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3 years ago

A neutral object develops an electric charge when it either gains or loses electrons

Physics

1 answer:

natima [27]3 years ago

8 0

Answer:

True

Explanation:

A neutral object is an object whose net charge is zero, so the sum of the positive charges is equal to the sum of negative charges:

$Q=Q_{pos}+Q_{neg}=0\\Q_{pos} = -Q_{neg}$

If the neutral object develops an electric charge (= different from zero), it means that this balance has changed. In particular, usually electric charge is carried by electrons (negative charges), so the object has either gained or lost electrons.

In particular:

- if the object has gained electrons, it has became negatively charged

- If the object has lost electrons, it has became positively charged

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When charged objects are brought close (not touching) to uncharged objects, what occurs?

seraphim [82]

They will become equally negative charges, I believe.

6 0

3 years ago

Calculate the radiative and collisional energy losses (in keV/micron) for a 1.9 MeV electron in lead and determine the rad./coll

Andrew [12]

Answer:

Explanation:

During an energy transfer, the collision loss for an electron can be determined by using the formula:

$Q = \dfrac{4mME }{(m+M)^2}$

However; from the total stopping power & power loss of the electron;

$\dfrac{radiational \ energy \ loss}{colisional \ energy \ loss } = \dfrac{ZE}{800}$

where;

Z = atomic no. for lead = 82

E = 1.9 MeV

∴

radiational energy loss = collisional energy loss $=\dfrac{82 \times 1.9}{800}$

= 0.19475

Normally, the traditional lead shielding in its pure shape contains high brittleness. However, the functionality of this carbon group chemical element is useful for protection because it has an excessive density.

Initially, the conventional lead protection however reduces the mild clarity at the same moment as plexiglass is useful for light transmittance and readability.

Moreover, the traditional lead with its high density and thickness reduces observation features, in the meantime, the plexiglass is a whole lot higher than the stated.

Finally, plexiglass contains a high dimensional balance with an excessive dielectric constant.

4 0

3 years ago

Electrically neutral objects cannot exert an electrical force on each other, but they can exert a gravitational force on each ot

Olenka [21]

The correct answer is
A) true

In fact, the electrical force can be present only if the two objects have a net charge different from zero, because the electrical force is proportional to the value of the charges:
$F_e = k_e \frac{Q_1 Q_2}{r^2}$
so, if they are neutral they cannot exert an electrical force between each other, because their charge Q is zero.

Viceversa, the gravitational force is always present when the objects have masses, since the force is proportional to the value of the masses:
$F_g = G \frac{m_1 m_2}{r^2}$
so, even if the two objetcs are neutral, if they have mass, they exert a gravitational force to each other.

4 0

4 years ago

_______is used as a magnifying glass to read very small print.

Alexus [3.1K]

Answer:

The answer is B

Explanation:

Concave lens

7 0

2 years ago

Read 2 more answers

13.An airplane starts from rest at the end of a runway and accelerates at a constant rate. In the first second, the airplane tra

Murrr4er [49]

Answer:

The answer is 3.33m

Explanation:

The acceleration "a" is constant.

Acceleration is the variation of velocity over time,

$\frac{dv}{dt} = a$ .

solving the last equation

$\int_{v_0}^v dv = a\int_0^t dt \rightarrow v-v_0 = at$ ,

where $v_0=0$ because the airplane starts from rest.

Once again, velocity is the variation of distance over time.

$\frac{dx}{dt} = at \rightarrow \int_{x_0}^x dx = a\int_0^t t\ dt$

then

$x- x_0 = \frac{1}{2}at^2$

where $x_0=0$ if we consider the end of the runway as the initial point (this step is for simplicity but you can let it expressed, it's going to cancel anyway).

If $x=1.11\ m$ at $t=1s$ , then

$a = \frac{2x}{t^2} = 2.22\ m/s^2$

and the final expression for the distance is

$x = 1.11 t^2$ .

If t = 2s, x = 4.44 m. Which means thad the additional distance is

$x(2s) - x(1s) = 4.44 - 1.11 = 3.33\ m$

8 0

4 years ago