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Allisa [31]
2 years ago
5

Which components of an atom are found outside of the nucleus

Physics
2 answers:
butalik [34]2 years ago
3 0
Electrons are found outside of the nucleus.
irina [24]2 years ago
3 0

Answer:

electron

Explanation:

I took the test  

You can see other question and answer in my article.  

Just search in search engine with: Learningandassignments diy4pro  

Click on my site and find these related article post:  

Atoms Quiz- Grade 8 Unit 2 Leson 2

Hope it helps.  

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a car traveling at a velocity of 2 m/s undergoes an acceleration of 4.5 m/s^2 over a distance of 340 m. How fast will it be goin
ra1l [238]
Vi = 2m/s
a= 4.5 m/s 
d= 340 m
vf= ?

use this equation ...  vf^2=vi<span>^2+2ad

you should get vf = 55.3
hope this helps </span>
3 0
3 years ago
I Need answer ASAP
anzhelika [568]

Answer: Convection and conduction

Tell me that I got it right??

Explanation

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3 0
3 years ago
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Una varilla de 5m de longitud y 1.5 cm^2 de sección transversal se alarga 0.10 cm al someterla a una tensión de 700 N. Determina
KengaRu [80]
Bzzhbzaijsb she’s aha ha ha s she’s has shah a sus abuse she d dvhbx
4 0
2 years ago
The energy content of a certain food is to be determined in a bomb calorimeter that contains 3 kg of water by burning a 2-g samp
GuDViN [60]

Answer:

20.179 x 10⁶ J /kg

Explanation:

The food after the reaction gives out heat which increases the temperature of water and air in the reaction chamber . The heat absorbed by water and air gives the estimate of energy content of the food.

Heat absorbed by water = mass x specific heat x rise in temperature

=  3 x 4.18 x 10³ x 3.2

= 40.128 x 10³ J

Heat absorbed by air  = mass x specific heat x rise in temperature

0.1 x 3. 2 x .718 x 10³

= 0.23 x 10³

Total heat energy evolved

= 40.358 x 10³ J

This energy is evolved by 2 x 10⁻³ kg of food

energy content per kg of food

= 40.358 x 10³ / 2 x 10⁻³

= 20.179 x 10⁶ J /kg

7 0
2 years ago
Calculate the maximum absolute uncertainty for R if:
Radda [10]

Answer:

ΔR = 9 s

Explanation:

To calculate the propagation of the uncertainty or absolute error, the variation with each parameter must be calculated and the but of the cases must be found, which is done by taking the absolute value

           

The given expression is      R = 2A / B

the uncertainty is                 ΔR = | \frac{dR}{dA} | ΔA + | \frac{ dR}{dB} | ΔB

we look for the derivatives

     \frac{dR}{dA} = 9 / B

     \frac{dR}{dB} = 9A ( - \frac{1}{B^2 } )

we substitute

     ΔR = \frac{9}{B}  ΔA + \frac{9A}{B^2}  ΔB

the values ​​are

     ΔA = 2 s

     ΔB = 3 s

 

     ΔR = \frac{9}{11}   2 + \frac{9 \ 32}{11^2 }  3

     ΔR = 1.636 + 7.14

     ΔR = 8,776 s

the absolute error must be given with a significant figure

     ΔR = 9 s

3 0
3 years ago
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