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4 years ago

How many electrons can fit in the fourth energy level

Physics

2 answers:

lidiya [134]4 years ago

8 0

The answer is 2 electrons because 2 + 6 + 10 + 14 = 32 electrons.

luda_lava [24]4 years ago

4 0

Hello User,

Approximately 32 electrons can be fit in the fourth energy level.

Solution:
2+4+6+10+10=32

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In a coal-burning power plant, a condenser is used to cool down the steam and convert it back to liquid water true or false

balandron [24]

False because they let it in to the atmosphere and then its condenced

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4 years ago

Read 2 more answers

?/1 Jorge traveled 5 miles north to school. He then traveled 3 miles west to the store. Then he left the store and traveled 5 mi

aleksklad [387]

Answer:

He's 3 miles west of school.

Explanation:

He went 5 miles up and 5 miles down which means that he really didn't go up or down. In between that, he went 3 miles west so if the 5 milers don't count, this puts him at 3 miles west of school.

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3 years ago

A force of 50 n acts on abody of mass 5 kg .calculate the acceleration produced

Leokris [45]

Answer:

a = 10 m/s²

Explanation:

Given:

Force = F = 50 N

Mass = m = 5 kg

Required:

Acceleration = a = ?

Formula:

F = ma

Solution:

Rearranging the formula for a

=> a = F/m

=> a = 50/5

=> a = 10 m/s²

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3 years ago

What is the weight in (newton's ) of a bowling ball which has a mass of 3 kg

Wewaii [24]

Check Google it might have the answer sorry I couldn't be much help

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3 years ago

A woman is 1.6 m tall and has a mass of 50 kg. She moves past an observer with the direction of the motion parallel to her heigh

eduard

To solve this problem it is necessary to apply the concepts related to linear momentum, velocity and relative distance.

By definition we know that the relative velocity of an object with reference to the Light, is defined by

$V_0 = \frac{V}{\sqrt{1-\frac{V^2}{c^2}}}$

Where,

V = Speed from relative point

c = Speed of light

On the other hand we have that the linear momentum is defined as

P = mv

Replacing the relative velocity equation here we have to

$P = \frac{mV}{\sqrt{1-\frac{V^2}{c^2}}}$

$P^2 = \frac{m^2V^2}{1-\frac{V^2}{c^2}}$

$P^2 = \frac{P^2V^2}{c^2}+m^2V^2$

$P^2 = V^2 (\frac{P^2}{c^2}+m^2)$

$V^2 = \frac{P^2}{\frac{P^2}{c^2}+m^2}$

$V^2 = \frac{(2.1*10^10)^2}{\frac{(2.1*10^10)^2}{(3.8*10^8)^2}+50^2}$

$V = 2.81784*10^8m/s$

Therefore the height with respect the observer is

$l = l_0*\sqrt{1-\frac{V^2}{c^2}}$

$l = 1.6*\sqrt{1-\frac{(2.81*10^8)^2}{(3*10^8)^2}}$

$l = 0.56m$

Therefore the height which the observerd measure for her is 0.56m

8 0

3 years ago