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3 years ago

Which does not describe the relationship between the nucleus and the electron cloud of an atom?

1 answer:

4 0

Choices 'a', 'c', and 'd' are true.

In choice 'b', I'm not sure what it means when it says that masses
are 'balanced'. To me, masses are only balanced when they're on
a see-saw, or on opposite ends of a rope that goes over a pulley.

Maybe the statement means that the mass of the nucleus and the
mass of the electron cloud are equal. This is way false. It takes
more than 1,800 electrons to make the mass of ONE proton or
neutron, and the most complex atom in nature only has 92 electrons
in it. So there's no way that the masses of the nucleus and the electrons
in one atom could ever be anywhere near equal.

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3. A bottle of vitamin C contains 100 tablets and weighs 80 g. If the

Elan Coil [88]

Answer:

0.57g

20.52

28.5

Explanation:

The bottle weighs 80g with tablets

If the bottle alone weighs 23g, the tablets weigh 57g

100 tablets weigh 57g, 1 tablet weighs 0.57g

0.57g is one tablet, so to achieve 36 tablets we must multiply by 36

0.57 multiplied by 36 is 20.52

To find 50 tablets we can use the same method we used before or a slightly faster method

100 tablets is 57g, so all we have to do is halve to find 50

57 divided by 2 is 28.5

4 0

2 years ago

Ina shoots a large marble (Marble A, mass: 0.08 kg) at a smaller marble (Marble B, mass: 0.05 kg) that is sitting still. Marble

stich3 [128]

From the calculations, the final momentum of B is 8.16 m/s

<h3>What is conservation of linear momentum?</h3>

According to the principle of the conservation of linear momentum, the momentum before collision is equal to the total momentum after collision.

This implies that;

MaUa + MbUb = MaVa + MaVa

Substituting values;

(0.08 kg * 0.5 m/s) + (0.05 kg * 0 m/s) = (0.08 kg * −0.1 m/s) + (0.05 kg * v)

0.4 = -0.008 + 0.05v

v = 8.16 m/s

Learn more about more about momentum: brainly.com/question/24030570:

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3 0

2 years ago

Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field st

Black_prince [1.1K]

Answer:

Part A:

$E_{midpoint}=0$

Part B:

$E_{center}=2711.7558 N/C$

Explanation:

Part A:

Formula of Electric Field Strength:

$E=\frac{1}{4\pi\epsilon}\frac{xQ}{(x^2+R^2)^{3/2}}$

Where:

x is the distance from the ring

R is the radius of the ring

$\epsilon$ is constant permittivity of free space=8.854*10^-12 farads/meter

Q is the charge

For right Ring E at the midpoint can be calculated as:

x for right plate=25/2=12.5 cm=0.125 m

Radius=R=10/2=5 cm=0.05 m

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{right}=9208.1758 N/C$

For Left Ring E at the midpoint can be calculated as:

Since charge on both plates is +ve and same in magnitude, the electric field will be same for both plates.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{left}=9208.1758 N/C$

Electric Field at midpoint:

Both rings have same magnitude but the direction of fields will be opposite as they have same charge on them.

$E_{midpoint}=E_{left}-E_{right}\\E_{midpoint}=9208.1758-9208.1758\\E_{midpoint}=0$

Part B:

At center of left ring:

Due to left ring Electric field at center is zero because x=0.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0)*(20*10^{-19})}{((0)^2+(0.05)^2)^{3/2}}\\E_{left}=0 N/C$

Due to right ring Electric field at center of left ring:

Now: x=25 cm= o.25 m (To the center of left ring)

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.25)*(20*10^{-19})}{((0.25)^2+(0.05)^2)^{3/2}}\\E_{right}=2711.7558 N/C$

Electric Field Strength at center of left ring is same as that of right ring.

$E_{center}=2711.7558 N/C$

5 0

4 years ago

A 5.93 kg ball is attached to the top of a vertical pole with a 2.35 m length of massless string. The ball is struck, causing it

Delvig [45]

Answer

given,

mass of ball = 5.93 kg

length of the string = 2.35 m

revolve with velocity of 4.75 m/s

acceleration due to gravity = 9.81 m/s²

T cos θ = mg

T cos θ = $5.93\times 9.81$

T cos θ = 58.17

$T sin \theta =\dfrac{mv^2}{r}$

$T sin \theta =\dfrac{5.93\times 4.75^2}{2.35 sin \theta}$

$T sin^2 \theta =56.93$

$sin^2 \theta = 1 - cos^2 \theta$

$T (1 - cos^2 \theta) =56.93$

$T (1 - (\dfrac{58.17}{T})^2) =56.93$

T² - 56.93T - 3383.75 = 0

T = 93.22 N

$cos \theta = \dfrac{58.17}{93.22}$

θ = 51.39°

6 0

4 years ago

Examine the roller coaster track above. Assume there is negligible friction as the roller coaster moves from position A to posit

anyanavicka [17]

At point E

the kinetic energy of the rollercoaster is small compared to the potential energy
the potential energy is greater than the kinetic energy
the total energy is a mixture of potential and kinetic energy

<h3>What is the energy of the roller coaster at point E?</h3>

The energy of a roller coaster could either be potential energy, kinetic energy or a combination of both potential and kinetic energy.

Using analogies, the energy of the roller coaster at point E can be compared to a falling fruit from a tree which falls onto a pavement and is the rolling towards the floor. Point E can be compared to the midpoint of the fall of the fruit.

At point E

the kinetic energy of the rollercoaster is small compared to the potential energy
the potential energy is greater than the kinetic energy
the total energy is a mixture of potential and kinetic energy

In conclusion, the energy of the rollercoaster at E is both Kinetic and potential energy,

Learn more about potential and kinetic energy at: brainly.com/question/18963960

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5 0

2 years ago