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Readme [11.4K]
3 years ago
10

Which statement best describes the electrons in this illustration ?

Physics
2 answers:
Nonamiya [84]3 years ago
7 0
A. electrons are negatively charged outside nucleus
koban [17]3 years ago
5 0
A is obviously true
the electrons are located around the nucleus (nucleus is made up of protons and neutrons.)
electrons are negatively charge particles
protons are positively charged particles
neutrons are neutral
hope this helps
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it is about 384,750 kilometers from earth to the moon. it took the apollo astronauts about 2 days and 19.5 hours to fly to the m
Jet001 [13]
We know, speed = Distance / Time
d = 384,750 Km
t = 2 days, 19.5 hours = 48+19.5 = 67.5 hour

Substitute their values, 
s = 384,750 / 67.5
s = 5700 Km/h

In short, Your Answer would be 5700 Km/h

Hope this helps!
7 0
3 years ago
Read 2 more answers
The relationship between an object's acceleration and its mass
Shalnov [3]

Answer:

Newton's second law

Explanation:

The relationship between mass and acceleration is described in Newton's Second Law of Motion. His Second Law states that the more mass an object has, more force is necessary for it to accelerate.

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3 years ago
What does 'change only one variable at a time' mean
matrenka [14]

Answer: to only change one factor in an experiment or test

8 0
3 years ago
A car travels a distance of 540km in 6 hours. What speed did it travel at?
maria [59]

Speed v = distance travelled / time taken

v = d / t

v =  540 / 60h

v =   9 km /h

4 0
3 years ago
Read 2 more answers
A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the
irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

x=V_{o} cos \theta t (1)

y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2} (2)

Where:

x is the horizontal distance between the cannon and the ball

V_{o}=23 m/s is the cannonball initial velocity

\theta=0\° since the cannonball was shoot horizontally

t is the time

y=0 is the final height of the cannonball

y_{o}=1.96 m is the initial height of the cannonball

g=9.8 m/s^{2} is the acceleration due gravity

Isolating t from (2):

t=\sqrt{-\frac{2(y-y_{o})}{g}} (3)

t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}} (4)

t=0.63 s (5)

Substituting (5) in (1):

x=(23 m/s) cos(0\°) 0.63 s (6)

Finally:

x=14.49 m

5 0
3 years ago
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