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3 years ago

Which components of an atom are found outside of the nucleus

Physics

2 answers:

butalik [34]3 years ago

3 0

Electrons are found outside of the nucleus.

irina [24]3 years ago

3 0

Answer:

electron

Explanation:

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Atoms Quiz- Grade 8 Unit 2 Leson 2

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An astronaut finds herself in a predicament in which she has become untethered from her shuttle. She figures that she could get

Blizzard [7]

In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.

The equation that defines the linear moment is given by

$mV_i = (m-m_O)V_f - m_OV_O$

where,

m=Total mass

$m_O =$ Mass of Object

$V_i =$ Velocity before throwing

$V_f =$ Final Velocity

$V_O =$ Velocity of Object

Our values are:

$m_1=5.3kgm_2=7.9kg\\m_3=10.5kg\\m_A=75kg\\m_{Total}=m=98.7Kg$

Solving to find the final speed, after throwing the object we have

$V_f=\frac{mV_0+m_TV_O}{m-m_O}$

We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.

That way during each section the equations should be modified depending on the previous one, let's start:

A) $5.3Kg\rightarrow 15m/s$

$V_{f1}=\frac{mV_0+m_TV_O}{m-m_O}$

$V_{f1}=\frac{(98.7)*0+5.3*15}{98.7-5.3}$

$V_{f1}=0.8511m/s$

B) $7.9Kg\rightarrow 11.2m/s$

$V_{f2}=\frac{mV_{f1}+m_TV_O}{m-m_O}$

$V_{f2}=\frac{(98.7)(0.8511)+(7.9)(11.2)}{98.7-5.3-7.9}$

$V_{f2} = 2.0173m/s$

C) $10.5Kg\rightarrow 7m/s$

$V_{f3}=\frac{mV_{f2}+m_TV_O}{m-m_O}$

$V_{f3}=\frac{(98.7)(2.0173)+(10.5)(7)}{98.7-5.3-7.9-10.5}$

$V_{f3} = 3.63478m/s$

Therefore the final velocity of astronaut is 3.63m/s

7 0

3 years ago

Melissa's favorite exercise equipment at the gym consists of various springs. in one exercise, she pulls a handle grip attached

konstantin123 [22]

The formula for force exerted on/by a spring is

F = k*e where k is the spring constant and x is the distance stretched from unstrained position. This should allow you to find what you need.

Using F = k x e,

where k is the spring constant,

and e is the extension,

The F is her weight = 45 X 0.80

= 36 N

6 0

3 years ago

A 200kg ball on the end of string is swung in horizontal circle with radius of 0.5m . The ball makes revolution every 2second th

ANEK [815]

Answer: $\dfrac{\pi}{2} ms^{-1}$

Explanation:

Let $T$ be the time required to make one revolution.

Let $r$ be the radius of the circular path.

Let $d$ be the distance travelled by ball in one revolution.

As we know,the distance travelled in one revolution is the circumference of the circle.

So, $d=2\pi r$

Given, $d=0.5m\\T=2sec$

$d=2\times \pi \times 0.5=\pi m$

Speed of an object moving is circular path is define as the ratio of distance travelled in one revolution to the time taken by the object to complete one revolution.

Let $s$ be the speed of the ball.

$s=\frac{d}{T}=\frac{\pi }{2}ms^{-1}$

So,the speed of the ball is $\frac{\pi }{2}ms^{-1}$

5 0

3 years ago

How we can best share the observation of the light through one hole the paper

zhenek [66]

Drawing is an easy way to connect to your inner child

5 0

3 years ago

Why are strong forces able to hold atomic nuclei together?

Galina-37 [17]

The nucleons(protons and neutrons) are held together by means of this strong force. If this strong never existed, all the nucleus will blow themselves due to strong repulsive force between protons(neutron has no charge).
Thats it!
If I explain beyond, it will surely bounce off your head. Anyways, if you wanna know more bout it, ping me. (:

4 0

3 years ago