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2 years ago

How does a positive ion differ from an uncharged atom of the same element?

2 answers:

8 0

Atoms vs. Ions. Atoms are neutral; they contain the same number of protons as electrons. By definition, an ion is an electrically charged particle produced by either removing electrons from a neutral atom to give a positive ion or adding electrons to a neutral atom to give a negative ion.

Leto [7]2 years ago

8 0

Answer: The ion has fewer electrons.

Explanation:

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One form of energy can be _____ another type of energy.

kotegsom [21]

We can answer this one very quickly. From the <em>Law of Conservation of Energy</em>, we know that "Energy can't be created or destroyed.".

So that only leaves us one way to complete the sentence in this question:

"One form of energy can be <em>transformed into</em> another type of energy. " <em>(B)</em>

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2 years ago

A 0.10 g honeybee acquires a charge of +23 pC while flying.

kari74 [83]

Answer:

a) $\frac{F}{w} =2.347\times 10^{-6}\ N$

b) $E=4.2609\times 10^7\ N.C^{-1}$ parallel to the earth surface.

In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.

Explanation:

Given:

mass of the bee, $m=10^{-4}\ kg$

charge acquired by the bee, $q_2=23\times 10^{-12}\ C$

Electrical field near the earth surface, $E=100\ N.C^{-1}$

Now the electric force on the bee:

we know:

$F=\frac{1}{4\pi.\epsilon_0} \times \frac{q_1.q_2}{r^2}$

$F=E.q_2$

$F=100\times 23\times 10^{-12}$

$F=23\times 10^{-10}\ N$

The weight of the bee:

$w=m.g$

$w=10^{-4}\times 9.8$

$w=9.8\times10^{-4}\ N$

Therefore the ratio :

$\frac{F}{w} =\frac{23\times 10^{-10}}{9.8\times10^{-4}}$

$\frac{F}{w} =2.347\times 10^{-6}\ N$

The condition for the bee to hang is its weight must get balanced by the electric force acing equally in the opposite direction.

So,

$F=9.8\times10^{-4}\ N$

$E.q_2=9.8\times10^{-4}\ N$

$E\times 23\times 10^{-12}=9.8\times10^{-4}\ N$

$E=4.2609\times 10^7\ N.C^{-1}$ parallel to the earth surface.

In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.

3 0

2 years ago

How is the tesla coil used today

marshall27 [118]

H<span>igh voltage demonstrations!</span>

6 0

2 years ago

Read 2 more answers

The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s. (a) What

Neko [114]

Complete Question

The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s.

(a) What is its angular acceleration in revolutions per minute-squared

(b) How many revolutions does the engine make during this 20 s interval?

rev

Answer:

$\alpha = 6261 \ rev/minutes^2$

$\theta = 613 \ revolutions$

Explanation:

From the question we are told that

The initial angular speed is $w_i = 1120 \ rev/minutes$

The angular speed after $t = 13.8 s = \frac{13.8}{60 } = 0.23 \ minutes$ is $w_f = 2560 \ rev/minutes$

The time for revolution considered is $t_r = 20 \ s = \frac{20}{60} = 0.333 \ minutes$

Generally the angular acceleration is mathematically represented as

$\alpha = \frac{w_f - w_i }{t}$

=> $\alpha = \frac{2560 - 1120 }{0.23}$

=> $\alpha = 6261 \ rev/minutes^2$

Generally the number of revolution made is $t_r = 20 \ s = \frac{20}{60} = 0.333 \ minutes$ is mathematically represented as

$\theta = \frac{1}{2} * (w_i + w_f)* t$

=> $\theta = \frac{1}{2} * (1120+ 2560 )* 0.333$

=> $\theta = 613 \ revolutions$

5 0

2 years ago

What cloud extending to the gravitational limits of the solar system would comets come from?.

ser-zykov [4K]

The Oort cloud extends to the gravitational limits of the solar system would comets come from.

<h3>What is oort cloud?</h3>

It is a hypothetical idea of a cloud of mostly frozen planetesimals that would orbit the Sun at distances between 2,000 and 200,000 AU.

The Oort cloud reaches the solar system's outer gravitational boundaries, where comets originate.

Hence cloud extending to the gravitational limits of the solar system will be oorto cloud.

To learn more about the oort cloud refer;

brainly.com/question/23368033

#SPJ1

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1 year ago