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3 years ago

What has research determined about the orbit of an electron around a nucleus?

2 answers:

8 0

The one that research has determined about the orbit of an electron around nucleus is : Each sub-level electron type has a unique path where it will likely to be found
Here are the sub levels of an electron :
-sub level s, maximum number of 2 electrons
- sub level p, maximum number of 6 electrons
- sub level d, maximum number of 10 electrons
- sub level f, maximum number of 14 electrons

melamori03 [73]3 years ago

4 0

Answer:

Each sub-level electron type has a unique path where it will likely be found.

Explanation:

The quantum atomic model is the actual model of an atom. Electrons are present in certain regions called as orbitals. Each orbitals corresponds to some sub levels of energy. These are identified with letters s,p,d and f.

Each orbitals have different shapes like s orbitals are spherical, p orbitals have dumbbell shape etc.

The orbit of an electron around a nucleus is like unique path where it will likely to found. So, the correct option is (c).

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Joy uses 20n of force to shovel the snow 10 meters. how much work does she do?

GarryVolchara [31]

Work = Force * distance

Work = 20 N * 10 m = 200 Nm

Work = 200 Joules.

7 0

3 years ago

The velocity of the transverse waves produced by an earthquake is 8.9 km/s, and that of the longitudinal waves is 5.1 km/s. A se

Brrunno [24]

Answer: The distance is 723.4km

Explanation:

The velocity of the transverse waves is 8.9km/s

The velocity of the longitudinal wave is 5.1 km/s

The transverse one reaches 68 seconds before the longitudinal.

if the distance is X, we know that:

X/(9.8km/s) = T1

X/(5.1km/s) = T2

T2 = T1 + 68s

Where T1 and T2 are the time that each wave needs to reach the sesmograph.

We replace the third equation into the second and get:

X/(9.8km/s) = T1

X/(5.1km/s) = T1 + 68s

Now, we can replace T1 from the first equation into the second one:

X/(5.1km/s) = X/(9.8km/s) + 68s

Now we can solve it for X and find the distance.

X/(5.1km/s) - X/(9.8km/s) = 68s

X(1/(5.1km/s) - 1/(9.8km/s)) = X*0.094s/km= 68s

X = 68s/0.094s/km = 723.4 km

6 0

4 years ago

A crate is pushed up each of the two ramps shown in the diagram below. Based on the concept that simple machines make work easie

Naddik [55]

Answer:

ramp b requires less force than ramp a

Explanation:

7 0

3 years ago

When a certain air-filled parallel-plate capacitor is connected across a battery, it acquires a charge of magnitude 172 μC on ea

crimeas [40]

Answer:

k = 2.279

Explanation:

Given:

Magnitude of charge on each plate, Q = 172 μC

Now,

the capacitance, C of a capacitor is given as:

C = Q/V

where,

V is the potential difference

Thus, the capacitance due to the charge of 172 μC will be

C = $\frac{(172\ \mu C)}{V}$

Now, when the when the additional charge is accumulated

the capacitance (C') will be

C' = $\frac{(172+220)\ \mu C)}{V}$

C' = $\frac{(392)\ \mu C)}{V}$

now the dielectric constant (k) is given as:

$k=\frac{C'}{C}$

substituting the values, we get

$k=\frac{\frac{(392\ \mu C)}{V}}{\frac{(172)\ \mu C)}{V}}$

k = 2.279

6 0

4 years ago

When tightening a bolt, you push perpendicularly on a wrench with a force of 165 N at a distance of 0.140 m from the center of t

choli [55]

Answer:

Part a)

$\tau = 23.1 Nm$

Part b)

$\tau = 17.05 Foot pound force$

Explanation:

As we know that torque is defined as the product of force and its perpendicular distance from reference point

so here we have

$\tau = \vec r \times \vec F$

now we have

$\tau = (0.140)(165)$

$\tau = 23.1 Nm$

Part b)

Now we know the conversion as

1 meter = 3.28 foot

1 N = 0.225 Lb force

now we have

$\tau = 23.1 Nm$

$\tau = 23.1 (0.225 Lb)(3.28 foot)$

$\tau = 17.05 Foot pound force$

3 0

3 years ago