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Alja [10]
3 years ago
11

What is the name of the particle found in the shell

Physics
2 answers:
elena-s [515]3 years ago
7 0
If you’re talking about atoms, I hope this helps
Atoms consist of three basic particles: protons, electrons, and neutrons. The nucleus (center) of the atom contains the protons (positively charged) and the neutrons (no charge). The outermost regions of the atom are called electron shells and contain the electrons (negatively charged)
FrozenT [24]3 years ago
3 0

That's a "pistachio" .

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If your brakes give out, why can't you just pull the keys out of the ignition?
Pachacha [2.7K]

There should be a small amount of play in the wheel when the steering is locked. Gently pull the key from the ignition while you slowly jiggle the steering wheel back and forth. If this is the cause of the problem, the key should come out after a little effort.

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Organisms that live on or burrow into the ocean floor are
creativ13 [48]
Benthos 

Option b is the answer



8 0
3 years ago
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The density of nuclear matter is about 1018 kg/m3. Given that 1 mL is equal in volume to 1 cm3, what is the density of nuclear m
Sonbull [250]

Answer:

density is 10^{6} Mg/µL

Explanation:

given data

density of nuclear = 10^{18} kg/m³

1 ml = 1 cm³

to find out

density of nuclear matter in Mg/µL

solution

we know here

1 Mg = 1000 kg

so

1 m³ is equal to 10^{6} cm³

and here 1 cm³ is equal to  1 mL

so we can say 1 mL is equal to 10³ µL

so by these we can convert density

density = 10^{18} kg/m³

density = 10^{18} kg/m³ × \frac{10^{-3} }{10^{6} }  Mg/µL

density =  10^{6} Mg/µL

8 0
3 years ago
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The rate at wich energy is used in a circuit
Vaselesa [24]
Power is the rate at which energy is used in a circuit. 
3 0
3 years ago
What is the heat capacity of an object at 25.5∘C that absorbs 45 kJ of heat and is heated to 28.2∘C?
sergey [27]

Answer:

16.6 kJ/°C

Explanation:

given,

Amount of heat absorbed = 45 kJ

initial temperature, T₁ = 25.5°C

final temperature, T₂ = 28.2°C

change in temperature = T₂ - T₁

                                       = 28.2 - 25.5  = 2.7° C

Heat\ capacity = \dfrac{Heat\ absorbed}{\Delta T}

Heat\ capacity = \dfrac{45\ kJ}{2.7}

Heat\ capacity = 16.6\ kJ/^0C

Heat capacity of the object is equal to 16.6 kJ/°C

4 0
4 years ago
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