How many prtons neutrons electrons are in helium

Answers

Answer 1
Answer:
There are a truly huge, enormous number of all of them in
any sample of helium, no matter how small the sample is.

In just ONE ATOM of helium, there are 2 protons and 2 neutrons
inside the nucleus, and two electrons in the 'cloud' of probability
outside the nucleus.


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Earth has a mass of 5.97 x 1024kg, and a mean radius of 6.38 x 10m. What would be the orbitasatellite in orbit 1.44 x 108 m above Earth?
a. v, = 630m/s,
c. v, - 1630m/s,
T = 2.78 x 10 s
T=5.78 x 10s
b. V = 1820m/s,
d. V, - 1260m/s,
T = 6.78 x 106 s
T=5.78 x 10's
Please select the best answer from the choices provid​

Answers

Answer:

v = 1630 m/s

T = 5.78 x 10^5 s

Explanation:

The tangential speed of the satellite can be found by requiring that the gravitational force on the satellite is equal to the centripetal force:

G(Mm)/((R+h)^2)=m(v^2)/(R+h)

where

G is the gravitational constant

M=5.97 x 1024kg is the Earth's mass

m is the satellite's mass

R=6.38 \cdot 10^6 m is the Earth's radius

h=1.44\cdot 10^8 m is the altitude of the satellite

v is the speed of the satellite

Solving for v,

v=\sqrt{(GM)/(R+h)}=\sqrt{((6.67\cdot 10^(-11))(5.97\cdot 10^(24)kg))/(6.38\cdot 10^6 m+1.44\cdot 10^8 m)}=1627 m/s \sim 1630 m/s

And the period of the orbit is equal to the ratio between the distance covered during one revolution (the circumference of the orbit) and the speed:

T=(2 \pi (R+h))/(v)=(2\pi (6.38\cdot 10^6 m+1.44\cdot 10^8 m))/(1630 m/s)=5.79\cdot 10^5 s

So the correct answer is

v = 1630 m/s

T = 5.78 x 10^5 s

When does new characteristics appear before?

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New characteristics appear before the character speaks for the first time .


I hope that's help !

Which can be used to determine if an electromagnet is working? A. compass B. a permanent magnet C. a switch D. galvanometer

Answers

The device used to datermine a electromagnet is a /compass

Simple Machine Lever: A ________ that pivots on top of a fixed point called a _______.

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A bar that pivots on top of a fixed point is called a lever.

Two examples of an insulator

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Some examples of insulators are plastic, glass and rubber.

What is the kinetic energy of a 1 kg ball thrown into the air with an initial velocity of 30 m/s

Answers


Welll ... we can do the math and the physics and get an answer
for you, but then we'll really want to step back into the real world
and see how feasible all of this is.

      Kinetic Energy = (1/2) (mass) (speed)²

                              = (1/2)  (1 kg)  (30 m/s)²

                              = (1/2)  (1 kg)  (900 m²/s²)

                              =     450 kg-m²/s²  =  450 joules .

That's the KE of the ball at the instant it leaves your hand,
moving at 30 m/s .  Immediately after that, we don't know
what happens to it.  It may lose speed, if you tossed it upward,
or it may gain speed, if you tossed it downward or horizontally. 
As soon as its speed changes, so does its KE.

And now, a word from the real world:
' 30 m/s ' is about 67 miles per hour, and the ball weighs a little
over 2 pounds.  Do you really think you could make that toss ?