k}\cdot N = 0$ (Ec. 1)