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yulyashka [42]
4 years ago
10

An atom of sodium has a larger mass than an atom of neon. How is this possible when neon has so many more valence electrons?

Physics
1 answer:
jeka944 years ago
5 0

The electrons contribute just about zero to the mass of an atom.
It takes more than 1,800 electrons to make the mass of one
proton or neutron.

The naturally occurring element with the most complex atom is
Uranium. That's element #92 , so a neutral uranium atom has
92 electrons.  It would take almost exactly 20 times that many
electrons to add the mass of one proton or neutron to the atom!
(And no other element has that many electrons in an atom of it.)

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The control group in an experimental study will __________.
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You took a running leap off a high diving platform. You were running at 3.1 m/s and hit the water 2.4 seconds later. How high wa
rusak2 [61]

The distance you free-fall from rest is  D = (1/2) (g) (T²) <== memorize this

Height of the platform = (1/2) (9.8 m/s²) (2.4 sec)²

Height = (4.9 m/s²) (5.76 s²)

Height = (4.9/5.76) meters

Height = 28.2 meters (a VERY high platform ... about 93 ft off the water !)

Without air-resistance, your horizontal speed doesn't change.  It's constant.  Traveling 3.1 m/s for 2.4 sec, you cover (3.1 m/s x 2/4 s) = 7.4 m horizontally.

7 0
4 years ago
A dolphin in an aquatic show jumps straight up out of the water at a velocity of 15.0 m/s. (a) List the knowns in this problem.
astra-53 [7]

Answer:

a)

Y0 = 0 m

Vy0 = 15 m/s

ay = -9.81 m/s^2

b) 7.71 m

c) 3.06 s

Explanation:

The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards

Y(0) = 0 m

Vy(0) = 15 m/s

ay = -9.81 m/s^2 (negative because it points down)

Since acceleration is constant we can use the equation for uniformly accelerated movement:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

To find the highest point we do the first time derivative (this is the speed:

V(t) = Vy0 + a * t

We equate this to zero

0 = Vy0 + a * t

0 = 15 - 9.81 * t

15 = 9.81 * t

t = 0.654 s

At this time it will have a height of:

Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m

The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.

0 = Y0 + Vy0 * t + 1/2 * a * t^2

0 = 0 + 15 * t - 1/2 * 9.81 t^2

0 = 15 * t - 4.9 * t^2

0 = t * (15 - 4.9 * t)

t1 = 0 This is the moment it jumped into the air

0 = 15 - 4.9 * t2

15 = 4.9 * t2

t2 = 3.06 s This is the moment when it falls again.

3.06 - 0 = 3.06 s

5 0
3 years ago
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